quantum mechanics

Lecture notes on Quantum Physics I (Phys. 591)

Kirill Tuchin1

1 Department of Physics and Astronomy, Iowa State University, Ames, IA 50011(Dated: November 5, 2014)

Contents

I. Basic principles of Quantum Mechanics 3§1. The uncertainty principle 3§2. Wave function 4§3. Wave function of a free particle 4

A. Wave packet 4B. Time evolution of wave packet 6C. Particle in a box 6

§4. Operators 7A. Expectation values of coordinate and momentum 7B. Hermitian operators 8C. Examples 9

§5. Eigenfunctions and eigenvalues of operators 10A. Properties of discrete spectrum 11B. Properties of continuous spectrum 13

§6. Examples of eigenvalue problems 14A. Operator of momentum p. 14B. Operator of position r. 14C. Operator of orbital angular momentum L. 14

§7. Description of quantum states 16§8. Heisenberg relations 16

A. Example 17§9. Classical limit of quantum mechanics 18

II. Schrodinger equation 19§1. Hamiltonian 19§2. Stationary states 20§3. Time derivative of operators 22§4. Schrodinger and Heisenberg pictures of time-evolution 24

A. Schrodinger picture 24B. Heisenberg picture 25C. Interaction picture 26

§5. Symmetries and conserved quantities 26A. Translations 27B. Rotations 27C. Time evolution 29D. Space inversion 29

III. Motion in one-dimension 30§1. Delta-potential 30

A. Continuity of the wave function 30B. Discrete spectrum 30C. Continuous spectrum 31

§2. Rectangular potential well 32A. Discrete spectrum 32B. Three-dimensional well 34C. Continuous spectrum 34

Kirill Tuchin Phys. 591 Lecture Notes 2

§3. Infinite one-dimensional crystal 35§4. Harmonic oscillator 37

A. Energy spectrum 37B. Ladder operators 39

§5. General properties of one-dimensional motion 41§6. Integral form of Schrodinger equation 43

A. Discrete spectrum 43B. Continuous spectrum 44

IV. Representation theory 46§1. Representations of states 46

A. Hilbert space 46B. Discrete (energy) representation 46C. Continuous (momentum) representation 47D. General case 47

§2. Representations of operators 48§3. Eigenvalue problem in matrix form 51§4. Unitary operators 52§5. Schrodinger equation in momentum representation 54§6. Occupation number representation of harmonic oscillator 56

V. Motion in central potential 59§1. General properties of motion in central potential 59§2. Spherical waves 61§3. Spherical potential well 62§4. Spherical harmonic oscillator 63§5. Coulomb potential 64

A. Discrete spectrum 64B. Continuous spectrum 66

§6. Effective electric potential of the hydrogen atom 67§7. Origin of degeneracy of the energy spectrum 68

VI. Angular momentum 69§1. Angular momentum operator 69§2. Spin 72§3. Addition of angular momenta 75§4. Matrix elements of vector operators 77

VII. 80

A. Fourier analysis 81

B. Dirac delta function 81

C. Levi-Civita symbol 82


I. BASIC PRINCIPLES OF QUANTUM MECHANICS

§1. The uncertainty principle

Quantum mechanics considers microscopic systems with length scales . 10−6 cm. Physical laws that govern suchsystems are very different form those that govern macroscopic systems, which are described by the classical mechanics.Moreover, the microscopic phenomena can be observed only by means of a macroscopic apparatus that translates theaction of microscopic objects into the macroscopic language (examples: Geiger counter, bubble chamber). Quantummechanics is a coherent mathematical framework that describes the laws of microscopic systems and translates theminto the classical language. The ultimate success of quantum mechanics is evident in a great number of experiments.However, its interpretation is still being debated. The goal of this course is to develop the mathematical frameworkof quantum mechanics and to illustrate how it can be applied to number of important problems.

Historically, the first indications that the classical theory is not adequate for description of microscopic systemscame in the early 20th century. It was experimentally established that electromagnetic radiation possesses both waveand corpuscular character. In particular, it is absorbed and emitted in separate portions, quanta, which we now callphotons. Photon energy E turned out to be proportional to its frequency ω as

E = ~ω , (1.1)

where ~ = 1.054 · 10−27 erg·sec is Planck’s constant. In free space photons move with the velocity of light c and havethe momentum

p = ~k , (1.2)

where k is the wave-vector with the length k = |k| = 2π/λ = 1/λ. We know from the electromagnetic theory thatω = kc. It follows that p = ~ω/c = E/c, which is the relation between the energy and momentum of a relativisticparticle of zero mass.

Similarly to the electromagnetic radiation, matter particles also posses both wave and corpuscular character. Thiswas most clearly seen in the electron diffraction experiment where a homogeneous beam of electrons pass through acrystal. The emergent beam exhibits a pattern of alternate maxima and minima of intensity similar to the diffractionof electromagnetic waves. In the two-slit experiment one considers two screens: one impermeable to electrons in whichto slits are cut and behind it another continuous screen. When electrons pass through the first slit, while the secondone is covered, one observes a certain distribution of intensity on the continuous screen. A different distribution isobtained when the first slit is opened and the second is covered. However, the intensity distribution that is obtainedwhen both slits are open is different from a simple sum of the two intensity distribution in contrast to expectation fromthe classical physics. One consequence of this observation is that electrons have no trajectory. Using this diffractiveexperiments one can assign to a free particle a wave length λ that is uniquely determined by its momentum

k =p

~, k =

2π

λ, (1.3)

in analogy with the photon. In this context λ is called the de Broglie wavelength.An important consequence of the electron diffraction experiment is that the classical notion of trajectory does not

apply to microscopic objects. Had electron beams moved along a certain trajectory they would have not interfered.This absence of a trajectory is the essence of the uncertainty principle. Later in this course we will give a mathematicalformulation of this fundamental principle. Suppose now that we measure the electron position with an apparatus(a classical devise). The more accurate is the measurement, the stronger apparatus affects the electron, so thatconsecutive measurements would give a discontinuous and disorderly results indicating absence of trajectory. Only atvery low accuracy (e.g. in bubble chamber), the results can be approximated by a smooth curve – classical electrontrajectory. As a result, the classical definition of velocity as a time derivative of the particle position is meaninglessin quantum theory. Later we will construct a reasonable definition of velocity in quantum mechanics that has correctclassical limit.

In classical mechanics a full description of a system is achieved by providing the coordinates and velocities of allparticles at a given time. The uncertainty principle implies that such description is impossible in quantum mechanics.Quantum description is less detailed than the classical one, and its predictions are less certain. For instance, givenan electron in a certain initial state, a subsequent measurement can yield various results. The problem in quantummechanics consists in determining the probability of obtaining of these different results in a measurement.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §1, Merzbacher, “Quan-tum Mechanics” 3rd edition, Ch. 1.


§2. Wave function

The properties of microscopic systems are described in quantum mechanics by means of an auxiliary quantity – thewave function ψ. The wave function is a complex continuous function of coordinates and time and determines theprobabilities of a different outcomes of a measurement. This is known as the statistical or Copenhagen interpretationof the wave function. Let q denote the set of coordinates of a quantum system (often referred to as the configurationspace) and dq the product of differentials of these coordinates (i.e. element of volume of the configuration space).For example, for one particle q = (x, y, z) and dq = dx dy dz, for two particles q = (x1, y1, z1, x2, y2, z2) and dq =dx1dy1dz1dx2dy2dz2, etc. According to the statistical interpretation, quantity

|ψ(q)|2dq = ψ(q)ψ∗(q)dq (1.4)

is proportional to the probability that we find as the result of our measurement that values of coordinates lie withinthe interval [q, q + dq]. The sum of the probabilities of all possible value of the coordinates of the system must beequal to unity implying that

∫|ψ(q)|2dq = 1 . (1.5)

This equation is called the normalization condition. If a wave function is normalized according to (1.5), then ρ = |ψ(q)|2is known in mathematics as the probability density. In some idealized systems integral over the configuration spacedoes not converge, viz.

∫|ψ(q)|2dq =∞ . (1.6)

In this case |ψ(q)|2 cannot be interpreted as the probability density. Instead, it is meaningful to talk about the relativeprobabilities at different q’s.

Wave function satisfies the following fundamental superposition principle: if a system can be in states describedby wave functions ψ1 and ψ2, it can also be in all states described by the linear combination a1ψ1 + a2ψ2, for anytime-independent complex numbers a1 and a2. In particular, in view of the normalization condition, wave functionsψ and aψ, with non-zero a, describe the same state. Note, that (1.5) requires a = eiα with any real α. Thisambiguity does not affect any of the physical results, because ρ does not depend on it. Note that the superpositionprinciple in quantum mechanics is essentially different from the superposition of vibrations in classical physics, wherea superposition of a vibration onto itself gives a new vibration with a larger or smaller amplitude.

Apart of its dependence on coordinates, the wave function also depends on time. Its time-dependence, or evolution,is determined by a differential equation that we will discuss in the next chapter. That equation must be linear so thatits solutions satisfy the superposition principle.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §2, Merzbacher, “Quan-tum Mechanics” 3rd edition, Ch. 1.

§3. Wave function of a free particle

A. Wave packet

It is instructive to consider a simple example before we start developing mathematical framework of quantummechanics. Consider a free non-relativistic particle of mass m. Its energy E = p2/2m and momentum p are conserved,which reflects uniformity of space and time (will discuss this in more detail later). Therefore, its wave function mustdepend on E and p. According to our discussion in I §1 this particle behaves in diffractive experiments as a wave withwave-vector k = p/~ and energy ω = E/~. In optics, a wave with given k and ω is a plane wave

ψ(r, t) = Nei(k·r−ωt) , (1.7)

where N is a constant. Therefore, it is reasonable to assume that (1.7) describes motion of a free particle. Thisargument is apparently an educated guess, which cuts many corners, but leads to the right answer as will be shownlater. A state of free particle of mass m, momentum p′ and energy E′ is described by

ψ′(r, t) = Nei(k′·r−ω′t) . (1.8)


According to the superposition principle Ψ = aψ + a′ψ′ also describes a possible state of a particle. In this statehowever, the particle does not have a definite momentum.

Generally, one can consider a linear combination of any number of states of form (1.7) with different momenta.Suppose for simplicity that particle moves along the z-axis. Then its state is described by the following wave function

ψ(z, t) =

∫ +∞

−∞ake

i(kz−ωt)dk (1.9)

with an arbitrary a(k). In practice, particle momentum, say k0, is known with a certain accuracy ∆k, where ∆k k0

(otherwise we cannot say that particle momentum is known). In this case, (1.9) is called the wave packet and can bewritten down as

ψ(z, t) =

∫ k0+∆k

k0−∆k

akei(kz−ωt)dk . (1.10)

To estimate this integral, introduce a new variable l = k − k0, and expand ω as a function of k near k = k0:

ω(k) ≈ ω0 +

(dω

dk

)

0

l , (1.11)

where ω0 = ω(k0). In this approximation integration in (1.10) produces

ψ(z, t) =

∫ +∆k

−∆k

ak0ei(l+k0)z−iω0t−i( dωdk )

0ltdl

= ak0ei(k0z−ω0t)

∫ +∆

−∆

ei[z−( dωdk )0t]ldl

= 2ak0sin[z −

(dωdk

)0t]

∆k

z −(dωdk

)0t

︸︷︷︸“amplitude”

ei(k0z−ω0t)︸︷︷︸fast-oscillating

(1.12)

The wave packet is a product of two factors: slowly changing “amplitude” and rapidly oscillating plane wave. Att = 0 the wave packet has global maximum at z = 0 and then decreases as 1/z while oscillating with period 2π/∆k.Zeros of the amplitude are located at zn = πn/∆k, where n = ±1,±2, . . .. The spatial extent of the packet can beestimated as a distance between the two zeros around the maximum at z = 0:

∆z =2π

∆k. (1.13)

In view of (1.3) we conclude that

∆z∆p = 2π~ . (1.14)

In other words, uncertainty in the particle positions is inversely proportional to the uncertainty of its momentum.This is consistent with the uncertainty principle that a microscopic particle does not have a trajectory.

According to (1.12) the maximum of the wave packet moves along the z-axis with group velocity

vg =

(dω

dk

)

0

. (1.15)

Since ω = p2/2m~ = ~k2/2m we find

vg =p0

m, (1.16)

as it should be for a free particle. Note, that the phase velocity vp = ω/k = E/p = p/2m is not velocity of particle.


B. Time evolution of wave packet

Consider a wave packet that at t = 0 is given by

ψ(z, 0) = Ne− z2

2ρ2 . (1.17)

Here ρ is a parameter that specifies the width of the Gaussian (1.17). Using (1.5) we find the normalization constant

ψ(z, 0) =1

π1/4√ρe− z2

2ρ2 . (1.18)

Using (1.9) we can write

∫ +∞

−∞ake

ikzdk =1

π1/4√ρe− z2

2ρ2 . (1.19)

Inverting the Fourier transform (see App. A) we get

ak =

∫ +∞

−∞

dz

2πe−ikz

1

π1/4√ρe− z2

2ρ2 =

√ρ√

2π3/4e−

k2ρ2

2 . (1.20)

Employing (1.9) again and recalling that ω = ~k2/2m we have

ψ(z, t) =

√ρ√

2π3/4

∫ +∞

−∞ei(kz−ωt)e−

k2ρ2

2 dk =1

π1/4√ρ+ i~t/mρ

e− z2

2(ρ2+i~t/m) . (1.21)

The probability density reads

|ψ(z, t)|2 =1

√π√m2ρ4 + ~2t2

e− z2

ρ2+~2t2/m2ρ2 . (1.22)

This shows that the width of the wave packet grows as a function of time, so that the particle position becomes lesscertain at later time. This can be easily understood as follows. At t = 0 the width of the packet is ∆z ∼ ρ andaccording to (1.14) ∆p ∼ ~/ρ. Components of the wave packet with different k move with different velocities, theuncertainty in velocity being ∆v = ∆p/m ∼ ~/mρ. Thus, the packet width grows as ∆z(t) ∼ ~t/mρ in agreementwith (1.22). This effect is important at t > t0 = mρ2/~, whereas at t t0 the diffusion of the wave packet canbe neglected. To put this into perspective, take ρ = 10−8 cm as a typical atomic scale. Then for electron we gett0 = 10−16 sec. In contrast, for m = 1 gr, t0 = 104 years.

C. Particle in a box

Wave function of a free particle (1.7) is an example of a wave function that cannot be normalized with condition(1.5), but rather its normalization integral is divergent as in (1.6). This divergence appears because the notion of freeparticle is an idealization. In reality, we can only talk about a particle that can be considered approximately free ina sufficiently large box, beyond which its interactions cannot be neglected. So, consider a free particle confined in acubic box of edge L 10−6 cm. On the surface of the box ψ must satisfy certain boundary conditions. Since L ismuch larger than the particle de Broglie wavelength, these boundary conditions have very little effect on the particlemotion. Therefore, let’s set the periodic boundary conditions because they are the simplest to deal with:

ψ(x, y, z) = ψ(x+ L, y, z) = ψ(x, y + L, z) = ψ(x, y, z + L) . (1.23)

A wave function that has the same form as (1.7) and satisfies the boundary condition (1.23) is (at t = 0)

ψk(r) =1√L3eik·r , kx =

2πnxL

, ky =2πnyL

, kz =2πnzL

, (1.24)

where nx,ny,nz are non-zero integers. This wave function satisfies the following relation∫ψ∗k′(r)ψk(r)d3r = δk′k , (1.25)


where the integration extends to the entire volume(1. Indeed, if k 6= k′ the integrand is rapidly oscillating functionthat averages to zero. Since momentum of a free particle is conserved, each function (1.24) corresponds to motionwith definite momentum p = ~k.

Consider now a particle in a box in an arbitrary state (i.e. not necessarily free) described by a wave function ψ(r).We know from the Fourier analysis (see Appendix A) that the set of wave functions (1.24) is a complete set. Namely,ψ(r) can be expanded as follows

ψ(r) =∑

k

akψk(r) . (1.26)

This formula tells us that if a particle is in a state ψ(r) and we measure its momentum, the result can be any numberp = ~k from a discrete infinite set specified in (1.24). Multiplying this formula by ψ∗k′(r) on both sides, integratingover the volume of the box and using (1.25) yields the Fourier coefficients

ak =

∫ψ(r)ψ∗k(r)d3r . (1.27)

Notice, that it follows from (1.5),(1.26) and (1.25) that

1 =

∫ψ∗(r)ψ(r)d3r =

∫ ∑

k′,k

aka∗k′ψk(r)ψ∗k′(r)d3r =

∑

k

|ak|2 . (1.28)

It seems reasonable to interpret |ak|2 as the probability that measurement of momentum of a particle in state ψ(r)gives p = ~k. Taking L→∞ corresponds to the free space limit.

A simple example developed in this section illustrates many important features of the general theory. I will use itas a reference in the forthcoming sections.

• Additional reading: Merzbacher, “Quantum Mechanics” 3rd edition, Ch. 2.

§4. Operators

A. Expectation values of coordinate and momentum

Suppose now that a particle is in a state described by ψ(r, t). Average (or expectation) value of the position vectoris

〈r〉 =

∫ψ∗(r, t)rψ(r, t)d3r . (1.29)

If F (r) is a function of r, than

〈F (r)〉 =

∫ψ∗(r, t)F (r)ψ(r, t)d3r . (1.30)

To determine the expectation value of momentum, we place the particle into a big box. We can always expand thewave function into a complete set of states with definite momentum, see (1.26). The probability to find the valueof particle momentum p = ~k is given by |ak|2, where ak is given by (1.27). Therefore, according to the statisticalinterpretation,

〈p〉 = ~∑

k

a∗kakk = ~∑

k

∫ψ∗(r)ψk(r)d3rk

∫ψ(r′)ψ∗k(r′)d3r′ (1.31)

(1 Kronecker delta δab is defined to be equal unity when a = b and zero otherwise.


It follows from (1.24) that i∇ψ∗k(r) = kψ∗k(r), so that (1.31) can be written as

〈p〉 = i~∑

k

∫ψ∗(r)ψk(r)d3r

∫ψ(r′)∇ψ∗k(r′)d3r′ (1.32)

Integrating by parts and using the periodic boundary conditions (1.23) implies∫ψ(r′)∇ψ∗k(r′)d3r′ =

∫∇[ψ(r′)ψ∗k(r′)]d3r′

︸︷︷︸=0

−∫

∇ψ(r′)ψ∗k(r′)d3r′ . (1.33)

Substituting (1.33) into (1.32) and employing the completeness relation (see home assignments)

∑

k

ψ∗k(r)ψk(r) = δ(r − r′) (1.34)

we obtain

〈p〉 =

∫ψ∗(r)(−i~∇)ψ(r)d3r . (1.35)

Taking limit L→∞ we derive that (1.35) is valid also for a particle in entire space. Generalization of (1.35) for anyrational function f(p) is straightforward:

〈F (p)〉 =

∫ψ∗(r)F (−i~∇)ψ(r)d3r . (1.36)

For example, expectation value of kinetic energy is⟨p2

2m

⟩=

∫ψ∗(r)

(−~2∇2

2m

)ψ(r)d3r . (1.37)

We can now generalize (1.30) and (1.37) as follows. Let F be a function of r and p such that

F (r,p) = F1(r) + F2(p) . (1.38)

Expectation value of F can be written in a symbolic form

〈F 〉 =

∫ψ∗(r)Fψ(r)d3r , (1.39)

where

F = F1(r) + F2(−i~∇) (1.40)

is a differential operator.

B. Hermitian operators

Eq. (1.40) is an example of an operator corresponding physical quantity F . We can can further generalize thisdefinition to include any physical quantity, not necessarily depending on r and/or p. (Quantities that have no

classical analogue cannot be expressed as a function of r and p, e.g. spin). Namely, an operator F can be put into

correspondence with any physical quantity F . By convention, Fψ means that operator F acts to the right on functionψ. Most of operators in quantum mechanics are linear, which means that

F (ψ1 + ψ2) = Fψ1 + Fψ2 , F (aψ) = aFψ , (1.41)

for any functions ψ1, ψ2 and for any constant a.Physical quantities are real, hence 〈F 〉 = 〈F 〉∗. In view of (1.39) this means that

∫ψ∗(q)Fψ(q)dq =

∫ψ(q)F ∗ψ∗(q)dq . (1.42)


In linear algebra operator F † is called adjoint of Hermitian conjugate of F if

∫ψ∗(q)Fϕ(q)dq =

∫ϕ(q)(F †ψ(q))∗dq . (1.43)

Linear operator is Self-adjoint or Hermitian if

∫ψ∗(q)Fϕ(q)dq =

∫ϕ(q)(Fψ(q))∗dq . (1.44)

A short-hand operator notation of (1.44) is

F = F † . (1.45)

Clearly (1.42) is a particular case of (1.44) with ϕ = ψ. Thus we can state that to any physical quantity therecorresponds a linear Hermitian operator.

Define the product F K of two operators F and K as a successive action of operator K and then of operator F .Note, that generally, F Kψ 6= KFψ; in operator notation: F K 6= KF . Even though F and K represent physicalquantities, their product does not necessarily do so. We are interested to find a condition under which F K may alsorepresent a physical quantity. In mathematical language, given Hermitian operators F and K, when is the operatorL = F K also Hermitian? To answer this question use (1.44) twice in the following integral

∫ψ∗Lϕdq =

∫ψ∗F (Kϕ)dq =

∫(Kϕ)F ∗ψ∗dq =

∫(F ∗ψ∗)Kϕdq =

∫ϕ(KFψ)∗dq . (1.46)

On the other hand, according to (1.43)

∫ψ∗Lϕdq =

∫ϕ(L†ψ)∗dq . (1.47)

Comparing (1.47) with (1.46) we can write down an operator equation(2

L† = (F K)† = KF . (1.48)

We see that operator F K is Hermitian iff operators F and K commute with one another, viz.

F K = KF . (1.49)

Commutator of two operators is defined as

[F , K] = F K − KF . (1.50)

With this notation (1.49) reads [F , K] = 0.

Out of any two Hermitian operators F and K we can always construct Hermitian operators S = KF + F K andG = i(KF − F K) (see homework assignments). In particular, [F , K] = iG.

C. Examples

1) It is convenient to denote components of vector A by Ai, where i = 1, 2, 3. For example, components of r are risuch that r1 = x, r2 = y, r3 = z; components of p are pi such that p1 = px, p2 = py, p3 = pz. We have the followingcommutators

[ri, rk] = 0 , [pi, pk] = 0 , (1.51)

(2 For any two operators A and B (1.46) and (1.47) imply (AB)† = B†A†.


where pi = −i~ ∂∂ri≡ −i~∂i. From the following identities

xpxψ(x) = x

(−i~∂ψ

∂x

)(1.52)

pxxψ(x) = −i~∂(xψ)

∂x= −i~∂ψ

∂x− i~ψ (1.53)

we derive

[x, px]ψ = i~ψ ⇒ [x, px] = i~ . (1.54)

In general,

[ri, pk] = i~δik . (1.55)

2) Orbital angular momentum of a particle is L = r × p. The corresponding operator:

L = −i~(r ×∇) . (1.56)

It is Hermitian as can be seen from (1.55) employing a theorem that we proved in the previous subsection. Crossproducts can be written using the Levi-Civita symbol εijk defined in Appendix C: Li = εijkrjpk, where summationover the repeated indices is implied. One can prove that (homework assignment)

[Li, Lk] = i~εiklLl , (1.57)

[L2, Li] = 0 , (1.58)

[Li, rk] = i~εiklrl , (1.59)

[Li, pk] = i~pl . (1.60)

• Additional reading: Merzbacher, “Quantum Mechanics” 3rd edition, 4.1–1.4, 3.4, 11.3–11.5, Landau and Lifshitz,“Quantum Mechanics: Non-relativistic theory”, §3,4.

§5. Eigenfunctions and eigenvalues of operators

As we have seen in I §3 a particle can be in a state with definite value of momentum p = ~k. In general, a systemcan be in a state with definite value of a certain physical quantity F . In such a state variation from the average〈F 〉 must vanish. A measure of such variation is variance, defined as the expectation value of the square of operator

∆F = F − 〈F 〉. Namely,

⟨(∆F )2

⟩=

∫ψ∗∆F ∆Fψdq =

∫ (∆F

†ψ)∗

∆Fψdq =

∫ (∆Fψ

)∗∆Fψdq =

∫ ∣∣∣∆Fψ∣∣∣2

dq , (1.61)

where I used that fact the operator F is Hermitian. Variance vanishes for states satisfying ∆Fψ = 0. Since in suchstates 〈F 〉 of course coincides with F we can write this condition as

Fψ = Fψ , (1.62)

Eq. (1.62) is usually supplemented by a boundary condition on ψ and is called the eigenvalue problem for operator Fwith a particular boundary condition. In mathematics it is a subject of the Sturm-Liouville theory. One of the resultsof this theory is that the eigenvalue problem has solutions only at certain discrete values of parameter F . Theseparticular values are called the eigenvalues of the operator F and the corresponding solutions the eigenfunctions. Aset of all eigenvalues is called its spectrum. I will denote eigenfunctions corresponding to the eigenvalue F as ψF .For example, functions (1.24) are eigenfunctions of operator of momentum corresponding to momentum p = ~k. Aswe have already seen before, it is often convenient to consider unbounded systems, in which case the spectrum iscontinuous.

The physical meaning of the spectrum is that measurement of physical quantity F can yield only it eigenvalues.If a system is in a state described by one of the eigenfunctions ψF , than the result of the measurement definitely


produces the corresponding eigenvalue F . Sometimes, several linear independent eigenvalues will correspond to oneeigenvalue of operator. The number of eigenvalues is called the degeneracy of the eigenvalue.

Since F corresponds to a physical quantity, its eigenvalues must be real numbers. To prove that this is indeed thecase, write Fψ = Fψ and its complex conjugate F ∗ψ∗ = F ∗ψ∗ and consider the following integrals

∫(ψ∗Fψ − ψF ∗ψ∗)dq =

∫(ψ∗Fψ − ψF ∗ψ∗)dq . (1.63)

Since F is Hermitian we can write this as

(F − F ∗)∫|ψ|2 =

∫(ψF ∗ψ∗ − ψF ∗ψ∗)dq = 0 . (1.64)

Thus, ImF = 0. Now let us discuss properties of discrete and continuous spectra.

• Additional reading: Arfken et. al. “Mathematical methods for physicists”, 7th edition, Ch.8.

A. Properties of discrete spectrum

1. Consider operator F that has a non-degenerate discrete spectrum Fn, i.e.

Fψn = Fnψn , ⇒ F ∗ψ∗m = F ∗mψ∗m = Fmψ

∗m , (1.65)

were I used (1.64). It follows that

∫ψ∗mFψndq = Fn

∫ψ∗mψndq , and

∫ψnF

∗ψ∗mdq = Fn

∫ψnψmdq . (1.66)

Subtracting these two equations yields

(Fn − Fm)

∫ψ∗mψndq = 0 . (1.67)

If m 6= n we arrive at the orthogonality property of the wave functions

∫ψ∗mψndq = 0 , m 6= n . (1.68)

It physical meaning is that measurement of F gives definitely either Fn in ψn state or Fm in ψm state for anym 6= n. In other words, measurement of F cannot result in a system being in a superposition of eigenfunctions.

2. As I mentioned above (see discussion after (1.62)), an eigenvalue problem has a discrete spectrum if it is restrictedto a finite domain. In this case wave functions ψn rapidly decrease at q →∞ implying that

∫|ψn|2dq is finite.

Thus, we can normalize these wave functions to unity. Together with (1.68) we obtain the orthonormalityproperty

∫ψ∗mψndq = δmn . (1.69)

3. An important result of the Sturm-Liouville theory is that eigenfunctions ψnn form a complete set. Namely,any function ψ(q) such that

∫|ψ(q)|2dq is finite can be expanded as follows

ψ(q) =∑

n

anψn(q) . (1.70)

Using the orthonormality property we can calculate the coefficients an:

an =

∫ψ(q)ψ∗n(q)dq . (1.71)

These equations generalize (1.25),(1.26),(1.27) for a particle in a box.


4. Eigenfunctions satisfy the completeness relation that reads (cp. (1.34))

∑

n

ψ∗n(q′)ψn(q) = δ(q − q′) . (1.72)

Indeed, using (1.70) and (1.71) we can write

δ(q − q′) =∑

n

an(q′)ψn(q) , where an(q′) =

∫δ(q − q′)ψ∗n(q)dq = ψ∗n(q′) . (1.73)

5. Suppose now that Fn is N -fold degenerate, i.e. functions ψnl with l = 1, . . . , N are eigenfunctions correspondingto the same eigenvalue Fn. Eigenfunctions with different n are still orthogonal, which can be proved as in (1.65)–(1.68). However, eigenfunctions with the same n, but different l are not necessarily so. Since it is convenientto deal with a set of orthogonal functions we can replace a set of N independent functions ψnl by another setof functions which are eigenfunctions of F and orthogonal. Let me show how this can be done in the simplestcase N = 2.

We start with two normalized eigenfunctions ψn1 and ψn2. Define two new functions ϕn1 = ψn1 and ϕn2 =a(ψn1 + λψn2). These two are also eigenfunctions of F . The idea is two choose λ such that

∫ϕ∗n1ϕn2dq = 0 ⇒ a

∫ψ∗n1(ψn1 + λψn2)dq = 0 ⇒ λ−1 = −

∫ψ∗n1ψn2dq , (1.74)

and choose a such that∫ϕ∗n2ϕn2dq = 1 . (1.75)

This procedure is called orthogonalization or Gram-Schmidt process. It can be easily generalized to arbitraryN . The resulting spectrum satisfies the following relations:

∫ψ∗ml(q)ψnk(q)dq = δmnδkl , (1.76)

ψ(q) =∑

n,l

anlψnl(q) , where anl =

∫ψ(q)ψ∗nl(q)dq , (1.77)

∑

n,l

ψ∗nl(q′)ψnl(q) = δ(q − q′) . (1.78)

6. Expectation value of F can be expressed through an as follows

〈F 〉 =

∫ψ∗Fψdq =

∫ (∑

n

anψn

)∗F∑

m

amψmdq =∑

m

∑

n

a∗nam

∫ψ∗nFmψm

=∑

m

∑

n

Fmδmna∗nam =

∑

n

|an|2Fn . (1.79)

In particular, for F = 1 we have

∑

n

|an|2 = 1 . (1.80)

Compare this with (1.28). As in I §3 we can interpret |an|2 as probability to find a system in state n.

It is important to note that knowledge of probabilities |an|2 does not completely specify the wave function of aquantum state. The wave function is not known unless all relative phases between different states ψn are also known.States for which only probabilities are known are called the mixed states, whereas states with a wave function arecalled the pure states. Instead of wave function, the mixed states are described by the density matrix. Such descriptionis not full even in quantum mechanical sense. Mixed states play in quantum statistical physics the same role as thewave function in quantum mechanics.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §3.


B. Properties of continuous spectrum

1. Consider operator F that has continuous spectrum: FψF = FψF , where F is continuous. In this case∫|ψF |2dq = ∞ because the system (or its part) can go to infinity. We can derive properties of continuous

spectrum by taking the limit L → ∞ as we did in the case of a particle in box. As a result we obtain thatψF F is a complete set, i.e. any ψ(q) can be expanded as

ψ(q) =

∫aFψF (q)dF . (1.81)

If ψ(q) is normalized, then aF must be normalized as well due to (1.80). Namely

∫ψ∗(q)ψ(q)dq =

∫a∗FaF dF = 1 . (1.82)

Let us show that this allows us to invert (1.81) similarly to (1.71). Write

∫ψ∗(q)ψ(q)dq =

∫dqψ(q)

∫a∗Fψ

∗F (q)dF =

∫dFa∗F

[∫dqψ∗Fψ(q)

](1.83)

and subtract from it (1.82). This gives

0 =

∫dFa∗F

[∫dqψ∗Fψ(q)− aF

], (1.84)

which is true if

aF =

∫ψ∗Fψ(q)dq . (1.85)

2. Substitute (1.81) into (1.85) to find

aF =

∫dqψ∗F (q)

∫aF ′ψF ′(q)dF

′ . (1.86)

This equation is self-consistent if the following delta-function normalization holds

∫ψ∗F (q)ψF ′(q)dq = δ(F − F ′) . (1.87)

Similarly, we can start with (1.81), substitute into it (1.85) and obtain that

∫ψ∗F (q′)ψF (q)dq = δ(q′ − q) . (1.88)

Thus, we learned how to normalize the wave functions of continuous spectrum.

Finally, some operators have both discrete and continuous spectrum. In this case the complete set of states includeseigenfunctions from both parts of the spectrum. Any function ψ(q) can be expanded as

ψ(q) =∑

n

anψn(q)dq +

∫aFψF (q)dq . (1.89)

Normalization∫|ψ(q)|2dq = 1 implies that

∑

n

|an|2 +

∫|aF |2dF = 1 . (1.90)



§6. Examples of eigenvalue problems

A. Operator of momentum p.

1. Eigenvalue problem for the operator px = −i~∂x reads

−i~∂ψ(x)

∂x= pxψ(x) . (1.91)

Solution to this equation is

ψpx(x) = Neipxx/~ . (1.92)

This is the eigenfunction of px corresponding to eigenvalue px. Spectrum of this operator is continuous −∞ <px <∞. ψpx(x) describes motion of a particle along x-axis with definite momentum px.

The normalization constant N can be fixed using (1.87) and (B11) to be N = 1/(2π~)1/2. Indeed,

∫ψ∗px(x)ψp′x(x′)dx =

1

2π~

∫ ∞

−∞ei(px−p

′x)x/~dx = δ(px − p′x) . (1.93)

2. In three dimensions p = −i~∇. The corresponding eigenvalue problem is solved by eigenfunctions

ψp(r) =1

(2π~)3/2eip·r/~ (1.94)

corresponding to eigenvalues p. Normalization is similar to the one-dimensional case if one uses (B16).

B. Operator of position r.

This is another example of continuous spectrum. The eigenvalue problem reads

rψr0(r) = r0ψr0(r) . (1.95)

Action of r on a function of r is a simple product. If r = r0, then (1.95) is identity, if r 6= r0, then (1.95) is satisfiedonly if ψr0 = 0. Therefore, the properly normalized solution to (1.95) is

ψr0(r) = δ(r − r0) . (1.96)

We can expand any wave function in complete set of eigenfunctions ψr0(r) as follows

ψ(r) =

∫ar0ψr0(r)d3r0 , (1.97)

where

ar0 =

∫ψ(r)δ(r − r0)d3r = ψ(r0) . (1.98)

We observe that |ar|2 = |ψ(r)|2 is the probability density to find a particle at position r. This agrees with theinterpretation we gave in the previous section.

C. Operator of orbital angular momentum L.

1. Consider operator Lz – projection of orbital angular momentum on z-axis. In Cartesian coordinates it has form(see (1.56))

Lz = −i~ (x∂y − y∂x) . (1.99)


In spherical coordinates

Lz = −i~∂φ , (1.100)

where 0 ≤ φ ≤ 2π. The eigenvalue problem reads

−i~∂φψLz (φ) = LzψLz (φ) , (1.101)

with the periodic boundary condition

ψLz (φ) = ψLz (φ+ 2π) . (1.102)

Since this eigenvalue problem is restricted to unit circle, its eigenvalues Lz are discrete. Solution to the differentialequation (1.101) is

ψLz (φ) = NeiLzφ/~ . (1.103)

Boundary condition implies that Lz = m~, with any integer m. We can thus label eigenvalues and eigenfunctionsof Lz by m. Imposing the normalization condition

∫ 2π

0

ψ∗mψmdφ = 1 (1.104)

we finally obtain the orthonormal set of eigenfunctions

ψm(φ) =1√2πeimφ . (1.105)

2. Another operator that we will use extensively in this course is L2 =∑3k=1 L

2k. In spherical coordinates it reads

L2 = −~2

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

. (1.106)

The eigenvalue equation L2ψlm = L2ψlm becomes

−~2

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2+L2

~2

ψlm(θ, φ) = 0 . (1.107)

One of the boundary conditions is periodicity in φ, as in (1.102). Another requires that the eigenfunction befinite at θ = 0 and θ = π. The resulting eigenfunctions are spherical harmonics Ylm(θ, φ) and the correspondingeigenvalues are L2 = ~2l(l + 1), where

l = 0, 1, 2, . . . and m = 0,±1,±2, . . . ,±l . (1.108)

Explicit form of the spherical harmonics is

Ylm(θ, φ) =

√(2l + 1)(l −m)!

4π(l +m)!Plm(cos θ)eimφ , (1.109)

where Plm are associated Legendre polynomials. A list of spherical harmonics can be found in all books onquantum mechanics. Normalization condition

∫Y ∗lm(θ, φ)Yl′m′(θ, φ)dΩ = δll′δmm′ . (1.110)

Here dΩ = sin θdθdφ is an element of solid angle.

In addition to being eigenfunctions of L2, spherical harmonics are also eigenfunctions of Lz:

LzYlm = −i~∂φYlm = ~mYlm . (1.111)

We make two important observations: (i) Eigenvalue l of operator L2 is 2m+ 1-fold degenerate because 2m+ 1different values of Lz correspond to the same l, see (1.108). (ii) Two physical quantities can have definite valuesin the same quantum state, i.e. two operators can share the same eigenfunction.

• Merzbacher, “Quantum Mechanics” 3rd edition, 11.3,11.4.


§7. Description of quantum states

Since, as we have seen in the previous section, two different physical quantities can simultaneously have definitevalues, let us find a relationship between the corresponding operators. Consider operators F and G that have definitevalues in state ψn. Formally, this means that

Fψn = Fnψn , Gψn = Gnψn . (1.112)

Acting on this equations with F and G we get

GFψn = FnGψn = FnGNψn , (1.113)

F Gψn = GnFψn = GnFnψn . (1.114)

Subtracting (1.113) from (1.114) gives

(GF − F G)ψn = 0 . (1.115)

Now consider any function ψ:

(GF − F G)ψ = (GF − F G)∑

n

anψn =∑

n

an(GF − F G)ψn = 0 . (1.116)

The conclusion is

[F , G] = 0 . (1.117)

Converse statement is also true, viz. if two operators commute, they have the same eigenfunctions. Indeed, let[F , G] = 0 and Gψn = Gnψn. Then

G(Fψn) = F (Gψn) = Gn(Fψn) . (1.118)

This implies that function Fψn is also an eigenfunction of Gn corresponding to eigenvalue Gn. If n is not degenerate,than Fψn ∝ ψn. Denoting the proportionality factor by Fn we obtain Fψn = Fnψn. If n is degenerate, we canconstruct linear combinations ϕnl =

∑k alkψnk that are eigenfunctions of F .

In summary, two physical quantities can have simultaneously definite values iff the corresponding operators com-mute. Based on this result, one can decide which set of physical quantities can be chosen for a description of aquantum system. If we know the values of all independent quantities with definite values in a given state, then thewave function in that state must be an eigenfunction of all corresponding operators. We have already discussed twoexamples: states with definite value of momentum p (its three projections commute with each other), see (1.94),(1.51) and states with definite values of square of orbital angular momentum and its projection along some direction,see (1.109), (1.58). More examples will be considered in forthcoming chapters.


§8. Heisenberg relations

Let K and F be Hermitian operators. Their commutator can be written as [K, F ] = iG, where G is also a Hermitianoperator (see below (1.50)). Define

∆F = F − 〈F 〉 , ∆K = K − 〈K〉 . (1.119)

It is easy to see using this definition that

KF − F K = ∆K∆F − ∆F ∆K = iG . (1.120)

Consider now the following auxiliary function I(α) of a real parameter α:

I(α) =

∫ ∣∣∣(α∆K − i∆F )ψ∣∣∣2

dq ≥ 0 . (1.121)


Using the fact that F and K are Hermitian operators, and hence obey (1.44), we can cast this function into a differentform(3:

I(α) =

∫(α∆K

∗+ i∆F

∗)ψ∗ (α∆K − i∆F )ψ =

∫ψ∗(α∆K + i∆F )(α∆K − i∆F )ψdq

=

∫ψ∗[α2(∆K)2 + (∆F )2 + αG]ψdq = α2

⟨(∆K)2

⟩+⟨(∆F )2

⟩+ α 〈G〉 ≥ 0 . (1.122)

I(α) is a parabola with its branches pointing up (because α2 > 0). Therefore, (1.122) is satisfied iff the minimumvalue of I is positive. Minimum of I occurs at α = α0 given by

α0 = − 〈G〉2 〈(∆K)2〉 , (1.123)

hence the minimum of I is

I(α0) =⟨(∆F )2

⟩− 〈G〉2

4 〈(∆K)2〉 ≥ 0 . (1.124)

We arrived at the Heisenberg uncertainty relation between the uncertainties of two Hermitian non-commuting opera-tors:

⟨(∆F )2

⟩ ⟨(∆K)2

⟩≥ 〈G〉

2

4. (1.125)

A system can be in a certain state that minimizes the uncertainty relation; for such a state inequality (1.121)becomes an equation. This happens when

I(α0) =

∫ ∣∣∣∣(− 〈G〉

2 〈(∆K)2〉∆K − i∆F)ψ

∣∣∣∣2

dq = 0 , (1.126)

implying that( 〈G〉

2 〈(∆K)2〉∆K + i∆F

)ψ = 0 . (1.127)

States that satisfy this equation are called the coherent states.

A. Example

Let K = x, F = px, then G = ~ since [x, px] = i~. The uncertainty relation (1.125) reads

⟨(∆x)2

⟩ ⟨(∆p)2

⟩≥ ~2

4. (1.128)

Denote 〈x〉 = x0 and 〈px〉 = p0, so that

∆K = x− x0 , ∆F = −i~∂x − p0 . (1.129)

Plugging into (1.127) yields(

~(x− x0)

2 〈(x− x0)2〉 + i(−i~∂x − p0)

)ψ(x) = 0 . (1.130)

Look for a solution in form ψ(x) = NeS(x), where N is a constant. We obtain an equation for S:

(x− x0)

2 〈(∆x)2〉 +dS

dx− ip0

~= 0 . (1.131)

(3 Caution: operator i∆F is not Hermitian.


Its solution (modulo an additive constant)

S =ip0x

~− (x− x0)2

4 〈(∆x)2〉 . (1.132)

Constant N is fixed by the normalization condition. The final result is

ψ(x) =1

(2π 〈(∆x)2〉)1/4exp

ip0x

~− (x− x0)2

4 〈(∆x)2〉

. (1.133)

It is not difficult to verify that for this state (1.128) turns into equation.


§9. Classical limit of quantum mechanics

We know that macroscopic objects consists of microscopic ones: molecules, atoms, electrons etc. Microscopicsystems are governed by quantum mechanics, while the macroscopic ones by classical physics. Therefore, one shouldbe able to obtain classical equations of motion as a limiting case of quantum mechanics. This statement is known as thecorrespondence principle. Consider application of this principle to the wave packet discussed in I §3. The uncertaintiesof particle position and its momentum (1.14) decouple (i.e. become independent as in classical mechanics) in the formallimit ~→ 0. In other words, in this limit one can in principle measure the position and momentum with any accuracy.To better understand what is meant by this limit (after all ~ is a physics constant) note that ~ has units of action S.So when for a given system S ~, quantization can be neglected.

Consider the wave function of a free particle with given energy and momentum (1.7):

ψ(r, t) = Nei(p·r−Et)/~ . (1.134)

In classical mechanics, action of a free particle S = p · r − Et, so we can write

ψcl(r, t) = NeiS/~ . (1.135)

The classical limit corresponds to large values of the phase in (1.135). This is similar to the geometrical optics limitof electromagnetic theory. There any component of electromagnetic wave can be written as u = Neiφ with real Nand φ. In geometrical optics limit phase φ becomes large corresponding to the limit of very short wave length λ→ 0.In such a case propagation of electromagnetic wave can approximated by straight lines (rays). Moreover, both S andφ satisfy the variational principle that says that they must take the least possible value. Since any wave function canbe expanded in plane waves (1.134), it seems reasonable that (1.135) represents the form of the wave function in theclassical limit. We will discuss a more accurate derivation later in this course.

For future reference, recall a few facts form the classical mechanics. A mechanical system described by action S(q, t)satisfies the Hamilton-Jacobi equation

−∂S∂t

= H , (1.136)

where H = H (q, p, t) is the Hamiltonian function with the canonical momenta given by

p =∂S

∂q. (1.137)

If the Hamiltonian function does not explicitly depend on time, then energy E = H is a conserved quantity. Depen-dence of E on coordinates and momentum of a free particle follows from general physical requirements of uniformityand isotropy of space and time and invariance under Galileo transformations, which leads to

Hfree =p2

2m(1.138)

for one free particle.


II. SCHRODINGER EQUATION

§1. Hamiltonian

Wave function ψ contains all information about a given physical system at any time (see for example I §2). Therefore,∂ψ/∂t at any given time must be determined by the value of ψ at the same time. The most general such relationreads

i~∂ψ

∂t= Hψ . (2.1)

Due to the superposition principle H must be a linear operator and i~ is extracted from H for convenience. Eq. (2.1)describes the time-evolution of the wave function.(4

To find out the physical meaning of H let us take the classical limit by substituting the wave function (1.135) into(2.1). We have

Hψcl = −∂S∂tψcl = Hψcl , (2.2)

i.e. operator H in the classical limit becomes the Hamiltonian function, see (1.136). On account of this correspondence

operator H is called the Hamiltonian operator or simply the Hamiltonian. Its operator form for a free particle followsfrom (1.138):

Hfree =p2

2ma= − ~2

2m∇2 . (2.3)

Interaction of a system of non-relativistic particles is described by potential energy U which is a function coordinatesof particles. Thus, the Hamiltonian of a system of non-relativistic particles reads

H =∑

a

p2a

2m+ U(r1, . . . , ra) . (2.4)

In obtaining (2.4) we relied on the correspondence principle, which is supported by a great number of experiments.Substituting (2.4) into (2.1) we obtain the Schrodinger equation for a single particle

i~∂ψ

∂t= − ~2

2m∇2ψ + U(r)ψ , (2.5)

where U is the potential energy of the particle in an external field. For a free particle it has solution

ψ(r, t) = Nei(p·r−Et)/~ , (2.6)

which agrees with (1.7).A possible time-dependence of the wave function ψ is restricted by the requirement of the total probability conser-

vation

d

dt

∫|ψ(q, t)|2dq = 0 . (2.7)

Let us verify that Schrodinger equation satisfies (2.7). From (2.1) it follows that

i~ψ∗∂ψ

∂t= ψ∗Hψ − i~ψ∂ψ

∗

∂t= ψH∗ψ∗ . (2.8)

(4 This time-evolution of the wave function is a result of the influence of forces acting on the system. It has nothing to do with the changesin the system introduced by the measurement process. The later replaces one wave function by another, which is known as the wavefunction collapse.


Subtracting these equations gives

i~∂

∂t(ψ∗ψ) = ψ∗Hψ − ψH∗ψ∗ . (2.9)

Integrating over q and using the fact that H is a Hermitian operator we derive (2.7).While the total probability is conserved, the probability density ρ = |ψ(q, t)|2 is not. Indeed, substituting (2.4) (for

a single particle) into (2.9) yields

i~∂

∂tρ = − ~2

2m

(ψ∗∇2ψ − ψ∇2ψ∗

). (2.10)

Introduce the probability current density

j =~

2mi(ψ∗∇ψ − ψ∇ψ∗) . (2.11)

Then, (2.10) can be cast in the form of the continuity equation:

∂ρ

∂t= −∇ · j . (2.12)

Since ψ is a complex function, it can represented in term of two real functions f and ϕ as

ψ(r, t) = f(r, t)eiϕ(r,t) . (2.13)

In this notation we have

ρ = f2 , j = ρ∇(~ϕm

). (2.14)

In particular, if the phase ϕ is independent of coordinates, then the probability current vanishes j = 0.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §8,17, Merzbacher,“Quantum Mechanics” 3rd edition,

§2. Stationary states

If the Hamiltonian function of a classical system is not explicitly time-dependent, its energy is conserved. To seehow energy conservation is expressed in quantum mechanics, consider a system describe by a Hamiltonian that doesnot explicitly depend on time, i.e.

∂H

∂t= 0 . (2.15)

For such a system time-dependence of the wave function can be computed from (2.1) employing the method ofseparation of variables:

ψ(q, t) = T (t)ψE(q) . (2.16)

This yields

i~∂T (t)

∂t

1

T (t)=HψE(q)

ψE(q)= E , (2.17)

where E is a constant. Indeed, a function of t can be equal a function of q only if both are constants. As a result wehave two equations

HψE(q) = EψE(q) , (2.18)

i~∂T (t)

∂t= ET (t) (2.19)

(2.20)


Eq. (2.19) immediately gives

T (t) = e−iEt/~ . (2.21)

States with definite energy are called the stationary states. The time dependence of their wave functions reads

ψ(q, t) = ψE(q)e−iEt/~ . (2.22)

Eq. (2.19) becomes the time-independent Schrodinger equation :

− ~2

2m∇2ψ + (U − E)ψ = 0 . (2.23)

In fact, since one is usually interested in stationary states, (2.23) as simply referred to as the Schrodinger equation .Consider properties of the stationary states.

1. Time-dependence of a stationary state is uniquely determined by its energy.

2. ρ = const.(t) and j = const.(t).

3. If F does not explicitly depend on time, then its expectation value in a stationary state is also time-independent:

〈F 〉 =

∫ψ∗(q, t)Fψ(q, t)dq = const.(t) , (2.24)

which means that F is a conserved.

Note, that although 〈F 〉 is time-independent, the wave function of a stationary state does depend on timeaccording to (2.21).

4. Consider a system in a stationary state ψ(q, t) and let ψn(q) be a complete set of eigenfunctions of F . Wecan expend ψ(q, t) =

∑n an(t)ψn(q). The probability to find the system in the n’th eigenstate is

|an|2 =

∣∣∣∣∫ψ(q, t)ψ∗n(q)dq

∣∣∣∣2

= const.(t) (2.25)

5. Functions ψE(q), being eigenfunctions of Hamiltonian, form a complete set. Therefore, any solution of theSchrodinger equation can be expanded in them as ψ(q, t) =

∑E TE(t)ψE(q). Owing to (2.21) we obtain a

general form of the expansion of an arbitrary wave function ψ(q, t) in a complete set of stationary states:

ψ(q, t) =∑

n

anψn(q)eiEnt/~ , (2.26)

for discrete spectrum and

ψ(q, t) =

∫aEψE(q)eiEt/~dE , (2.27)

for the continuous one.

Eqs. (2.26),(2.27) indicate that in order to calculate the time-evolution of a wave function one needs to (i) findthe spectrum of the Hamiltonian, (ii) fix coefficients an using the initial condition at t = 0 and (iii) sum orintegrate over the energy spectrum according to (2.26), (2.27).

6. We proved in (1.64) that values of E are real. This is also seen in (2.23) which is a real differential equation.Moreover, the wave functions of the stationary states can also be chosen to be real (except for a system inmagnetic field). Indeed, ψ and ψ∗ satisfy the same equation. For a non-degenerate energy level it implies thatthey are the same functions up to a non-important phase. For a degenerate level, one can form real linearcombinations of the wave functions.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §8,10,17,19,Merzbacher, “Quantum Mechanics” 3rd edition, 3.1.


§3. Time derivative of operators

Time derivative of a physical quantity F cannot be defined as in classical mechanics since in quantum mechanics thevalue of F may not be known at two arbitrarily close time instants. Instead we will consider an observable quantity:the expectation value of F . I pointed out in (2.24) that if F does not explicitly depend on time, its expectationvalue in a stationary state is time-independent. Consider now the expectation value of a Hermitian operator F inan arbitrary state ψ: 〈F 〉 =

∫ψ∗Fψdq. Taking its time derivative and using the Schrodinger equation ψ = Hψ/i~,

ψ∗ = −H∗ψ∗/i~ we derive

d 〈F 〉dt

=

∫ ψ∗∂F

∂tψ +

∂ψ∗

∂tFψ + ψ∗F

∂ψ

∂t

dq (2.28)

=

∫ ψ∗∂F

∂tψ − 1

i~(H∗ψ∗)Fψ +

1

i~ψ∗F Hψ

dq (2.29)

=

∫ψ∗∂F

∂t+

1

i~[F , H]

ψdq , (2.30)

where in the last step I used (1.44) with the replacements ϕ→ Fψ and F → H. Define now an operator dF /dt suchthat its expectation value obeys the following condition

d 〈F 〉dt

=

⟨dF

dt

⟩=

∫ψ∗dF

dtψdq . (2.31)

In view of (2.30) we obtain an operator expression of the time derivative

dF

dt=∂F

∂t+

1

i~[F , H] . (2.32)

This implies that if ∂F∂t = 0 (no explicit time dependence) and [F , H] = 0, then F does not change with time, i.e.

F is a conserved quantity. We also conclude that any conserved quantity can be measured concurrently with theHamiltonian.

Consider application of (2.32) to the motion of a single particle. Replacing F first by p and then by r we get twooperator equations

dp

dt=

1

i~[p, H] , (2.33)

dr

dt=

1

i~[r, H] . (2.34)

Commutators are computed as follows:

[p, H]ψ = [p, U ]ψ = −i~[∇, U ]ψ = −i~[∇(Uψ)− U∇ψ] = −i~∇Uψ , (2.35)

[r, H]ψ =1

2m[r, p2

k]ψ =1

2mei[ri, pk]pk + pk[eiri, pk] = i~

p

m, (2.36)

where ei is a unit vector. In the last line summation over k and i is implied and (1.55) is used. Plugging (2.35),(2.36)into (2.33) and (2.34) gives

dp

dt= −∇U , (2.37)

v =dr

dt=p

m, (2.38)

in agreement with the correspondence principle (which justifies our definition (2.31) a posteriori). These equationsare knows as the Ehrenfest theorem. Taking another time derivative of (2.38) we derive using (2.32)

dv

dt=

1

m

dp

dt= − 1

m∇U . (2.39)


Formally, this formula has the same form as the classical equation of motion. However, it must be understood as arelationship between the corresponding expectation values:

md2

dt2

∫ψ∗rψd3r = −

∫ψ∗∇Uψd3r . (2.40)

To better understand the difference between the operator formula (2.39) and its classical counterpart, it is worthexamining the classical limit at this point. Namely, we would like to know what are the conditions that allow us toreplace operators in (2.39) with the corresponding classical quantities. For the sake of simplicity we consider only thex-components. Denote x = 〈x〉, ∆x = x− x and expand at small ∆x (in the classical limit ∆x vanishes):

∂U(x)

∂x=∂U(x)

∂x+∂2U

∂x2∆x+

1

2

∂3U

∂x3(∆x)2 + . . . (2.41)

Substituting into (2.40) and noting that∫|ψ|2∆xdx = 〈x〉 − 〈x〉 = 0 we have

md2x

dt2= −∂U(x)

∂x− 1

2

∂3U

∂x3

⟨(∆x)2

⟩+ . . . (2.42)

In the classical approximation we must require∣∣∣∣∂U

∂x

∣∣∣∣1

2

∣∣∣∣∂3U

∂x3

∣∣∣∣⟨(∆x)2

⟩(2.43)

In addition to the requirement that uncertainty in x be small, we also have to find a similar requirement for themomentum. Consider the kinetic energy

⟨p2x

2m

⟩≡⟨

[〈px〉+ (px − 〈px〉)]22m

⟩=〈px〉22m

+2 〈px〉〈p− 〈px〉〉

2m+

⟨(∆px)2

⟩

2m. (2.44)

The second term obviously vanishes. In order that the third term be small the following condition must be satisfied

〈px〉2 ⟨(∆px)2

⟩=

~2

4 〈(∆x)2〉 . (2.45)

Inequalities (2.43) and (2.45) indicate that the classical limit applies when external field is slowly varying and momentaare large. We can combine (2.43) and (2.45) into one condition:

⟨p2x

⟩ ~2

8

∣∣∣∣U ′′′

U ′

∣∣∣∣ . (2.46)

As in classical mechanics, we can prove the quantum virial theorem. On one hand, in a stationary state expectationvalue of p · r is time-independent, see (2.24), i.e.

d

dt〈p · r〉 = 0 . (2.47)

On the other hand,

d

dt(p · r) =

1

i~[p · r, H] =

1

i~p · [r, H] + [p, H] · r

=p2

m− r∇U . (2.48)

Employing the definition (2.31) we cast this in form

d

dt〈p · r〉 =

⟨p2

m

⟩− 〈r∇U〉 (2.49)

We thus arrive at the required relation⟨p2

m

⟩= 〈r∇U〉 . (2.50)

For illustration, suppose U ∝ rn. Then, 〈r∇U〉 = n 〈U〉 and the virial theorem implies a relation between theexpectation values of kinetic and potential energies:

⟨p2

2m

⟩=n

2〈U〉 . (2.51)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §9,19, Merzbacher,“Quantum Mechanics” 3rd edition, 3.3,3.5,3.6.


§4. Schrodinger and Heisenberg pictures of time-evolution

In practical application it is sometimes convenient to use different representations or “pictures” of time-evolution.Observed quantities do not depend on a choice of a picture, of course.

A. Schrodinger picture

Thus far we have followed the Schrodinger picture of time-evolution in which wave functions depend on time,whereas the operators corresponding the physical quantities usually do no explicitly depend on time. We can expressthe time-evolution of a wave function ψ(q, 0) (at t = 0) as a result of action of the time-evolution operator S(t) asfollows

ψ(q, t) = S(t)ψ(q, 0) . (2.52)

The normalization of the wave function should not change with time, i.e.

∫ψ∗(q, t)ψ(q, t)dq =

∫ψ∗(q, 0)ψ(q, 0)dq = 1 . (2.53)

On the other hand,

∫ψ∗(q, t)ψ(q, t)dq =

∫[S∗ψ∗(q, 0)][Sψ(q, 0)]dq =

∫ψ∗(q, 0)S†Sψ(q, 0)dq , (2.54)

where I used the definition (1.43) with F → S† and ϕ→ Sψ. We conclude that

S†S = 1 . (2.55)

Operators satisfying (2.60) are unitary operators. To find the explicit form of operator S, substitute (2.52) to theSchrodinger equation :

i~∂S(t)

∂tψ(q, 0) = HS(t)ψ(q, 0) . (2.56)

In operator form:

i~∂S(t)

∂t= HS(t) . (2.57)

For closed systems Hamiltonian does not explicitly depend on time. In such a case one can integrate (2.57) to obtain

S(t) = e−i~ Ht . (2.58)

How does the operator (2.58) act on a wave function? Since S is a function of the Hamiltonian, it is helpful to

expand ψ(q, 0) in a set of stationary states ϕn(q)n and act on the result by the evolution operator S. Namely, using(2.52) we have

ψ(q, t) =S∑

n

anϕn(q) =∑

n

an∑

k

1

k!

(− iHt

~

)kϕn(q)

=∑

n

an∑

k

1

k!

(− iEnt

~

)kϕn(q) =

∑

n

anϕne−iEnt/~ , (2.59)

in agreement with (2.26). One can think of (2.58) as a compact way of expressing the time evolution of (2.26) and(2.27).


B. Heisenberg picture

In the Heisenberg picture wave functions do not evolve with time, but the operators do. Let ψ(q, t) be a wavefunction in the Schrodinger picture describe a certain state and ψH(q, t) a wave function in the Heisenberg picturedescribing the same state. According to (2.52), in order that ψH(q, t) be time-independent it must satisfy

ψH(q, t) = S−1ψ(q, t) (2.60)

Consider now an expectation value of a physical quantity F . It must be the same regardless of the picture we chooseto work in. Thus,

〈F 〉 =

∫ψ∗Fψdq =

∫(SψH)∗F SψHdq =

∫ψ∗H S

†F SψHdq . (2.61)

Let us call F an operator in the Schrodinger picture and append a subscript S to it. Then the same operator in theHeisenberg picture reads

FH = S−1F S , (2.62)

where I used (2.55) to write S† = S−1.

Suppose that at some initial time t FH(t) = F . After evolving over a short time-interval ∆t with (2.62) we have

FH(t+ ∆t) = S−1(∆t)FH(t)S(∆t) . (2.63)

Expanding in Taylor series we get

FH(t) +dFHdt

∆t =

(1 +

i

~H∆t

)FH(t)

(1− i

~H∆t

). (2.64)

Keeping only linear terms in ∆t we obtain

dFHdt

=1

i~[FH , H] . (2.65)

This formula is similar to (2.32), but has a different meaning. Eq. (2.32) is a definition of the time-derivative of

F , while in dFH/dt in (2.65) is the derivative of the physical quantity (whose operator is time-dependent in theHeisenberg picture).

∗ As an example, consider motion of a free particle. Its Hamiltonian is H = p2/2m. In the Schrodinger picturep = −i~∇. In the Heisenberg picture

pH = ei~ Ht(−i~∇)e−

i~ Ht = p+

it

~[H, p] = p , (2.66)

where I used the following identity:

eABe−A = B +1

1![A, B] +

1

2![A, [A, B]] + . . . (2.67)

Position operator can be calculates similarly using (2.36):

rH = ei~ Htre−

i~ Ht = r +

it

~[H, r] +

(it)2

2~2[H, [H, r]] + . . . = r +

p

mt . (2.68)

Alternatively, we can solve equations of motion

d

dtrH =

i

~[H, rH ] =

p

m,

d

dtpH =

i

~[H, pH ] = 0 (2.69)

which leads to (2.68).


C. Interaction picture

Another popular representation of the time-evolution is the interaction picture. Suppose we can split the Hamilto-nian into two parts:

H = H0 + V , (2.70)

where H0 describes a system without taking interactions into account, while V describes interactions. Introduce aunitary operator

S = eiH0t/~ (2.71)

and define the wave function in this picture as follows

ψI(q, t) = Sψ(q, t) . (2.72)

Substituting this into the Schrodinger equation i~ψ = Hψ yields

i~ψI = SV S−1ψI . (2.73)

Define operator VI in the interaction picture

VI = SV S† = eiH0t/~V e−iH0t/~ . (2.74)

Then we can rewrite (2.73) as

i~ψI = VIψI . (2.75)

In other words, the wave function ψI evolves in time only according to the interaction part of the Hamiltonian. Atthe same time an operator F evolves with the non-interacting part of the Hamiltonian as in (2.74):

FI = eiH0t/~F e−iH0t/~ . (2.76)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §13, Merzbacher,“Quantum Mechanics” 3rd edition, Ch.14.

§5. Symmetries and conserved quantities

In classical mechanics uniformity of space implies momentum conservation, its isotropy implies the orbital angularmomentum conservation and uniformity of time intimates energy conservation. Since the Hamilton function of asystem carries all information about its properties, it must exhibit the same symmetry properties as the system. Thesame is true in quantum mechanics were we have to replace the Hamilton function with the corresponding operator H.Let S be an operator performing a symmetry transformation. Invariance of Hamiltonian under such transformationmeans that [S, H] = 0 (5.

(5 Although this is pretty evident, let me illustrate this in the case of infinitesimal continuos transformation. Such a transformation canbe written as S = 1 + is, where s is Hermitian operator. This follows from the requirement that

∫ψ∗ψdq =

∫(Sψ)∗Sψdq up to small

terms of order s2. It is easy to check that in order that 〈H〉 stay invariant under ψ → Sψ we need to require that [s, H] = 0, implying

that [S, H] = 0. This equation is true for discrete transformations as well.


A. Translations

Consider translation (displacement as a whole) of a closed system of particles by a small constant vector δ, viz.ra → r′a = ra + δ. Under this transformation the wave function transforms as follows

ψ(r1 + δ, r2 + δ . . .) = ψ(r1, r2, . . .) + δ ·∑

a

∇aψ(r1, r2, . . .) =(1 +

∑

a

δ ·∇a

)ψ(r1, r2, . . .) . (2.77)

Operator 1 +∑a δ ·∇a is an operator of an infinitesimal translation. Since the Hamiltonian is invariant (i.e. do not

change) under translations we arrive at the statement

[∑

a

∇a, H]

= 0 . (2.78)

Recalling that pa = −i~∇a, we conclude that uniformity of space implies the total momentum conservation[∑a pa, H] = 0. In fact, we guessed the form of the momentum operator in I §4. Instead, we could have defined

momentum as the infinitesimal translation operator (up to a constant factor). Consider the operator of finite trans-lations

Taψ(r) = ψ(r + a) , (2.79)

where a is a finite constant. Expanding the right-hand-side in Taylor series we get

Taψ(r) =

[1 +

i

~a · p+

1

2

(i

~a · p

)2

+ . . .

]ψ(r) = e−ia·p/~ψ(r) . (2.80)

Thus,

Ta = eia·p/~ . (2.81)

The generator of translations is the following operator

GT =∂

∂aTa

∣∣∣∣a=0

=i

~p = ∇ . (2.82)

Notice that the generator does not depend on a.

B. Rotations

Invariant of a closed system under rotations follows from the isotropy of space. The corresponding operator mustbe proportional to the operator of the total angular momentum. To see that this is indeed the case, consider aninfinitesimal rotation by angle δφ:

ra → r′a = ra + δφ× ra . (2.83)

Under this rotation the wave function ψ(r) changes as follows

ψ(ri + [δφ× r]i

)= ψ(ri + εijkδφjrk) = ψ(ri) + ∂iψ(ri)εijkδφjrk = ψ(ri) + δφ · (r ×∇)ψ(ri) , (2.84)

ψ(r1 + δφ× r1, r2 + δφ× r2, . . .) = (1 + δφ ·∑

a

(ra ×∇a)

)ψ(r1, r2, . . .) (2.85)

Operator 1 + δφ ·∑a

(ra ×∇a) is the operator of infinitesimal rotations. Since the closed system is invariant under

rotations it commutes with the Hamiltonian. In particular,

[∑

a

(ra ×∇a), H] = 0 . (2.86)


It follows that∑a

(ra ×∇a) must be proportional to the total angular momentum of the system, which justifies our

definition (1.56).Operator of rotation by a finite angle φ around axis with direction n can be computed by analogy with (2.79)-(2.81)

with the result

Rnφ = eiL·nφ/~ . (2.87)

Generator of rotations around the n-direction is

Gn =∂

∂φRnφ

∣∣∣∣φ=0

=i

~L · n . (2.88)

If represent vectors as columns

A =

AxAyAz

, (2.89)

then rotation operators can be represented as 3× 3 matrices. For example, operator of rotation around x-axis reads

Rxφ =

1 0 00 cosφ − sinφ0 sinφ cosφ

(2.90)

The corresponding generator of rotations around x-axis is

Gx =∂

∂φRxφ

∣∣∣∣φ=0

=

0 0 00 0 −10 1 0

(2.91)

By the same token,

Gy =

0 0 10 0 0−1 0 0

, Gz =

0 −1 01 0 00 0 0

(2.92)

It is easy to verify that [Gx, Gy] = Gz etc. Since Gx = iLx/~, Gy = iLy/~, Gz = iLz/~, we confirm the commu-tation relations (1.57). We will see later that this only one of infinite number of possible representations of angularmomentum.

Operators of all physical quantities can be classified according to their transformational properties under rotations.Scalar operator S is an operator that commutes with the operator of rotation:

[S, Rnφ ] = 0 , ⇒ [S, Li] = 0 . (2.93)

For example, r2, p2, p · r, L2 are all scalar operators.Vector operator V is an operator whose components satisfy

[Li, Vk] = iεiklVl . (2.94)

Examples of vector operators: r, L, (p · r)p.

Tensor operator of second rank Tik is an operator whose components satisfy

[Li, Tkl] = i(εikpδnl + εilnδkp)Tpn . (2.95)

Examples of tensor operators: rirk, ripk.


C. Time evolution

For infinitesimal time shift t→ t′ = t+ τ

ψ(t+ τ) = ψ(t) +∂

∂tψ(t)τ =

(1 + τ

∂

∂t

)ψ(t) . (2.96)

Due to uniformity of time, the Hamiltonian of a closed system must commute with the operator ∂t. Therefore weidentify the time derivative operator as the energy operator (up to a multiplicative factor). But energy operator isthe Hamiltonian, see (2.1). It is clearly commutes with itself and so energy is conserved.

Operator of finite time shift can is defined and calculated as in (2.79) and (2.80). Using ∂t = −iH/~ we have

Sτψ(t) = ψ(t+ τ) = e−iHτ/~ψ(t) . (2.97)

Operator Sτ evolves quantum system forward in time. This agrees with (2.58).

Translation, rotation and time evolution are examples of continuous transformations. Invariance of a system(and hence the Hamiltonian) under such transformations leads to conservation laws both in classical and quantummechanics.

D. Space inversion

Unlike the previous continuous symmetry transformations, space inversion is a discrete transformation. In classicalmechanics it does not lead to any conservation laws, however in quantum mechanics it does. Space inversion reflectsthe position vector r → r′ = −r, i.e. the right-handed coordinates system goes into the left-handed one. Thecorresponding operator is called the parity operator P and is defined as

Pψ(r) = ψ(−r) . (2.98)

Among systems conserving parity are those that interact through electromagnetic and strong forces, and systems incentral potentials. Conservation of parity means that [H, P ] = 0.

Consider the eigenvalue problem for the parity operator:

Pψ(r) = Pψ(r) (2.99)

Applying P on both side again and noting that P 2 = 1 (unit operator) we derive ψ(r) = P 2ψ(r), so P 2 = 1 and

P = ±1 (2.100)

are the eigenvalues of the parity operator. The corresponding eigenstates ψ± satisfy

Pψ± = ±ψ± . (2.101)

State ψ+ (ψ−) is said to be an even (odd) parity state.Suppose that the Hamiltonian of a system is an even function of its coordinates, viz. H(r) = H(−r). Then,

[H, P ]ψ(r) = H(r)Pψ(r)− P H(r)ψ(r) = H(r)ψ(−r)− H(−r)ψ(−r) = 0 , (2.102)

i.e. the parity operator commutes with the Hamiltonian implying that that the parity is conserved. Wave functionsof the stationary states of a system invariant under the space inversion can be chosen to have a definite parity.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §15,26,30, Merzbacher,“Quantum Mechanics” 3rd edition, 17.1,17.2,17.9.


III. MOTION IN ONE-DIMENSION

§1. Delta-potential

A. Continuity of the wave function

Wave function must be a continuous function of coordinates in order that its probabilistic interpretation makesense. Moreover, its first derivative must also be a continuous function, provided that the potential is a non-singularfunction. To prove this statement, consider a particle of mass m moving in the following singular one-dimensionalpotential

U(x) = V (x)− αδ(x− x0) , (3.1)

where α and x0 are real numbers, V (x) is a non-singular function. Schrodinger equation reads

− ~2

2mψ′′E(x) + [V (x)− α(x− x0)]ψE(x) = EψE(x) . (3.2)

Integrating over x0 −∆x ≤ x ≤ x0 + ∆x:

∫ x0+∆x

x0−∆x

dx

− ~2

2mψ′′E(x) + [V (x)− α(x− x0)]ψE(x)

= E

∫ x0+∆x

x0−∆x

dxψE(x) . (3.3)

The discontinuity of the wave function at x is

∆ψ′E(x) = ψ′E(x+ ∆x)− ψ′E(x−∆x) . (3.4)

Taking ∆x→ 0 we derive from (3.3)

∆ψ′E(x0) = −2mα

~2ψE(x0) . (3.5)

This indicates, in particular, that if the potential is continuous, the first derivative of the wave function is continuousas well.

B. Discrete spectrum

Consider a particle moving in the delta potential U = −αδ(x). The corresponding Schrodinger equation

− ~2

2mψ′′E(x)− α(x)δ(x)ψE(x) = EψE(x) . (3.6)

Suppose E < 0 and introduce a parameter κ that has dimension of inverse distance (same as wavenumber):

κ =

√−2mE

~2> 0 . (3.7)

Solution to the Schrodinger equation that is finite at x→ ±∞ is

ψ(x) = Ae−κx x > 0 , (3.8)

ψ(x) = Beκx , x < 0 . (3.9)

To fix constants A and B we need two conditions. First, we use (3.5):

∆ψ′(0) = A(−κ)−Bκ = −2mα

~2A . (3.10)

Second, the continuity of the wave function itself at x = 0 implies A = B. Thus,

κ =mα

~2(3.11)


There are two possibilities: (i) if α < 0, then according to (3.11) κ < 0 which contradicts (3.7). In this case there areno bound states. (ii) If α > 0, there is one bound state

E0 = −κ2~2

2m= −mα

2

2~2. (3.12)

The corresponding wave function is

ψ0(x) =√κ0e−κ0|x| , κ0 =

mα

~2. (3.13)

C. Continuous spectrum

Continuous spectrum lie at E > 0. It is convenient to introduce

k =

√2mE

~2> 0 . (3.14)

Solution to the Schrodinger equation is

ψE(x) = A1eikx +A2e

−ikx , x > 0 , (3.15)

ψE(x) = B1eikx +B2e

−ikx , x < 0 . (3.16)

It corresponds to a free particle traveling in positive and negative directions along the x-axis. A usual problem is thatgiven a flux of particles incident on the potential from, say, x = −∞, find what fraction of particles is reflected fromthe potential and what fraction is transmitted. In such a case at x > 0 there are only particles that move towardx = +∞, meaning that A2 = 0. It is also convenient to normalize the incident flux of particles to one particle perunit volume. To this end, we compute the probability current density of incident particles using (2.11):

ji = |B1|2~km

= |B1|2v . (3.17)

The required normalization is B1 = 1.Using the continuity of ψ and (3.5) at x = 0 we can now compute A1 and B2:

A1 =1

1 + mαi~2k

, (3.18)

B2 = − 1

1 + i~2kmα

. (3.19)

Fluxes of transmitted and reflected particles are

jt = |A1|2v , (3.20)

jr = |B2|2v . (3.21)

(3.22)

The corresponding fractions are transmitted and reflected

T =jtji

= |A1|2 =1

1 + mα2

2~2E

, (3.23)

R =jrji

= |B2|2 =1

1 + 2~2Emα2

. (3.24)

T and R are called the transmission and reflection coefficients respectively. The probability conservation impliesR+ T = 1, which is indeed seen in (3.23),(3.24).

Notice that the transmission and reflection coefficients do not depend on the sign of α.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §25, Merzbacher,“Quantum Mechanics” 3rd edition, 6.1.


1 2 3

U0

U(x)

x−a

2a2

FIG. 1:

§2. Rectangular potential well

A. Discrete spectrum

In the case of the rectangular potential shown in Fig. 1 all solutions have E ≥ 0. Indeed, equation 〈E〉 =⟨p2/2m

⟩+ 〈U〉 implies that 〈E〉 ≥ 〈U〉 ≥ 0. Thus, we can define

k =

√2mE

~(3.25)

in place of E. Furthermore, bound states correspond to energies E ≤ U0, at which the particle cannot escape tox→ ±∞.

Another important observation is that the potential is invariant under the inversion U(−x) = U(x) implying that

H(−x) = H(x). In this case the stationary states can be chosen to have definite parity either even or odd, which willbe denoted by plus or minus superscript, see (2.102). As a consequence, it is sufficient to find the wave functions onlyat x ≥ 0. Schrodinger equation in the region 2 reads

ψ′′2 (x) + k2ψ2 = 0 , 0 ≤ x ≤ a/2 . (3.26)

It has two independent solutions with definite parity

ψ+2 = B cos(kx) , P = +1 , (3.27)

ψ−2 = C sin(kx) , P = −1 . (3.28)

In the region 3,

ψ′′3 − κ2ψ3 = 0 , x ≥ a/2 , (3.29)

where

κ =

√2m(U0 − E)

~. (3.30)

Its solution finite at x→∞ is

ψ3(x) = Ae−κx , x ≥ a/2 . (3.31)

Matching (boundary) conditions for even states:

ψ+2 (a/2) = ψ3(a/2) , ψ′+2 (a/2) = ψ′3(a/2) , ⇒ (3.32)

B cos(ak/2) = Ae−κa/2 , −kB sin(ak/2) = −Aκe−κa/2 . (3.33)

Taking the ratio gives

k tan(ka/2) = κ =

√2mU0

~2− k2 . (3.34)


Bound states exist only for those values of k = kn, n = 0, 1, . . . , N that satisfy (3.34). There is a finite number ofsolutions because according to (3.34)

0 ≤ ka ≤√

2mU0a2

~. (3.35)

The larger is U0a2, i.e. the deeper and wider the the well, the more bound states it contains. This can be seen in

Fig. 2 where I illustrated a numerical solution to this equation. Intersection of blue and red lines give the values of kthat solve (3.34). The red line terminates at the maximal ka satisfying (3.35). Using (3.25) they can be converted tothe corresponding energy levels.

Π 2 Π 3 Π 4 Π 5 Π 6 Πka

-5

5

Π 2 Π 3 Π 4 Π 5 Π 6 Πka

-5

5

FIG. 2: Values of parameters are a = 1 and mU0/~2 = 100. Left panel (even solutions): red line is tan(k/2), blue panel

is√

2mU0/k2~2 − 1. Right panel (odd solutions): red line is cot(k/2), blue panel is −√

2mU0/k2~2 − 1. Intersections aresolutions to (3.34) and correspondingly. In this figure there are three even and two odd states.

Similarly, for odd states, boundary conditions yield

C sin(ak/2) = Ae−κa/2 , C cos(ak/2) = −κkAe−κa/2 . (3.36)

Thus,

k cot(ka/2) = −κ . (3.37)

Solution to this equation is illustrated in the right panel of Fig. 1.

Consider the limit U0a2 →∞ at fixed ka, which corresponds to infinitely deep well. In such case, (3.34) for even

states becomes

tan(ka/2) ≈√

2mU0

k2~2 1 . (3.38)

The corresponding solutions

kna = π(2n+ 1) , ⇒ E+n =

~2k2n

2m=π2(2n+ 1)2~2

2ma2. (3.39)

Even eigenfunctions:

ψ+n (x) =

√2

acos

(π(2n+ 1)x

a

), 0 ≤ x ≤ a/2 . (3.40)

At x > a/2 eigenfunction vanishes because its argument is

κna ≈√

2mU0

~a→∞ . (3.41)

For odd states

− cot(ka/2) 1 ⇒ kna = 2πn ⇒ E−n =π2(2n)2~2

2ma2. (3.42)


The corresponding eigenfunctions are

ψ−n (x) =

√2

asin

(π(2n)x

a

), 0 ≤ x ≤ a/2 . (3.43)

Thus, there is an infinite number of even and odd levels in the infinitely deep potential well.Notice that ψ±(±a/2) = 0, which means that particle cannot penetrate the region |x| ≥ a/2 because of the infinite

potential. At x = ±a/2 the derivative ψ′ is discontinuous.

In the opposite limit a√U0 ~/

√m, which corresponds to shallow and/or narrow well there is only one even

level because k ≈ √2mU0/~ and ka 2. The corresponding energy is

E =~2k2

2m= U0 . (3.44)

B. Three-dimensional well

In three dimensional rectangular potential well U(x, y, z) with sides a, b, c, solution to the Schrodinger equationcan be found employing separation of variables:

ψn1,n2,n3(r) = ψn1(x)ψn2

(y)ψn3(z) . (3.45)

In the infinite well the corresponding energy levels are

En1,n2,n3=π2~2

2m

(n2

1

a2+n2

2

b2+n2

3

c2

), n1, n2, n2 = 1, 2, 3, . . . (3.46)

and the wave functions

ψn(x) =

√2a cos

(πnxa

), n = 1, 3, 5, . . .√

2a sin

(πnxa

), n = 2, 4, 6, . . .

(3.47)

If a 6= b 6= c each energy value corresponds to one wave function, i.e. there is no degeneracy. Potential U is invariantunder rotations by π around the three coordinate axes and inversion. If a = b = c, then

En1,n2,n3=

π2~2

2ma2(n2

1 + n22 + n2

3) , n1, n2, n2 = 1, 2, 3, . . . (3.48)

Additional symmetry of this case is the symmetry of a cube. This results in degeneracy of the energy levels. For

example, quantum numbers (n1, n2, n3) = (5, 1, 1), (1, 5, 1), and (1, 1, 5) correspond to the same level 27π2~2

2ma2 . Interest-ingly, there is another level (3, 3, 3) that has the same energy. This additional degeneracy is caused by the particularway in which potential depends on the coordinates, but not on the symmetry of the field and is called the accidentaldegeneracy.

C. Continuous spectrum

Continuous spectrum lies at E > U0. In this case we are interested to know which fraction of the flux incidenton the well is reflected and which one is transmitted. Accordingly we write down the solution to the Schrodingerequation in the following way

ψ1 = Aeik0x +Be−ik0x , x < −a/2 , k0 =√

2m(E − U0)/~ , (3.49)

ψ2 = αeikx + βe−ikx , −a/2 < x < a/2 , k =√

2mE/~ , (3.50)

ψ3 = Ceik0x , x > a/2 . (3.51)

Matching conditions at x = a/2 are

Ceik0a/2 = αeika/2 + βe−ika/2 ,k0

kCeik0a/2 = αeika/2 − βe−ika/2 . (3.52)


Solving for α and β we findα = C

2

(1 + k0

k

)ei(k0−k)a/2

β = C2

(1− k0

k

)ei(k0+k)a/2

(3.53)

Matching conditions at x = −a/2 are

Ae−ik0a/2 +Beik0a/2 = αe−ika/2 + βeika/2 , (3.54)

Ae−ik0a/2 −Beik0a/2 =k

k0

(αe−ika/2 − βeika/2

), (3.55)

which together with (3.53) yield

A =C

4eik0a

[(1 +

k

k0

)(1 +

k0

k

)e−ika +

(1− k

k0

)(1 +

k0

k

)eika

](3.56)

B =C

4

[(1− k

k0

)(1 +

k

k0

)e−ika +

(1 +

k

k0

)(1− k0

k

)eika

]. (3.57)

Transmission coefficient

T =1

1 + 14

(k0k − k

k0

)2

sin2(ka), (3.58)

and the reflection coefficient R = 1− T .When sin(ka) = 0, i.e. ka = πn, the flux of particles is completely transmitted T = 1, R = 0. The corresponding

values of energy

En =π2~2n2

2ma2, where n is integer, (3.59)

are called the resonance energies or virtual levels. Since En ≥ U0

n2 ≥ 2a2mU0

π2~2. (3.60)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §22,25,50. Merzbacher,“Quantum Mechanics” 3rd edition, 6.4,7.4. Messiah, “Quantum mechanics”, VI.10.

§3. Infinite one-dimensional crystal

-3 -2 -1 1 2 3

5

10

15

20

25

30

FIG. 3: a = 1. Delta functions are approximated by (B12) with ε = 0.01.

Consider motion of a particle in a periodic one-dimensional potential

U(x) = α

∞∑

n=−∞δ(x− na) . (3.61)


that serves as an elementary model of an ideal infinite crystal. Points xn = na correspond to the lattice sites. Todetermine the wave function we need to solve the Schrodinger equation between any two lattice sites, say an < x <a(n + 1) and apply the matching conditions at the points x = a(n − 1) and x = an. Solution to the Schrodingerequation reads

ψ(x) = Aneik(x−na) +Bne

−ik(x−na) , k =

√2mE

~. (3.62)

Here An, Bn are constants and constant phases e±ina are written for further convenience (they could have beenabsorbed into the An and Bn). Since the potential is a periodic function of x with the period a, the wave function ψmust also be periodic with the same period:

ψ(x+ a) = µψ(x) , (3.63)

where µ is a complex constant. Suppose that |µ|2 > 1, then ψ diverges at x→∞ and cannot be normalized. Similarly,if |µ|2 < 1, it diverges at x→ −∞. Therefore, |µ|2 = 1, i.e. constant µ is a phase. We can parametrize it as

µ = eiqa , −π ≤ qa ≤ π . (3.64)

States with fixed q are called Bloch states, while ~q is called the quasi-momentum (don’t confuse it with the particlemomentum ~k).

From (3.62) and (3.63) it follows that

An = µAn−1 , Bn = µBn−1 . (3.65)

Now we apply the boundary conditions at x = na. For δ → 0 we have

ψ(na+ δ) = ψ(na− δ) , (3.66)

ψ′(na+ δ)− ψ′(na− δ) =2mα

~2ψ(na) . (3.67)

The wave functions on the each side of x = na are

ψ(na− δ) = An−1eik(x−(n−1)a) +Bn−1e

−ik(x−(n−1)a) , (3.68)

ψ(na+ δ) = Aneik(x−na) +Bne

−ik(x−na) , (3.69)

Using (3.66),(3.67) we get

An−1eika +Bn−1e

−ika = An +Bn , (3.70)

ik(An−1eika −Bn−1e

−ika)− ik(An −Bn) =2mα

~2(An +Bn) , (3.71)

Replacing An−1 and Bn−1 with An and Bn according to (3.65) we arrive at the following system of equations

An(eika − µ) +Bn(e−ika − µ) = 0 , (3.72)

An

[eika − µ

(1 +

2mαi

~2k

)]−Bn

[e−ika − µ

(1− 2mαi

~2k

)]= 0 . (3.73)

Dividing the second of these equations by the first one we obtain a condition on µ:

µ2 − 2µ[cos(ka) +

mα

~2ksin(ka)

]+ 1 = 0 . (3.74)

Being a quadratic equation it has two solutions µ1 and µ2:

µ1,2 = f(E)±√f2(E)− 1 , f(E) = cos(ka) +

mα

~2ksin(ka) . (3.75)

In view of (3.64) f2 ≤ 1 for otherwise µ is not a pure phase. It is easy to verify that µ1µ2 = µ1µ∗1 = 1.

We can rewrite (3.74) using (3.64) as follows

f =µ2 + 1

2µ= cos(qa) . (3.76)


1st zone 2nd zone 3rd zone

cos(qa)

Π 2 Π 3 Πka

-1

1

2

3

fHEL

FIG. 4: Solution to (3.77) with cos(qa) = 0.6 (red line) and mα/~2 = 2. f(E) is a blue line. Intersection of blue and red linesare solutions to (3.77). Physical values of f satisfy f2 ≤ 1 and lie between the two horizontal black lines which divide possiblevalues of ka into the Brillouin zones. In each zone there is one solution kna (red circles) corresponding to the energy level En.

Now employing (3.75) we obtain

cos(qa) = cos(ka) +mα

~2ksin(ka) . (3.77)

Solutions to this equations is a set of discrete values kn and the corresponding energy levels En. They are depictedin Fig. 4.

Solution to the Schrodinger equation have form (3.61) even at x → ±∞ indicating that a particle is not localizedand can move in the entire crystal with quasi-momentum ~q.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §, Merzbacher, “Quan-tum Mechanics” 3rd edition,

§4. Harmonic oscillator

A. Energy spectrum

In many physical problems a physical system moves near a minimum of potential well U(x). Suppose that thisminimum is at x = 0 and expand

U(x) ≈ U(0) +1

2U ′′(0)x2 . (3.78)

The corresponding force is −U ′′(0)x. In classical mechanics motion of a particle under such force is governed by theequation mx = −U ′′(0)x describing periodic motion with frequency ω = (U ′′(0)/m)1/2. In quantum mechanics weare interested to solve the following Schrodinger equation

− ~2

2m

d2

dx2+ω2m

2x2 − E

ψ(x) = 0 . (3.79)

Introduce dimensionless variables

ξ = x

√mω

~, ε =

2E

~ω. (3.80)


We haved2

dξ2− ξ2 + ε

ψ(ξ) = 0 . (3.81)

At ξ → ∞, we can neglect ε. The resulting equation ψ′′ = ξ2ψ has asymptotic solution ψ ∼ e−ξ2/2. Thus, we can

look for a solution in the form

ψ(ξ) = v(ξ)e−ξ2/2 . (3.82)

Substituting and denoting ε− 1 = 2n we get

v′′ − 2ξv′ + 2nv = 0 . (3.83)

We can find a solution to this equation in the form of the power series

v(ξ) =

∞∑

k=0

akξk . (3.84)

Upon substitution

∞∑

k=2

akk(k − 1)ξk−2 − 2

∞∑

k=0

kakξk + 2n

∞∑

k=0

akξk = 0 . (3.85)

Changing the summation variable in the first sum as k′ = k − 2 and then renaming it again to k′ → k we obtain

∞∑

k=0

ak+2(k + 2)(k + 1)ξk − 2

∞∑

k=0

kakξk + 2n

∞∑

k=0

akξk = 0 . (3.86)

This equations is satisfied if coefficients in front of same powers vanish, viz.

ak+2(k + 2)(k + 1)− 2kak + 2nak = 0 , ⇒ ak+2 =2(k − n)

(k + 2)(k + 1)ak . (3.87)

If we start with a finite a0 and a1 = 0 we get a polynomial containing only even powers of ξ. However, if we startwith a finite a1 and a0 = 0 we get an odd polynomial. This reflects the fact that the Hamiltonian of the harmonicoscillator is invariant under the spatial inversion, and thus we can choose its stationary states to have definite parity.

It can be shown (see Merzbacher’s book) that ψ(ξ) diverges at ξ → ±∞ unless the polynomials are finite, whichrequires n to be integer. Since E ≥ 0, n ≥ −1/2 implying that n = 0, 1, . . .. For illustration consider n = 2:

a2 =2 · (−2)

2 · 1 a0 = −2a0 , a4 = a6 = . . . = 0 . (3.88)

The corresponding polynomial is H2(ξ) = −(2ξ2 − 1)a0. Consider n = 3.

a3 =2 · (1− 3)

3 · 2 a1 = −2

3a1 , a5 = a7 = . . . = 0 . (3.89)

The corresponding polynomial is H3(ξ) = (− 23ξ

3 + 1)a1. Coefficients a0 and a1 are chosen for each n in such a waythat the resulting polynomials satisfy the following formula

Hn(ξ) = (−1)neξ2 dne−ξ

2

dξn, (3.90)

and are known as the Hermit polynomials. The lowest polynomials are

H0 = 1 , H1 = 2ξ , H2 = 4ξ2 − 2 , H3 = 8ξ3 − 12ξ , (3.91)

Let me note for the future reference the following recurrence relations

ξHn(ξ) = nHn−1(ξ) +1

2Hn+1(ξ) , (3.92)

dHn

dξ= 2nHn−1(ξ) . (3.93)


To wave functions normalized as∫ +∞

−∞ψn(x)ψm(x)dx = δmn (3.94)

read

ψn(x) =(mωπ~

)1/4 1√2nn!

Hn

(x

√mω

~

)e−mωx

2/(2~) . (3.95)

The corresponding energy levels

E = ~ω(n+

1

2

). (3.96)

There is only one eigenfunction for each n, so there is no degeneracy. Also note that at n = 0 the oscillator has energyE0 = ~ω/2 known as the zero-point energy and is interpreted as the energy of vacuum fluctuations.

Since the Hamiltonian of the harmonic oscillator is invariant under spatial inversion, the stationary states havedefinite parity, viz. states with even n are even functions of x, while states with odd n are odd.

∗ Knowledge of the stationary states of the harmonic oscillator allows us to calculate various observable quantities.As an example, let us compute the expectation value of the particle position in the n’th stationary state.

〈x〉n =

∫ ∞

−∞ψ2nxdx = 0 (3.97)

because x is an odd function, while ψ2n is even. Variance of the position is

⟨∆x2

⟩n

=⟨x2⟩n

=

∫ ∞

−∞ψ2nx

2dx . (3.98)

The easiest way to take this integral is to employ the orthonormality condition (3.94). To this end we substitute(3.95) into (3.92) and obtain after some simple algebra

ξψn =

√n

2ψn−1 +

√n+ 1

2ψn+1 . (3.99)

Using this equations again we get

ξ2ψn =1

2

√n(n− 1)ψn−2 +

(n+

1

2

)ψn +

1

2

√(n+ 1)(n+ 2)ψn+2 . (3.100)

Thus,

⟨∆x2

⟩=

~mω

∫ ∞

−∞ψ2nξ

2dx =~mω

(n+

1

2

). (3.101)

For future reference let me record a useful relation that follows from (3.93):

∂ψn∂ξ

=1√2

(√nψn−1 −

√n+ 1ψn+1

). (3.102)

B. Ladder operators

Eigenvalue problem for the harmonic oscillator can be reformulated in terms of the ladder operators(6

a =

√mω

2~

(x+

i

mωp

), (3.103)

a† =

√mω

2~

(x− i

mωp

). (3.104)

Properties of the ladder operators:

(6 In fact, this problem can be solved entirely employing these operators without solving first the Schrodinger equation , see Merzbacher’s“Quantum Mechanics”, 10.6.


1. a and a† are not Hermitian, i.e. they do not correspond to any physical quantities. Only their certain combina-tions, such as (3.108),(3.109) and (3.110) are Hermitian.

2. The ladder operators satisfy the following commutation relations:

[a, a†] = 1 , [a, a] = 0 , [a†, a†] = 0 , (3.105)

which can be derived using (3.103),(3.104) and the commutators [x, px] = i~ etc.

3. To verify the physical meaning of the ladder operators, apply them to the wave functions of the stationarystates. Using (3.95) and (3.93) we derive

aψn =

√mω

2~

(x+

~mω

d

dx

)ψn =

√nψn−1 , (3.106)

a†ψn =

√mω

2~

(x− ~

mω

d

dx

)ψn =

√n+ 1ψn+1 . (3.107)

The raising operator a† moves the particle to the next higher excited state, while the lowering operator a movesthe particle one state lower.

Equations (3.103),(3.104) can be used to express x and px through the ladder operators:

x =

√~

2mω(a+ a†) , (3.108)

px = i

√mω~

2(a† − a) , (3.109)

In terms of the ladder operators the Hamiltonian reads

H =p2x

2m+ω2mx2

2= −~ω

4

[a2 + (a†)2 + aa† + a†a

]+

~ω4

[a2 + (a†)2 − aa† − a†a

]=

(a†a+

1

2

)~ω , (3.110)

where I used (3.105). Evidently, operator a†a commutes with the Hamiltonian. Employing (3.106) abd (3.107) wefind its eigenvalue in the n’th stationary state:

a†aψn = nψn . (3.111)

Thus we reproduce the spectrum (3.96): E = ~ω(n+ 1

2

).

The ladder operator method can be applied to calculate the eigenfunctions ψn as follows. Let ψ0 be the wavefunction of the ground state. Then, according to (3.106)

aψ0 =

√mω

2~

(x+

~mω

d

dx

)ψ0 = 0 . (3.112)

Solution to this differential equation is

ψ0 = N0e−mωx2/2~ . (3.113)

Constant N0 is fixed by the normalization condition. It is easy to see from (3.107) that the wave functions of theexcited states can be obtained as follows

ψn =1√n!

(a†)nψ0 . (3.114)

You can verify that it gives the same result as (3.95).

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §23, Merzbacher,“Quantum Mechanics” 3rd edition, Ch. 5, 10.6.


§5. General properties of one-dimensional motion

We can now generalize a few examples that we considered in the previous sections and derive the general propertiesof one-dimensional motion. We start with the Schrodinger equation

d2ψ

dx2+

2m

~2[E − U(x)]ψ = 0 . (3.115)

1. Energy levels of discrete spectrum are non-degenerate. Indeed, suppose the contrary is true, viz. there are twolinear independent eigenfunctions corresponding to the same energy levels:

ψ′′1ψ1

=2m

~2[U(x)− E] =

ψ′′2ψ2

, ⇒ ψ′′1ψ2 = ψ′′2ψ1 . (3.116)

Integating by parts

∫(ψ′′1ψ2 − ψ′′2ψ1)dx

︸︷︷︸const

= ψ′1ψ2 −∫ψ′1ψ

′2dx− ψ′2ψ1 +

∫ψ′2ψ

′1dx = ψ′1ψ2 − ψ′2ψ1 . (3.117)

Recall that for a bound state ψ1,2 → 0 as x→ ±∞, hence const=0 and

ψ′1ψ1

=ψ′2ψ2

. (3.118)

Integrating again

const =

∫ (ψ′1ψ1− ψ′2ψ2

)dx =

∫ (d

dxlnψ1 −

d

dxlnψ2

)dx = lnψ1 − lnψ2 . (3.119)

Thus, ψ1 ∝ ψ2 and upon normalization we obtain that ψ1 = ψ2.

2. Let the discrete spectrum of the Hamiltonian be E0 < E1 < . . . and the corresponding eigenfunctions ψ0, ψ1, . . .correspondingly. The oscillation theorem states that eigenfunction ψn has n zeros, i.e. it vanishes n times, notincluding the boundary values. You can find a proof in the book of Messiah. As an illustrate to this theorem Iplotted in Fig. 5 a few wave functions of the infinitely deep potential well.

-0.4 -0.2 0.2 0.4x

-1.0

-0.5

0.5

1.0

Ψn

FIG. 5: Wave functions of the three lowest states in the infinitely deep potential well: ψ0 =√

2/a cos(πx/a) (black), ψ1 =√2/a sin(2πx/a) (blue), ψ2 =

√2/a cos(3πx/a) (red). a = 1.

3. Consider potentail well of a general form shown in Fig. 6.

(a) Discrete spectrum corresponds to motion in a finite region of space. For a potential in Fig. 6 this is motionwith energies in the interval Umin < E < 0. (Motion with energies below Umin is not possible because〈E〉 = 〈U〉+

⟨p2/2m

⟩≥ 〈U〉 ≥ Umin).


U0

Umin

-3 -2 -1 1 2 3x

-0.5

0.5

1.0

UHxL

FIG. 6: An arbitary potential well U(x).

(b) In the region 0 < E < U0 the spectrum is continuous because particle can go to x → ∞. The eigenstatesare not degenerate, which can be proven as before using the fact ψ1,2 → 0 as x→ −∞.

at x→ +∞ : ψ′′ +2m

~2Eψ = 0 , ⇒ ψ = A cos(kx+ δ) , k =

√2mE

~. (3.120)

at x→ −∞ : ψ′′ − 2m

~2(U0 − E)ψ = 0 , ⇒ ψ = Beκx + Ce−κx , κ =

√2m(U0 − E)

~. (3.121)

The boundary condition requires that C = 0. Therefore, there is only one function that satisfies theboundary conditions at given E.

(c) At E > U0 the spectrum is continuous and two-fold degenerate:

at x→ +∞ : ψ = Aeikx +Be−ikx , k =

√2mE

~, (3.122)

at x→ −∞ : ψ = Ceik′x +De−ik

′x , k′ =

√2m(E − U0)

~. (3.123)

There are two solutions that satisfy the boundary conditions – one with particles incident from the leftand another from the right.

4. Consider two potentials U(x) and U(x) = U(x) + δU(x) with δU(x) ≥ 0. Let En and ψn(λ) be eigenvalues andeigenfunctions of the discrete spectrum of the following Hamiltonian

H(λ) =p2

2m+ U(x) + λδU(x) , (3.124)

where λ is a parameter. In one of the home assignments you proved the following theorem:

∂En(λ)

∂λ=

∫ψ∗n(λ)

∂H

∂λψn(λ)dx (3.125)

Plugging (3.124) into (3.125) we have

∂En(λ)

∂λ=

∫|ψn(λ)|2δU(x)dx ≥ 0 . (3.126)

Note that En(0) = En is an eigenvalue of H = p2/2m+U , while En(1) = En is an eigenvalue of H = p2/2m+U .

We conclude that En ≥ En. This means that a more shallow potential has higher lying energy levels. Thisstatement can be generalized to a multidimensional potentials.

As δU increases, energy levels go up, so that eventually the highest energy level (the one that has the smallest|En|) becomes En = 0. A further increase of δU will push this level into the continuum spectrum. Thus, inorder to determine at what values of parameters of the Hamiltonian a new bound states appears/disappears onehas to solve the Schrodinger equation with E = 0.


∗ Example: rectangular potential well shown in Fig. 1. Note, that the y-axis in Fig. 1 is shifted by U0 ascompared to Fig. 5. This is why the continuous spectrum starts at E = U0. The corresponding values of k andκ are k0 =

√2mU0/~ and κ0 = 0. Consider even states, (3.34) implies

k0 tan

(k0a

2

)= κ0 = 0 , ⇒ k0a = 2πn , n = 0, 1, 2, . . . (3.127)

The ground state exists if k0a ≥ 0, which is always satisfied. The first excited even state exists if k0a > 2π, etc.The n’th excited even state exists if k0a > 2πn. The total number of even states

N+ =

[ak0

2π

]+ 1 . (3.128)

Square brackets indicate the integer part. Similarly, odd states exist when k0a = π(1 + 2n), n = 0, 1, . . .. Ifk0a < π there are no odd states. The total number of odd states is

N− =

[ak0

2π− 1

2

]+ 1 =

[ak0

2π+

1

2

](3.129)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §21, Messiah, “Quan-tum Mechanics”, Sec. 12.

§6. Integral form of Schrodinger equation

Eigenvalue problem for the Hamiltonian can be written down as an integral equation for the wave function. I willconsider separately the discrete and continuous spectra.


We start with the Schrodinger equation

− ~2

2mψ′′(x) + Uψ(x) = Eψ(x) , (3.130)

and assume that U is a potential well such that U → 0 as x→ ±∞ and consider the bound states corresponding toE < 0. Introduce the Green’s function GE(x, x′) satisfying

− ~2

2m

d2

dx2GE(x, x′)− EGE(x, x′) = δ(x− x′) . (3.131)

Suppose that we solved (3.131) and found GE(x, x′). Then a solution to (3.130) reads

ψ(x) = Ae−κx +Beκx −∫ ∞

−∞GE(x, x′)U(x′)ψ(x′)dx′ , (3.132)

where A, B are constants and

κ =

√−2mE

~2. (3.133)

Indeed,(− ~2

2m

d2

dx2− E

)ψ(x) =

(− ~2

2m

d2

dx2− E

)(Ae−κx +Beκx)

−∫ +∞

−∞

(− ~2

2m

d2GE(x, x′)dx2

− EGE(x, x′)

)U(x′)ψ(x′)dx′

=−∫ +∞

−∞δ(x− x′)U(x′)ψ(x′)dx′

=− U(x)ψ(x) . (3.134)


Note, that although (3.132) is a formal “solution” of the Schrodinger equation , it is still an equation for ψ. Since weare interested in bound states which have wave functions localized at x→ ±∞ we set A = B = 0.

The Green’s function can be calculate as described in III §1B (7. We write

GE(x, x′) =

C(x′)eκ(x−x′) , x < x′ ,D(x′)e−κ(x−x′) , x > x′ .

(3.135)

and use boundary conditions (continuity of GE(x, x′) at x = x′ and discontinuity its derivative given by (3.5)) to fixC and D:

C = D , κ(D + C) =2m

~2, ⇒ C = D =

m

~2κ(3.136)

Thus,

GE(x, x′) =m

κ~2e−κ|x−x

′| . (3.137)

Finally, Schrodinger equation in the integral form is

ψE(x) = − m

κ~2

∫ +∞

−∞e−κ|x−x

′|U(x′)ψE(x′)dx′ . (3.138)

Subscript E indicates that we used the boundary condition for the bound states with E < 0.

∗ Example. For a particle moving in the delta-potential well U = −αδ(x) we have

ψE(x) =m

κ~2αψE(0)e−κ|x| , (3.139)

which is the eigenfunction of the only bound state. This equation must be consistent at x = 0 implying that

1 =m

κ~2α , (3.140)

which coincides with the result of III §1B.

B. Continuous spectrum

To obtain the Green’s function at E > 0 we notice that

κ =

√−2mE

~2= ∓ik , k =

√2mE

~2. (3.141)

Therefore, instead of (3.137) we obtain

G±E(x, x′) = ± im~2k

e±ik|x−x′| . (3.142)

Schrodinger equation for scattering problem where particles with momentum p = ~k are incident from the left isequivalent to the following integral equation

ψ+k (x) = eikx −

∫G+E(x, x′)U(x′)ψ+

k (x′)dx′ . (3.143)

The first term in the right-hand-side corresponds to the incident particles, while the second term describes the reflectedand transmitted waves. Indeed, U(x′) in the integrand is non-vanishing only at x . a, where a is the range of the

(7 The boundary condition (3.5) has to be slightly modified: ∆ψ′(x0) = −2mα/~2 because of the difference between (3.2) and (3.131).


potential. At |x| a we can approximate |x − x′| ≈ |x| so that asymptotic form of the solution to the Schrodingerequation becomes:

ψ+k (x) ≈

eikx −

im~2k

∫U(x′)ψ+

k (x′)dx′eikx = A1e

ikx , x→ +∞ ,

eikx −im~2k

∫U(x′)ψ+

k (x′)dx′e−ikx = eikx +B2e

−ikx , x→ −∞ .(3.144)

Compare this with (3.15),(3.16). Schrodinger equation in the integral form (3.143) is the starting point of the scatteringtheory.

∗ Example. In potential U = αδ(x) (3.143) reads

ψ+k (x) = eikx − imα

~2keik|x|ψ+

k (0) . (3.145)

At x = 0 this implies

ψk(0) =~2k

~2k + imα. (3.146)

This coincides with the wave functions given by (3.15),(3.16), (3.18),(3.19).


IV. REPRESENTATION THEORY

§1. Representations of states

A. Hilbert space

Consider a quantum system described by a set of quantum numbers a. Let the corresponding wave function beψa(q). Position q is an eigenvalue of a Hermitian position operator q. For a single particle q is the same as r. Froma mathematical standpoint there is nothing special about the operator q and its eigenvalues q. Therefore, one candescribe the same quantum system by different functions ϕa(p), Φa(En) etc., which depend on particle momenta,energy, etc. These are called representations and the mathematical framework that describes the relationships betweendifferent representations is called the representation theory.

In any representation, expectation values of physical quantities must be the same. It is therefore convenient tothink about an abstract state vector |a〉, called ket, that incorporates the physical information about the state of thesystem but does not dependent on a particular representation. All possible state vector form an abstract vector space.Superposition principle dictates that α|a〉 + β|b〉 is also a possible vector state. In other words, the vector space islinear.

For every ket |a〉 we introduce a dual state vector bra 〈a| such that |a〉† = 〈a|. We also define the scalar product oftwo state vectors |a〉 and |b〉 as 〈b|a〉. Clearly, 〈a|b〉† = 〈b|a〉. Linear vector space with defined scalar product is calledthe Hilbert space.

Similarly to the more familiar Euclidean space, we can project a state vector |a〉 onto any complete set of linearlyindependent basis vectors. In Euclidean space such projection are called coordinates; any vector A is completelydescribed by a set of ordered coordinates. It is convenient to choose the orthonormal basis vectors. By the sametoken, the infinite dimensional Hilbert space can be span by a set of orthonormal basis state vectors. Since spectrumof any Hermitian operator is complete and orthonormal, it can be chosen as a set of basis state vectors. Unlike theEuclidean space where projections are real numbers, projections in Hilbert space are functions.

For example, let’s take spectrum of the position operator r as a basis set in the Hilbert space. Denote the basisvectors by |r〉. Projections of a state vector |a〉 onto this basis states are 〈r|a〉. They specify the wave functions of asystem in state with quantum numbers a at any point r. In other words, 〈r|a〉 = ψa(r).

B. Discrete (energy) representation

Choose as a basis set of functions the eigenfunctions of the Hamiltonian. I will assume that the energy spectrum isdiscrete; each eigenstate can be labeled by a set of numbers En. Denote by ϕn(q) = 〈q|En〉 the stationary state in thecoordinate representation and by ϕ∗n(q) = 〈En|q〉 = 〈q|En〉† its complex conjugate. Orthonormality of ϕn(q)n, viz.

∫ϕ∗m(q)ϕn(q)dq = δmn (4.1)

implies∫dq〈Em|q〉〈q|En〉 ≡ 〈Em|En〉 = δmn . (4.2)

To change from the coordinate representation ψa(q) = 〈q|a〉 of a state vector |a〉 to energy representation we expandin a set of functions ϕn(q)n (in two equivalent notations)

ψa(q) =∑

n

ϕn(q)ψa(En) , (4.3)

〈q|a〉 =∑

n

〈q|En〉〈En|a〉 . (4.4)

Here ψa(En) = 〈En|a〉 is the wave function of state q in the energy representation. It has a simple physical meaning:the probability to find the system with energy En is Pn = |ψa(En)|2 = |〈En|a〉|2.

In q-representation the normalization condition reads∫dq〈a|q〉〈q|a〉 = 1 . (4.5)


Using

〈a|q〉 =∑

n

〈a|En〉〈En|q〉 , (4.6)

〈q|a〉 =∑

n

〈q|En〉〈En|a〉 , (4.7)

we obtain∑

n

〈a|En〉〈En|a〉 ≡∑

n

|ψa(En)|2 = 1 , (4.8)

i.e. wave functions in the energy representation are also normalized.Transformation inverse to (4.3) is (in two notations)

ψa(En) =

∫dqϕ∗n(q)ψa(q) , (4.9)

〈En|a〉 =

∫dq〈En|q〉〈q|a〉 (4.10)

C. Continuous (momentum) representation

The basis functions in momentum representation are the eigenfunctions of the momentum operator ϕp(q) = 〈q|p〉.We assume that the momentum spectrum is continuous. Orthonormality condition:

∫dqϕ∗p′(q)ϕp(q) = δ(p′ − p) , (4.11)

∫dq〈p′|q〉〈q|p〉 = 〈p′|p〉 = δ(p′ − p) . (4.12)

Expanding the wave function ψa(a) of a state a in in a complete set of momentum eigenstates we get:

ψa(q) =

∫dpϕp(q)ψa(p) , (4.13)

〈q|a〉 =

∫dp〈q|p〉〈p|a〉 , (4.14)

where we denoted ψa(p) = 〈p|a〉 the state vector |a〉 in the p-representation. Probability density reads

ρ(p) = |ψa(p)|2 = |〈p|a〉|2 . (4.15)

Inverse transformation:

〈p|a〉 =

∫dq〈p|q〉〈q|a〉 . (4.16)

D. General case

In summary, given state vector |a〉 in m-representation 〈m|a〉 we can transform it to ξ-representation as

〈ξ|a〉 =∑

m

〈ξ|m〉〈m|a〉 , (4.17)

where 〈q|m〉 are eigenfunctions of m in ξ-representation. Conversely,

〈m|a〉 =∑

ξ

〈m|ξ〉〈ξ|a〉 , (4.18)


As a more specific example, let

ψ(θ, φ) =

√3

4πsin θ sinφ. (4.54)

We can expand it in spherical harmonics as follows

ψ(θ, φ) =i√2

(Y1,−1 + Y1,1) . (4.55)

Hence, function ψ in the angular momentum representation is

〈1,−1|ψ〉 = 〈1, 1|ψ〉 =i√2, (4.56)

and all other 〈l,m|ψ〉’s vanish. A compact way to represent this state vector is to depict it as a column in anabstract the three-dimensional space span by the orthonormal state vectors |1,−1〉, |1, 0〉, |1, 1〉:

〈1,−1|ψ〉 =i√2

101

. (4.57)

• Additional reading: Merzbacher, “Quantum Mechanics” 3rd edition, 9.3–9.4.

§3. Eigenvalue problem in matrix form

In continuous representations eigenvalue problem for operator F can be written down as a differential equation.For example, in the q-representation

FψF (q) = FψF (q) . (4.58)

In a discrete representation the eigenvalue problem can be formulated as a matrix equation. To derive it expand aneigenfunction ψF (q) in a discrete, say, energy representation:

ψF (q) = 〈q|F 〉 =∑

n

〈q|En〉〈En|F 〉 . (4.59)

Substitute into (4.58):

F∑

n

〈q|En〉〈En|F 〉 = F∑

n

〈q|En〉〈En|F 〉 . (4.60)

Now multiply by the function 〈Em|q〉 and integrate over all q:∫dq〈Em|q〉F

∑

n

〈q|En〉〈En|F 〉 = F

∫dq〈Em|q〉

∑

n

〈q|En〉〈En|F 〉 . (4.61)

Using the definition (4.33) and (4.2) we can write

∑

n

Fmn〈En|F 〉 = F∑

n

δmn〈En|F 〉 , (4.62)

or, equivalently

∑

n

Fmn − Fδmn 〈En|F 〉 = 0 . (4.63)

Here ψF (En) = 〈En|F 〉 is the wave function in the E-representation. There are non-trivial solutions to (4.63) if

det Fmn − Fδmn = 0 . (4.64)


Since this determinant is infinite-dimensional, it has infinitely many solutions F1, F2, . . . which are eigenvalues of F .Once Fn’s are known, substitute them into (4.63) to determine the wave functions 〈En|F 〉.

It is convenient to represent functions 〈En|F 〉 as a column〈E1|F 〉〈E2|F 〉. . .

. (4.65)

Then (4.64) can be written as∣∣∣∣∣∣∣

F11 − F F12 F13 . . .F21 F22 − F F23 . . .F31 F32 F33 − F . . .. . . . . . . . . . . .

∣∣∣∣∣∣∣= 0 (4.66)

Thus, the eigenvalue problem in a discrete representation is equivalent to the problem of matrix diagonalization.As has been mentioned above, any operator is diagonal in its own representation. Indeed,

Fmn =

∫dq〈Fm|q〉F 〈q|Fn〉 =

∫dq〈Fm|q〉Fn〈q|Fn〉 = Fnδmn . (4.67)

∗ Example: Harmonic oscillator in the energy representation. Matrix elements of the position operator x in theenergy representation read according to (4.33)

xmn =

∫ +∞

−∞ψ∗m(x)xψn(x)dx , (4.68)

where ψn’s are given by (3.95). Using the identity (3.99) we obtain that the matrix elements are

xm,n =

√~

2mω(√nδm,n−1 +

√n+ 1δm,n+1) . (4.69)

Similarly,

(x2)mn =

∫ +∞

−∞ψ∗m(x)x2ψn(x)dx (4.70)

=~

2mω

[√n(n− 1)δm,n−2 + (2n+ 1)δm,n +

√(n+ 1)(n+ 2)δm,n+2

]. (4.71)

(p2)mn =

∫ +∞

−∞ψ∗m(x)(−i~∂x)2ψn(x)dx = −~2

∫ +∞

−∞ψ∗m(x)ψ′′n(x)dx (4.72)

= −~mω2

[√n(n− 1)δm,n−2 − (2n+ 1)δm,n +

√(n+ 1)(n+ 2)δm,n+2

]. (4.73)

Hmn =(p2)mn

2m+mω2(x2)mn

2= ~ω

(n+

1

2

)δm,n = Enδmn . (4.74)

Eq. (4.74) is a particular case of (4.67): the Hamiltonian is diagonal in the energy representation.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §11,23.

§4. Unitary operators

Transformation of the wave function between the representations can be thought of as an action of an certain oper-ator U (8. Consider, for example a transformation from a discrete F -representation to the discrete E-representation:

ψ(F ) = 〈Fm|ψ〉 =∑

n

〈Fm|En〉〈En|ψ〉 =∑

n

Umn〈En|ψ〉 = Uψ(E) . (4.75)

(8 Don’t confuse it with the operator of potential energy.


The second equations is a definition of Umn’s. In particular we can take |ψ〉 = |Fk〉 in which case (4.75) becomes

〈Fm|Fk〉 = δmk =∑

n

Umn〈En|Fk〉 =∑

n

UmnU∗kn =

∑

n

UmnU†nk , (4.76)

In matrix notation this reads as

U U† = 1 , (4.77)

or

U† = U−1 . (4.78)

Such operators are called unitary, see II §4. This definition can be generalized to continuous representations as well,in which case matrix elements Umn become kernel of integral transformation.

To find how an arbitrary operator P transforms from E to F -representation consider its action of some state:ψ′(E) = P (E)ψ(E). Substituting (4.75) we get

U−1ψ′(F ) = P (E)U−1ψ(F ) ⇒ ψ′(F ) = U P (E)U−1ψ(F ) . (4.79)

So, if we define operator P (F ) in F -representation as ψ′(F ) = P (F )ψ(F ), then

P (F ) = U P (E)U−1 . (4.80)

This guarantees that the expectation values of physical quantities are the same in different representations:

〈P (F )〉 = 〈ψ(F )|P (F )|ψ(F )〉 = 〈ψ(E)|U†P (F )U |ψ(E)〉 = 〈ψ(E)|P (E)|ψ(E)〉 = 〈P (E)〉 . (4.81)

We see, that every physical quantity can be represented by an infinite number of operators, differing from oneanother by unitary transformations.

Transformation between different representations in the Hilbert space is of course not the only example of unitaryoperators. For example, transformation between difference reference frames in Euclidean space and time evolutionare also described by the unitary operators as we saw in II §5 and II §4. The main property of the unitary operator isthat it preserves the “lengths” of vectors and the “angles” between them. In the Euclidean space these are literallylengths and angles, in the Hilbert space these are normalization and scalar products of the state vectors.

Here are some properties of the unitary transformations.

1. Invariance of the scalar product:

〈ψ(E)|ψ′(E)〉 = 〈ψ(F )|ψ′(F )〉 . (4.82)

2. Invariance of matrix elements

〈ψm(F )|P (F )|ψn(F )〉 = 〈ψm(E)|P (E)|ψn(E)〉 (4.83)

can be proved as (4.81).

3. Invariance of trace:

trP (F ) = trU P (E)U−1 = trP (E)U−1U = trP (E) . (4.84)

4. Invariance of the eigenvalue problem P (E)ψ(E) = Pψ(E)

U−1P (F )U U−1ψ(F ) = P U−1ψ(F ) ⇒ P (F )ψ(F ) = Pψ(F ) . (4.85)

5. If operator P (E) is Hermitian, then operator P (F ) is also Hermitian. Indeed,

P (F ) = U P (E)U−1 ⇒ P †(F ) = (U−1)†P †(E)U† = U P (E)U† = P (F ) . (4.86)


6. Transformation of commutation relations:

[P (E), Q(E)] = U−1U︸︷︷︸1

P (E) U−1U︸︷︷︸1

Q(E) U−1U︸︷︷︸1

− U−1U︸︷︷︸1

Q(E) U−1U︸︷︷︸1

P (E) U−1U︸︷︷︸1

= U−1P (F )Q(F )U − U−1Q(F )P (F )U = U−1[P (F ), Q(F )]U . (4.87)

7. Any unitary operator can be represented as

S = eiF , with F † = F , (4.88)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §12, Merzbacher,“Quantum Mechanics” 3rd edition, 9.4-9.6, 10.1-10.4.

§5. Schrodinger equation in momentum representation

Eigenvalue problem for the Hamiltonian can be formulated in the momentum representation. Formally, we can use(4.50). However, it is not very convenient, for although the kinetic term p2/2m is a constant, U may be a complicatedfunction of the operator ∂p. We need an explicit form of the U operator in the momentum space. I will show how toderive it in one dimension. Generalization to three dimensions is straightforward.

We start with the Schrodinger equation in the form that does not refer to any particular representation:

H|ψ〉 = E|ψ〉 . (4.89)

Expand |ψ〉 =∫dp′|p′〉〈p′|ψ〉:

∫dp′H|p′〉〈p′|ψ〉 = E

∫dp′|p′〉〈p′|ψ〉 . (4.90)

Now multiply on the left by the bra 〈p|:∫dp′〈p|H|p′〉〈p′|ψ〉 = E

∫dp′〈p|p′〉〈p′|ψ〉 . (4.91)

The matrix element of the Hamiltonian reads

〈p|H|p′〉 =p2

2mδ(p′ − p) + 〈p|U |p′〉 . (4.92)

where I used 〈p|p′〉 = δ(p− p′). We calculate the matrix element of U as follows

〈p|U |p′〉 =

∫dx′∫dx〈p|x〉〈x|U |x′〉〈x′|p′〉 =

∫dx′∫dx〈p|x〉U(x)δ(x′ − x)〈x′|p′〉

=1

2π~

∫U(x)ei(p

′−p)x/~dx . (4.93)

Assembling everything together and recalling that ϕ(p) = 〈p|ψ〉 is the wave function in the momentum representationwe arrive at the Schrodinger equation in the p-representation

p2

2mϕ(p) +

∫ +∞

−∞〈p|U |p′〉ϕ(p′)dp′ = Eϕ(p) . (4.94)

∗ Examples.

1. Bound state in the delta-well U = −αδ(x). Note that the potential is an even function of coordinate, so weexpect that the eigenstates can have definite parity.

〈p|U |p′〉 =1

2π~

∫eix(p′−p)/~(−α)δ(x) = − α

2π~. (4.95)


Substituting into (4.94) we get

p2

2mϕ(p)− α

2π~C = Eϕ(p) , (4.96)

where I denoted

C =

∫ +∞

−∞ϕ(p)dp . (4.97)

Solution to (4.96) is

ϕ(p) =mα

π~C

p2 + 2m|E| . (4.98)

I used that E = −|E| < 0. Now, inserting this into (4.97) and taking the integral yields

C =α

~

√m

2|E|C . (4.99)

Hence, the only bound state is

E = −mα2

2~2. (4.100)

Normalization is fixed from∫ϕ2(p)dp = 1 as

C =

√2πmα

~. (4.101)

2. Bound states in

U = −α[δ(x− a) + δ(x+ a)] , α > 0 . (4.102)

First we compute the Fourier image of the potential

〈p|U |p′〉 =1

2π~

∫eix(p−p′)/~(−α)[δ(x− a) + δ(x+ a)] = − α

2π~

[eia(p′−p)/~ + e−ia(p′−p)/~

]. (4.103)

Then write down the Schrodinger equation

p2

2mϕ(p)− α

2π~

[eiap/~C+ + e−iap/~C−

]= Eϕ(p) , (4.104)

where

C± =

∫ +∞

−∞e∓iap

′/~ϕ(p′)dp′ . (4.105)

Denote

κ2 = −2mE

~2> 0 , α =

mα

~2. (4.106)

Then

ϕ(p) =α~π

(eiap/~C+ + e−iap/~C−

) 1

p2 + ~2κ2. (4.107)

Substituting in (4.105) we derive a system of two linear equations with respect to C±:

C+ =α

κ

(C+ + e−2κaC−

), (4.108)

C− =α

κ

(e−2κaC+ + C−

). (4.109)


There is a non-trivial solution iff∣∣∣∣∣

1− ακ − ακ e−2κa

− ακ e−2κa 1− ακ

∣∣∣∣∣ = 0 . (4.110)

This is equivalent to the requirement that

1− α

κ= ± α

κe−2κa , ⇒ κ = α

(1± e−2κa

). (4.111)

Solution to this equation gives the energy levels of the system.

Consider first the ′+′ solution:

κ = α(1 + e−2κa

). (4.112)

We have

C+ =1

1 + e−2κa

(C+ + e−2κaC−

)⇒ C+ = C− ⇒ ϕ(p) = ϕ(−p) . (4.113)

We thus found an even stationary state, which is the only even state of this system. It is also the ground state,because the ground state cannot be odd.

In the limit κa 1 the two δ-wells are very close:

κ ≈ 2α ⇒ E+ = −2mα2

~2. (4.114)

In the opposite limit κa 1 we have e−2κa 1, so that in the first approximation κ ≈ α. In the secondapproximation κ = α(1− e−2αa). Therefore,

E+ = −mα2

2~2

(1 + e−2αa

). (4.115)

The ′−′ solution is a problem in one of the home assignments.

• Additional reading: Merzbacher, “Quantum Mechanics” 3rd edition, 2.2, 2.5.

§6. Occupation number representation of harmonic oscillator

In III §4B we represented the harmonic oscillator in terms of the ladder operators. The raising operator a† movesthe oscillator up to the next excited state, while the lowering operator a moves it down. We can give a differentinterpretation to these operator. Instead of counting the quantum states (ground state, first excited state, etc.) wewill count the number of quantum excitations, or simply quanta. Each quantum excitation of harmonic oscillatorhas energy E = ~ω, see (3.96). For example, harmonic quantum excitation of the lattice structure of a solid body isphonon, harmonic excitation of the electromagnetic field is photon. We will say that the n’th state contains n quantaand denote it by the ket |n〉. The physical meaning of the ladder operators is adding or removing a single quantumto the hormonic oscillator:

a|n〉 =√n|n− 1〉 , (4.116)

a†|n〉 =√n+ 1|n+ 1〉 . (4.117)

For this reason a and a† are also called the annihilation and creation operators.We can write the excitation number operator n in terms of the ladder operators:

n = a†a . (4.118)

Indeed, it is easy to check that

n|n〉 = n|n〉 . (4.119)


Hamiltonian of the harmonic oscillator is

H = ~ω(a†a+ 1/2) = ~ω(n+ 1/2) . (4.120)

Thus, states |n〉 are eigenstates of the Hamiltonian and the excitation number operator. State |0〉 is the ground statewhich we normalize as usual 〈0|0〉 = 1. The excited states can be obtain by a successive application of the creationoperator

|n〉 =1√n!

(a†)n|0〉 . (4.121)

You can verify that 〈n|n〉 = 1.State vectors |n〉 form a complete set of states because they are eigenstates of the Hamiltonian. Therefore, instead

of the coordinate representation that we discussed in III §4 one can introduce the occupation number representation.For any state

|ψ〉 =∑

n

|n〉〈n|ψ〉 , (4.122)

where |〈n|ψ〉|2 is interpreted as a probability to have n excitations. If a wave function is known in the coordinaterepresentation, then (4.122) implies that

〈q|ψ〉 =∑

n

〈q|n〉〈n|ψ〉 , (4.123)

In particular, a wave function describing a single particle in one dimension can be expanded as

ψ(x) = 〈x|ψ〉 =

∞∑

n=0

〈x|n〉〈n|ψ〉 =

∞∑

n=0

Cnψn(x) , (4.124)

where ψn(x) are eigenfunctions of the harmonic oscillator given by (3.95). The probability to have n excitations inthe state |ψ〉 is |Cn|2.

In matrix notation the state vectors |n〉 can be represented as follows

|0〉 =

1000. . .

, |1〉 =

0100. . .

, |2〉 =

0010. . .

, etc (4.125)

Ladder operators become matrices amn = 〈m|a|n〉 and a†mn = 〈m|a†|n〉:

a =

0√

1 0 0 . . .

0 0√

2 0 . . .

0 0 0√

3 . . .. . . . . . . . . . . . . . .

, a† =

0 0 0 . . .√1 0 0 . . .

0√

2 0 . . .

0 0√

3 . . .. . . . . . . . . . . .

(4.126)

By the way, we see that (a†)† = a as required. The number operator is a diagonal matrix

n =

0 0 0 . . .0 1 0 . . .0 0 2 . . .. . . . . . . . . . . .

(4.127)

Although the ladder operators are not Hermitian and therefore do not correspond to any physical quantity, theireigenstates have clear physical interpretation. The eigenvalue problem for harmonic oscillator reads

a|α〉 = α|α〉 . (4.128)


In the coordinate representation:

aψα(x) = αψα(x) . (4.129)

Employing (3.106) we can write it as a differential equation

~mω

ψ′ + (x− x0)ψ = 0 , with x0 = α

√2~mω

. (4.130)

Notice, that α is a complex number. Replacing x by x′ = x − x0 we obtain the same equation as in (3.112). Itssolution is given by (3.113):

ψα(x) =(mωπ~

)1/4

emω(x0−x∗0)x0/4~e−mω(x−x0)2/2~ . (4.131)

It satisfies the normalization condition∫|ψα|2dx = 1.(9 We can write it in terms of the expectation value

⟨(∆x)2

⟩=

~/(2mω) as follows:

ψα(x) = e(x0−x∗0)x0/8〈(∆x)2〉 1

(2π 〈(∆x)2〉)1/4e−(x−x0)2/4〈(∆x)2〉 . (4.132)

This is precisely the wave function of a coherent state generalized to complex x0, see (1.133) with p0 = 0. We observethat the eigenstates of the annihilation operator are the coherent states.

What is the probability that a coherent state contains n excitations of the harmonic oscillator? To answer thisquestion we can use (4.124) and expand (4.132) into the harmonic oscillator eigenstate basis:

ψα(x) =

∞∑

n=0

Cnψn(x) , (4.133)

where ψn are given by (3.95). Coefficients Cn can be computed using the orthonormality of the ψn’s:

C∗n =

∫ +∞

−∞ψ∗n(x)ψα(x)dx (4.134)

= emω(x0−x∗0)x0/4~(mωπ~

)1/2 1√2nn!

∫ +∞

−∞Hn

(x

√mω

~

)e−mωx

2/(2~)e−mω(x−x0)2/2~dx

= e(α−α∗)α/2 1√2nn!π

∫ +∞

−∞Hn(ξ)e−ξ

2/2−(ξ−√

2α)2/2dξ (4.135)

= e(α−α∗)α/2 1√2nn!π

∫ +∞

−∞(−1)neξ

2 dne−ξ2

dξne−ξ

2+√

2αξ−α2

dξ , (4.136)

where I used (3.90). Integrating by parts n times we get

C∗n = e(α−α∗)α/2 e−α2

√2nn!π

(√

2α)n∫ +∞

−∞e−ξ

2

e√

2αξdξ (4.137)

= e(α−α∗)α/2 e−α2

√2nn!π

(√

2α)neα2/2√π =

αn√n!e−|α|

2/2 . (4.138)

Thus, the probability to find a coherent state |α〉 with n harmonic oscillator excitations is

Pn = |Cn|2 =|α|2nn!

e−|α|2

=〈n〉nn!

e−〈n〉 , (4.139)

where 〈n〉 = 〈α|n|α〉 = |α|2 is the average number of excitations in the given coherent state. Distribution (4.139)is the Poisson distribution, which gives the probability to have n independent excitations in a coherent state withaverage excitation number 〈n〉. Independence of excitations is closely related to the fact that the coherent state is asclose to the classical limit as a quantum system can be.

• Additional reading: Merzbacher, “Quantum Mechanics” 3rd edition, 10.6.

(9 Function (4.131) can be multiplied by any phase. I used this freedom to fix the phase so that the coefficients Cn in (4.133) have simpleform (4.138).


V. MOTION IN CENTRAL POTENTIAL

§1. General properties of motion in central potential

Motion in central potential U(r) is described by the Hamiltonian:

H = −~2∇2

2M+ U(r) , (5.1)

where r is a distance form the center, M is mass of the particle. Laplacian in spherical coordinates

∇2 =1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

(5.2)

Using (1.106) and (5.2) in (5.1) we have

H = − ~2

2Mr2

∂

∂r

(r2 ∂

∂r

)+

L2

2Mr2+ U(r) . (5.3)

Since U(r) does not depend on angles [H, L2] = [H, Lz] = 0. Thus, systems in central potential can be in stationary

states with given values of L2 and Lz. Eigenfunctions of L2 and Lz in coordinate representation are spherical harmonicsYlm(θ, φ). The corresponding eigenvalues are L2 = ~2l(l+1), l = 0, 1, 2 . . . and Lz = ~m, m = 0,±1, . . . ,±l. Therefore,

we can look for a solution to the Schrodinger equation equation Hψ = Eψ in the form

ψElm(r, θ, φ) = REl(r)Ylm(θ, φ) . (5.4)

In place of the radial function R it is convenient to introduce a different function

χ(r) = rR(r) . (5.5)

Schrodinger equation becomes

− ~2

2M

d2χ

dr2+

[U(r) +

~2l(l + 1)

2Mr2

]χ = Eχ . (5.6)

Since R(r) must be finite at r → 0, we conclude that

limr→0

χ(r) = 0 . (5.7)

Normalization condition for the wave function of the discrete spectrum:∫ψ∗Elm(r, θ, φ)ψE′lm(r, θ, φ)d3r = δEE′

∫R2Elr

2dr = δEE′

∫χ2El(r)dr = δEE′ , (5.8)

while for the continuous spectrum:∫ψ∗klm(r, θ, φ)ψk′lm(r, θ, φ)d3r =

∫ ∞

0

χk′(r)χk(r)dr = δ(k − k′) , (5.9)

where k2 = 2ME/~2 ≥ 0 as usual.Some properties of the Schrodinger equation with the central potential:

1. For each l there is (2l + 1)-fold degeneracy corresponding to the values of m. States with l = 0, 1, 2, 3 . . . aredenoted by letters s, p, d, f, . . ..

2. Since [H, P ] = 0, all stationary states can be chosen to have definite parity. In fact the spherical harmonicsdo have definite parity. To see this, note that in spherical coordinates spatial inversion corresponds to thetransformation θ → θ′ = π − θ and φ → φ′ = φ + π. This is easy to verify by writing down the sphericalcoordinates of r and observing that r → r′ = −r. Under such transformation

P Ylm(θ, φ) = Ylm(π − θ, φ+ π) = (−1)lYlm(θ, φ) . (5.10)

Thus, spherical harmonics are eigenfunctions of P .


3. Eq. (5.6) looks like the Schrodinger equation for one dimensional motion in the effective potential

Ul = U(r) +~2l(l + 1)

2Mr2, (5.11)

with boundary condition (5.7). Hence, we can use the results of III §5 for one dimensional motion. In particular,the energy spectrum is non-degenerate.

4. Suppose that the potential is such that U → 0 as r → ∞ and consider particle with negative energy. Suchparticle is constrained to move in a volume around the origin and its energy spectrum is discrete. Multiplying(5.6) by χ and integrating

− ~2

2M

∫ ∞

0

χ′′χdr +

∫ ∞

0

Ulχ2dr =

∫ ∞

0

Eχ2dr . (5.12)

Since at χ→ 0 as r →∞ we can integrate the first term by parts

−∫ ∞

0

χ′′χdr =

∫ ∞

0

(χ′)2dr , (5.13)

which yields for the average energy

E =~2

2M

∫ ∞

0

(χ′)2 +

[l(l + 1)

r2+

2M

~2U(r)

]χ2(r)

dr . (5.14)

Let us estimate this integral in the case of the potential that is non-vanishing only at r ≤ a and has the formU(r) = −A/rn with positive A and n. By uncertainty relation χ′ ∼ χ/a, implying that

E ∼ ~2

2M

[1 + l(l + 1)

a2− 2M

~2

A

an

]. (5.15)

If n > 2, then E → −∞ is at a = 0, which describes “falling onto the center”. However, if n < 2, then theminimum of E is at a finite a, implying that no falling into the center occurs. In this case the energy spectrumstarts with the finite negative value. Recall that in the classical mechanics falling onto the center happens atany n > 0.

FIG. 7: falling onto the center happens at n < 2.

5. According to the oscillation theorem, if for a given l we arrange the energy eigenvalues in order of increasingmagnitude El0 < El1 < El2 < . . . and assign the corresponding states the radial quantum number nr = 0, 1, 2, . . .,then the number of zeros of χ(r) and of R(r) in the interval 0 < r <∞ is nr. (r = 0 is not counted).

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §32,18, Merzbacher,“Quantum Mechanics” 3rd edition, 12.1.


§2. Spherical waves

Free motion of a particle with given l and m is called the spherical wave. The corresponding solutions of Schrodingerequation can be found form (5.6) with U = 0:

d2χldr2

− ~2l(l + 1)

r2χl + k2χl = 0 . (5.16)

Consider first s-states l = 0:

χ′′0 + k2χ0 = 0 ⇒ χ0(r) = A sin(kr) +B cos(kr) . (5.17)

In view of the boundary condition (5.7), B = 0. Substituting into (5.9) we get

A2

∫ ∞

0

sin(k′r) sin(kr)dr =A2

2

∫ +∞

−∞

eik′r − e−ik′r

2i

eikr − e−ikr2i

dr

=A2

2(2i)2−2δ(k′ − k) + 2δ(k′ + k)︸︷︷︸

=0 (k>0, k′>0)

2π =πA2

2= 1 ⇒ A =

√2

π. (5.18)

Therefore,

Rk0(r) =χk0(r)

r=

√2

π

sin(kr)

r. (5.19)

In general, for l 6= 0 (5.16) has a solution in terms of the spherical Bessel and Neumann functions jl and ηl:

Rkl(r) = Ajl(kr) +Bηl(kr) . (5.20)

Spherical Bessel functions can be expressed through the Bessel functions of half-integer order:

jl(z) =

√π

2zJl+1/2(z) = (−1)l

(d

dz

)lsin z

z, (5.21)

ηl(z) = (−1)l+1

√π

2zJ−l−1/2(z) , (5.22)

Asymptotic behavior of these functions:

jl(z) ≈

zl

(2l+1)!! , z l

1z cos

[z − π

2 (l + 1)], z l

(5.23)

ηl(z) ≈

− (2l−1)!!zl+1 , z l

1z sin

[z − π

2 (l + 1)], z l

(5.24)

Evidently, only jl satisfies the boundary condition (5.7). Therefore, the general solution to the Schrodinger equationfor a free particle with given l and m reads

ψklm(r) = Ajl(kr)Ylm(θ, φ) . (5.25)

Eq. (5.16) describes one-dimensional motion of a particle in the effective potential Ul = ~2l(l+1)/2Mr2. Classically

particle moving with energy E is allowed to move only in the region E ≥ U , which corresponds to r ≥ rl =√l(l + 1)/k.

Thus, at r < rl the wave function χl(r) exponentially decays toward r → 0. In the classically accessible region r rl,we can neglect the effective potential:

χ′′l (r) + k2χl(r) = 0 , kr 1 ⇒ χl(r) ≈ A sin(kr + δl) , (5.26)

which is of course seen also in (5.23).

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §33, Merzbacher,“Quantum Mechanics” 3rd edition, 12.2.


§3. Spherical potential well

As an application of the result we derived in the previous section, consider motion of a particle in the infinitelydeep spherically symmetric potential well:

U(r) =

0 , r ≤ a

∞ , r > a(5.27)

Solution inside the well is given by (5.25). At r ≥ a the wave function vanishes. Thus ψl(ka) = 0. Let me denotezeros of the spherical Bessel function of the l’th order by Xln. Index n = 1, 2, . . . enumerates zeros starting from thelowest one. For example, the lowest four zeros are

Xs1 ≡ X01 = π , Xp1 ≡ X11 = 4.493 , Xd1 ≡ X21 = 5.763 , Xs2 ≡ X02 = 6.283 . (5.28)

The energy levels are follow from ka = Xln:

Enl =~2X2

ln

2Ma2. (5.29)

The radial quantum number that corresponds to the number of zeros in a given state is nr = n− 1.

FIG. 8: Spherical well of finite depth.

Consider now a well of finite depth, see Fig. 8:

U(r) =

−U0 , r ≤ a

0 , r > a(5.30)

Bounds states have energies in the interval −U0 ≤ E ≤ 0. Schrodinger equation equation for the s-state reads (l 6= 0states can be discussed along the same lines but involve a more tedious algebra):

χ′′ +2M

~2(U0 − |E|)χ = 0 , r ≤ a , (5.31)

χ′′ − 2M

~2|E|χ = 0 , r ≥ a (5.32)

Denote

k2 =2M

~2(U0 − |E|) , κ2 =

2M

~2|E| . (5.33)

Solution to (5.31) and (5.32) read:

χ = A sin(kr) , r < a , (5.34)

χ = Be−κr , r > a . (5.35)

The logarithmic derivative χ′/χ must be continuous across r = a implying that

k cot(ka) = −κ . (5.36)


0 π

2π 3π

22π 5π

2

0

1

2

3

4

5

k˜a

κa

FIG. 9: Solution to (5.36) is shown in blue. Orange (no bound states), green (single bound state) and red (two bound states)lines correspond to solution of (5.37) with 2MU0/~2 = 1, 5, 25.

Solutions to this transcendental equation are a set of discrete numbers kn corresponding to the energy levels in thewell. Notice that k and κ are related by (5.33) as follows:

k2 + κ2 =2MU0

~2. (5.37)

So, geometrically, k′ns lie at the intersections of curves given by (5.36) and (5.37), see Fig. 9.It is seen that there exist a solution only if κa ≥ π/2, which translates into a2U0 ≥ π2~2/8M . The first excited

state exists if κa ≥ 3π/2 etc.

• Additional reading: Merzbacher, “Quantum Mechanics” 3rd edition, 12.3.

§4. Spherical harmonic oscillator

Schrodinger equation for a particle moving in the three-dimensional harmonic oscillator potential

U(r) =Mω2r2

2, (5.38)

can be solved by separation of variables in Cartesian and in spherical coordinates. The former is one of the homeassignment. Clearly, physical observables such as the energy spectrum should not depend on the choice of the referenceframe, while the eigenfunctions of operators do depend on it. Let us confirm this by solving the Schrodinger equationfor 3D harmonic oscillator in spherical coordinates. We have

[− ~2

2M

d2

dr2+Mω2r2

2+

~2l(l + 1)

2Mr2− Enl

]χnl(r) = 0 . (5.39)

Introduce a length parameter

a =

√~Mω

(5.40)

and dimensional quantities ξ = r/a and ε = E/~ω. Then (5.39) can be cast in form[d2

dξ2− ξ2 − l(l + 1)

ξ2+ 2ε

]χnl(ξ) = 0 . (5.41)


Its solution can be expressed in terms of the confluent hypergeometric function F as follows

χnl(ξ) = Nnle−ξ2/2ξl+1F (−nr, l + 3/2, ξ2) , (5.42)

where

nr =ε

2− 1

2

(l +

3

2

). (5.43)

The only property of the confluent hypergeometric function that we need to know is that χ→ 0 when ξ →∞ only ifnr = 0, 1, 2, . . .. This gives energy spectrum as

Enrl = ~ω(

2nr + l +3

2

), nr, l = 0, 1, 2, . . . (5.44)

Wave functions of the corresponding eigenstates read

ψnrlm =1

rχnrl(r)Ylm(θ, φ) . (5.45)

Note, that energy levels (5.44) depend only on a combination of quantum numbers 2nr + l = n, n is called theprincipal quantum number. States are usually referred to by specifying its quantum numbers in form (n + 1)l. Forexample, 1s state has n = 0, l = 0, 1p state has n = 0, l = 1, 2s state has n = 1, l = 0 etc., while energy levels arelabeled by n as En.

Levels with n ≥ 2 are degenerate. For instance, E2 = (7/2)~ω is six-fold degenerate. One s state with l = 0, nr = 1and five d-states l = 2, nr = 0, m = ±2,±1, 0 have this energy. While degeneration of states with different m and thesame l is due to spherical symmetry, degeneration with different l’s is accidental. To degeneracy g(n) reads (homeassignment):

g(n) =1

2(1 + n)(2 + n) . (5.46)

For odd n the result is the same.

Schrodinger equation for 3D harmonic oscillator can be solved also in parabolic coordinates, which is helpful whendiscussing atom in external electric field.


§5. Coulomb potential

Schrodinger equation with Coulomb potential

U(r) = −e2Z

r(5.47)

has special physical significance as it describes motion of electrons in the field of atomic nucleus. In such case −e iselectron charge and Z is the nucleus electric charge in units of e.


It is conventional to introduce the atomic units: atomic units of length

a =~2

Me2, (5.48)

also called the Bohr radius and atomic unit of energy

Ea =e2

a=Me4

~2. (5.49)


For electron, these units are a = 5.29 · 10−9 cm and Ea = 27.21 eV.Introduce dimensionless quantities ρ = r/a and ε = E/Ea. Then Schrodinger equation for the radial component

takes form[d2

dρ2+ 2ε+

2Z

ρ− l(l + 1)

ρ2

]χ(ρ) = 0 . (5.50)

Bound states have E < 0 so we can denote α2 = −2ε > 0:[d2

dρ2− α2 +

2Z

ρ− l(l + 1)

ρ2

]χ(ρ) = 0 . (5.51)

At ρ→∞ the terms containing the inverse powers of ρ can be neglected yielding the asymptotic solution

χ(ρ) ≈ Ae−αρ +Beαρ , ρ→∞ . ⇒ B = 0 . (5.52)

Now we are looking for a solution in the form χ(ρ) = e−αρw(ρ), where w is unknown function that we expand inpower series as follows

w(ρ) = ργ∞∑

ν=0

βνρν . (5.53)

We also need to check the boundary condition at ρ → 0. In this case χ(ρ) ≈ ργβ0 which must satisfy (5.51). Thisgives

γ(γ − 1)ργ−2 −

:0

α2ργ +2Z

ρργ − l(l + 1)ργ−2 = 0 , (5.54)

where I dropped terms that vanish faster than others. Equation γ(γ − 1) = l(l + 1) has two solutions

γ = l + 1, −l . (5.55)

Only the first of them is finite at ρ→ 0. Thus, after applying the boundary conditions we have

χ(ρ) = e−αρρl+1∑

ν

βνρν . (5.56)

Loading this in (5.51) produces the following recurrence relation

βν+1 =2[α(ν + l + 1)− Z]

(ν + l + 2)(ν + l + 1)− l(l + 1)βν . (5.57)

In order that χ be finite at large ρ the power series in (5.56) must terminate at some ν = nr. This happens if

α(nr + l + 1)− Z = 0 , ⇒ α =Z

nr + l + 1=Z

n, (5.58)

where the principal quantum number is given by

n = nr + l + 1 . (5.59)

Energy levels read

En = −α2

2Ea = −M(e2Z)2

~2

1

2n2, n = 0, 1, 2, . . . (5.60)

It is evident from (5.60) that for a given n states with l = 0, 1, . . . , n− 1 and m = 0,±1,±2, . . .± l are degenerate.The degeneracy g(n) can be computed by summing the arithmetic series

g(n) =

n−1∑

l=0

(2l + 1) =1 + 2(n− 1) + 1

2n = n2 . (5.61)


The normalized wave functions can be expressed through the associated Laguerre polynomials L as follows

Rnl(r) =χnl(r)

r=

[(2Z

na

)3(n− l − 1)!

2n(n+ l)!

]1/2

e−Zr/na(

2Zr

an

)lL2l+1n−l−1

(2Zr

na

). (5.62)

A few lowest order radial functions:

R10(r) =

(4Z3

a3

)1/2

e−Zr/a , (5.63)

R20(r) =

(Z3

8a3

)1/2(2− Zr

a

)e−Zr/2a , (5.64)

R21 =

(Z3

24a3

)1/2Zr

ae−Zr/2a . (5.65)

The normalization condition is∫ ∞

0

R2nlr

2dr = 1 . (5.66)

For the future reference let me list here several useful formulas involving expectations values in states (5.62)

〈ρ〉 =1

2Z[3n2 − l(l + 1)] (5.67)

⟨ρ2⟩

=n2

2Z2[5n2 + 1− 3l(l + 1)] , (5.68)

⟨ρ−1

⟩=

Z

n2, (5.69)

⟨ρ−2

⟩=

Z2

n2(l + 1/2), (5.70)

⟨ρ−3

⟩=

Z3

n3(l + 1)(l + 1/2)l. (5.71)

B. Continuous spectrum

Continuous spectrum lies at E > 0. Introducing k2 = 2ε ≥ 0 we rewrite the Schrodinger equation (5.50) as follows[d2

dρ2+ k2 +

2Z

ρ− l(l + 1)

ρ2

]χ = 0 . (5.72)

At ρ→∞ the asymptotic behavior of the solution is

χ(ρ) ≈ Aeikρ +Be−ikρ . (5.73)

Thus we are looking for a solution in the form

χ(ρ) = e±ikρρl+1∑

ν

βνρν . (5.74)

Plugging into (5.73) we obtain

βν+1 =2[i(ν + l + 1)k − Z

(ν + l + 2)(ν + l + 1)− l(l + 1)βν . (5.75)

This power series describes a confluent hypergeometric function. The final solution to the Schrodinger equation reads

χkl(ρ) = e±ikρρl+1F

(l + 1± 2

ik; 2l + 2,∓2ikρ

). (5.76)



§6. Effective electric potential of the hydrogen atom

Hydrogen atom consists of positively charged proton and negatively charged electron orbiting around its commoncenter of mass. Since mass of proton is about two thousand times larger than that of electron the reduced massof hydrogen atom equals electron mass M with very good accuracy, so that the quantum mechanical problem is toquantize motion of electron in potential U(r). This potential obeys the Poisson equation

∇2U = −4π(−e)ρ , (5.77)

where ρ is the charge density of the hydrogen atom. It can be computed as

ρ(r) = eδ(r)− e|ψ(r)|2 . (5.78)

where ψ is the wave function on electron in the hydrogen atom. The first term in the right-hand-side of (5.78)described the proton charge, while the second term describes the electron “cloud”. Now, ψ satisfies the Schrodingerequation

p2

2Mψ + U(r)ψ = Eψ , (5.79)

where U is a solution to (5.77). We thus arrived at the system of coupled Schrodinger equation and Poisson equationsthat must be solved simultaneously.

We can proceed by the method of consecutive approximations. At first, we will neglect the electron cloud contri-bution in (5.78). Then, U = −e2/r is a Coulomb potential. Solution to (5.79) with such potential has been discussedin the previous section. The bound state solutions are a system of wave functions ψnlm(r). The second step is tosubstitute these wave functions into (5.77) and calculate the potential U . Clearly, the result depends on electronstate. For illustration, suppose that electron is in the ground state, viz.

ψ100(r) = 2a−3/2e−r/aY00(θ, φ) . (5.80)

Poisson equation

∇2U =1

r2

∂

∂r

(r2 ∂

∂rU

)= −4π(−e)ρ ⇒ (5.81)

is convenient to write for a auxiliary function f defined as U = −ef/r:

f ′′ = −4πr

[eδ(r)− 4ea−3e−2r/a 1

4π

]. (5.82)

At r > 0, the delta-function does not contribute, so that

f ′′ =4re

a3e−2r/a . (5.83)

At large distances U must decrease at least as fast as 1/r2. This is because atom is neutral (∫ρd3r = 0) and thus the

monopole contribution to the multipole expansion vanishes. Thus, the boundary conditions at r →∞ are f → 0 andf ′ → 0. Integrating (5.83) twice we get

f(r) =4e

a3

∫ ∞

r

dr′∫ ∞

r′r′′e−2r′′/adr′′ =

4e

a3

∫ ∞

r

dr′1

4a(a+ 2r′)e2r′/a =

e

a(a+ r)e−2r/a . (5.84)

Finally,

U(r) = −e2

(1

r+

1

a

)e−2r/a . (5.85)

We see that the Coulomb potential is modified at distances comparable to or larger than the Bohr radius a. Electroncloud screens the proton potential.

One can continue the iteration process and solve the Schrodinger equation with potential (5.85) to find the modifiedwave functions etc.


§7. Origin of degeneracy of the energy spectrum

We have seen many times that there is a connection between the degeneracy of energy levels of a quantum system andits symmetry. To make a general statement connecting degeneracy and symmetry consider two Hermitian operatorsF and K. Let [F , K] 6= 0, but [F , H] = [K, H] = 0, i.e. F and K correspond to conserved quantities, but are not

simultaneously measurable. Let ψE,F be a wave function of a stationary state: HψE,F = EψE,F and an eigenfunction

of F : FψE,F = FψE,F . This is possible because F commutes with H.

Now, ψE,F is not an eigenfunction of K because [K, F ] 6= 0, i.e. KψE,F /∝ψE,F . On the other hand,

H(KψE,F ) = K(HψE,F ) = EKψE,F . (5.86)

This indicates that KψE,F is an eigenfunction of H. Thus, we come to conclusion that there are at least two different

eigenfunctions of H corresponding to the same level E: ψE,F and KψE,F .

In summary, energy level is degenerated if there are operators that commute with H, but not with each other.Note, that while symmetry implies degeneracy, degeneracy does not necessarily imply symmetry.

∗ Examples.

1. Free one dimensional motion: H = p2/2m. [H, p] = 0 and [H, P ] = 0, where P is the inversion (parity) operator.

However [p, P ] 6= 0. Therefore, levels are two-fold degenerate. Indeed, there are two independent solution to theSchrodinger equation : ψ = e±ikx.

2. For a spherically symmetric potential U(r) the Hamiltonian is invariant under rotations:

[H, Lx] = [H, Ly] = [H, Lz] = 0 . (5.87)

However,

[Li, Lj ] = i~εijkLk 6= 0 . (5.88)

This is the source of the 2l + 1 degeneracy of l’th level.

∗ Examples of accidental degeneracy, i.e. degeneracy which is not related to a symmetry.

1. Spherical harmonic oscillator.

H =p2

2M+Mωr2

2. (5.89)

Operator

Tik =pipkM

+Mωrirk (5.90)

commutes with the Hamiltonian, but does not commute with L2.

2. Coulomb potential.

H =p2

2M− e2

r. (5.91)

Operator

K =1

2Me2

(L× P − p× L

)+r

r, (5.92)

commutes with the Hamiltonian, but [Li, Kj ] 6= 0 and [Ki, Kj ] 6= 0 if i 6= j.

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §36,§10, Merzbacher,“Quantum Mechanics” 3rd edition, 12.5.


VI. ANGULAR MOMENTUM

§1. Angular momentum operator

In I §4 and I §6 I introduced the orbital angular momentum operator L and discussed some of its properties in thecoordinate representation. In particular, we have seen that its components satisfy the commutation relations (1.57).In fact, a vector operator satisfying such commutation relations has much reacher mathematical structure than isreflected by the coordinate representation, which can be used to describe only those physical quantities that haveclassical analogues. This is why in this section we are going to study a generic operator satisfying the commutationrelations (1.57) without a reference to any particular representation (apart from a few examples at the end of thissection).

Define the angular momentum operator J whose components satisfy the commutation relation

[Ji, Jk] = i~εiklJl . (6.1)

Orbital angular momentum operator L is a particular example of such an operator. In mathematical physics relations(6.1) are referred to as the the Lie algebra. Introduce the Casimir operator

J2 =∑

k

J2k = J2

x + J2y + J2

z . (6.2)

This operator commutes with all components of J (summation over the repeated indexes is implied):

[J2, Ji] = [Jk, Jk, Ji] = Jk[Jk, Ji] + [Jk, Ji]Jk = i~(εkilJkJl + εkilJlJk) = i~ εkil︸︷︷︸sym

(JkJl + JkJk)︸︷︷︸anti-sym

= 0 . (6.3)

Since [J2, Jz] = 0 physical quantities J2 and Jz can simultaneously have definite values. Therefore, there exist acomplete set of states |jm〉 such that

J2|jm〉 = J2j |jm〉 , (6.4)

Jz|jm〉 = ~m|jm〉 . (6.5)

In other words, we denote the eigenvalues of J2 and Jz operators as J2j and ~m respectively; j denotes the largest value

of m. To determine the possible values of these eigenvalues it is convenient to introduce the following non-Hermitianoperators, which are similar to the ladder operators of harmonic oscillator:

J± =1√2

(Jx ± iJy) . (6.6)

Their properties:

[J2, J±] = 0 , (6.7)

[Jz, J+] =1√2

(i~Jy + i(−i~)Jx) = ~J+ , (6.8)

[Jz, J−] = −~J− , (6.9)

[J+, J−] = ~Jz , (6.10)

We can express J2 via J± and Jz. To this end write

2J+J− = (Jx + iJy)(Jx − iJy) = J2x + J2

y + iJyJx − iJxJy = J2x + J2

y + ~Jz = J2 − J2z + ~Jz . (6.11)

Thus,

J2 = 2J+J− + J2z − ~Jz . (6.12)

Simiarly,

J2 = 2J−J+ + J2z + ~Jz . (6.13)


Consider

JzJ+|jm〉 = (~J+ + J+Jz)|jm〉 = ~J+|jm〉+ ~mJ+|jm〉 = ~(m+ 1)J+|jm〉 ⇒ J+|jm〉 ∝ |j,m+ 1〉 . (6.14)

Note, that |j,m+ 1〉 is an eigenstate of Jz corresponding to the eigenvalue ~(m+ 1). By the same token,

JzJ−|jm〉 = ~(m− 1)J−|jm〉 ⇒ J−|jm〉 ∝ |j,m− 1〉 . (6.15)

We see that operators J+ and J− are diagonal with respect to j and increase/decrease m by one. Since −j ≤ m ≤ j itfollows that m and 2j must be integers. Therefore, j can be either a non-negative integer j = 0, 1, 2, . . . or a positivehalf-integer j = 1/2, 3/2, 5/2, . . ..

We are ready to compute J2j . Start from |j,m+ 1〉 ∝ J+|jm〉 and set m = j:

J+|jj〉 = 0 , (6.16)

because m ≤ j. Now, using (6.13)

J−J+|jj〉 =1

2(J2 − J2

z − ~Jz)|jj〉 =1

2(J2j − ~2j2 − ~2j)|jj〉 = 0 . (6.17)

Thus,

J2j = ~2j(j + 1) (6.18)

∗ Examples.

1. In applications it is often convenient to know the matrix elements of operators Jx and Jy. To compute them westart with the operator equation (6.12) and take its matric element:

〈jm|J2|jm〉 = ~2j(j + 1) = 2〈jm|J+J−||jm〉+ 〈jm|J2z − ~Jz|jm〉

= 2∑

m′j′

〈jm|J+|j′m′〉〈j′m′|J−|jm〉+ ~2(m2 −m)

= 2〈jm|J+|j,m− 1〉〈j,m− 1|J−|jm〉+ ~2m(m− 1) (6.19)

Since Jx and Jy are Hermitian operators

〈j,m− 1|J−|jm〉 = 〈j,m− 1| 1√2

(Jx − iJy)|jm〉 = 〈j,m| 1√2

(Jx + iJy)|j,m− 1〉∗ (6.20)

Substituting this into (6.19) we get

2|〈j,m|J+|j,m− 1〉|2 = ~2j(j + 1)− ~2m(m− 1) = ~2(j2 + j −m2 +m) (6.21)

Thus, the only non-vanishing matrix elements are

〈j,m|J+|j,m− 1〉 = 〈j,m− 1|J−|j,m〉 =~√2

√(j +m)(j −m+ 1) . (6.22)

Consequently,

〈j,m± 1|Jx|j,m〉 =~2

√(j ∓m)(j ±m+ 1) , (6.23)

〈j,m± 1|Jy|j,m〉 = ∓ i~2

√(j ∓m)(j ±m+ 1) . (6.24)

2. The choice of z-axis as a quantization direction is a matter of convention. We can instead choose, say, x-axis.For example, a state with the orbital angular momentum l = 1 and its projection on z-axis lz = ±1 is describedby the wave function

ψl=1,lz=±1 = ∓i√

3

8πsin θe±iφ = ∓i

√3

8π

x± iyr

. (6.25)


After a cycle permutation of x, y, z: x→ y, y → z, z → x, we get a wave function

ψl=1,lx=±1 = ∓i√

3

8π

y ± izr

= ∓i√

3

8π(sin θ sinφ± i cos θ) , (6.26)

describing a state with orbital angular momentum l = 1 and its projection on x-axis lx = ±1.

In general, transformation from one complete set |jm〉 to another complete set |jm〉′ can be done by the rotationoperator

Rnγ = eiJ·nγ/~ , (6.27)

which is a generalization of (2.87), as follows:

|jm〉′ = Rnγ |jm〉 . (6.28)

Acting on the left-hand-side with a unit operator 1 =∑j′m′ |j′m′〉〈j′m′| we can write this in a matrix form

|jm〉′ =∑

j′m′

|j′m′〉〈j′m′|Rnγ |jm〉 =∑

m′

D(j)m′m|jm′〉 , (6.29)

I introduced matrix elements

D(j)m′m = 〈jm′|Rnγ |jm〉 . (6.30)

D(j)m′m is called the Wigner matrix. This is the matrix element of the rotation operator (6.27) about the vector

n through the angle γ. In the case of the orbital angular momentum, projecting (6.30) onto the coordinaterepresentation we have (see e.g. IV §2):

Ylm(θ′, φ′) =

l∑

m′=−lD

(l)m′m Ylm′(θ, φ) . (6.31)

3. In the case j = 1 (6.23) can be represented as a 3× 3 matrix:

Jx =~2

0√

2 0√2 0

√2

0√

2 0

. (6.32)

Suppose that a particle is in a state with Jx = 0. Denote this state as

ψ0 =

abc

. (6.33)

Because Jx is definite, ψ0 must be an eigenstate of Jx, i.e.

Jxψ0 = 0 ⇒

0√

2 0√2 0

√2

0√

2 0

abc

=

√2

ba+ cb

= 0 . (6.34)

Thus, b = 0 and a = −c. Normalization requires that a = 1/√

2, so finally

ψ0 =1√2

10−1

. (6.35)


4. Consider a rotator (such as a diatomic molecule) with the moment of inertia I rotating in plane around itssymmetry axis. The Hamiltonian is

H =L2z

2I. (6.36)

Since [H, Lz] = 0, the wave functions of stationary states can be chosen to be also eigenfunctions of Lz = ~∂φ.The corresponding complete set is

ψm =1√2πeimφ , Em =

~2m2

2I, , m = 0,±1,±2, . . . . (6.37)

All states except the ground state (m = 0) are two-fold degenerate. This is because the Hamiltonian commutes

with the operator of inversion (reflection) with respect to the x-axis defined as Pxψ(x, y) = ψ(x,−y) (which is

equivalent to φ→ −φ), whereas [Px, Lz] 6= 0.

Instead of the states (6.37) that have definite Lz, one can choose states with definite Px = ±1:

ψ+m =

1√π

cos(mφ) , ψ−m =1√π

sin(mφ) . (6.38)

Suppose that the rotator is in a state described by the wave function

ψ = A cos2 φ , A =

√4

3π. (6.39)

We can expand

ψ =∑

m

Cm1√2πeimφ =

A

4

(e2iφ + 2 + e−2iφ

). (6.40)

It follows that the only three non-vanishing coefficients are

C0 =

√2

3, C±2 =

√1

6(6.41)

Expectation values of physical quantities

〈Lz〉 =∑

m

|Cm|2~m = 0 , (6.42)

⟨L2z

⟩=∑

m

|Cm|2~2m2 =4~2

3, (6.43)

〈E〉 =∑

m

|Cm|2Em =2~2

3I. (6.44)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §26-28,58, Merzbacher,“Quantum Mechanics” 3rd edition, Ch. 11,17.

§2. Spin

Thus far we considered only one example of the angular momentum operator, viz. the orbital angular momentumL. The corresponding quantum number l can be only integer, which as we will see, is intimately related to the factthat L has a classical analogue. In fact, we constructed L as a generalization of its classical expression.

I pointed out in VI §1 that j can be either integer or half-integer. The first evidence of half-integer angularmomentum can from the Stern-Gerlach experiment. In this experiment a beam of atoms passes through a magneticfield with fixed direction but varying absolute value in the direction x perpendicular to the beam velocity. After


passing through the magnetic field the beam is projected onto a screen. Atom in magnetic field is subject to a forceof magnitude

Fx = µB∂B

∂x

along the x-direction. Coefficient µB is a projection of the magnetic moment of a particle in the bean onto themagnetic field direction. If we compute the magnetic moment using the classical formula

µ = − e

2McL ,

where M is electron mass, and then quantize the orbital angular momentum along the magnetic field direction, thenthe possible values of the magnetic moment are

µB = − e~2Mc

m , m = 0,±1,±2, . . . ,±l .

Since l is integer, we expect to see 2l + 1 traces on the screen, which is an odd number. However, the experimentindicated that the incident beam of Ag and Na atoms produces an even number of traces. This can be explained onlyby existence of an additional half-integer contribution to the total angular momentum.

All empirical evidence points to the existence of an intrinsic angular momentum, or spin, of particles. Unlike theorbital angular momentum of a particle, which describes its motion in space, spin is an intrinsic property of a particle.Spin has no classical analogy and cannot be visualized as a rotation of a particle around its axis. Spin characterizestransformational property of the wave function under rotations.

Since spin S is an angular momentum operator, we can apply to it the formalism developed in VI §1. In particular,it satisfies the commutation relations

[Si, Sj ] = i~εijkSk . (6.45)

Let |sσ〉 be a complete set of eigenstates of operators S2 and Sz; the corresponding eigenvalues are S2 = ~2s(s + 1)and Sz = ~σ, σ = −s,−s+1, . . . , s−1, s. It serves as a basis of the spin representation. Spin s can be either integer orhalf-integer: s = 0, 1

2 , 1,32 , . . .. In the spin representation, a wave function of a particle with spin s can be represented

by a column of 2s+ 1 components called the spinor :

ψs(q)ψs−1(q). . .

ψ−s(q)

. (6.46)

The spin operator is represented by a square matrix that can obtained from (6.23),(6.24) after replacements J → S,j → s and m→ σ:

〈s, σ ± 1|Sx|s, σ〉 =~2

√(s∓ σ)(s± σ + 1) , (6.47)

〈s, σ ± 1|Sy|s, σ〉 = ∓i~2

√(s∓ σ)(s± σ + 1) , (6.48)

〈s, σ|Sz|s, σ〉 = ~σ . (6.49)

Of particular importance is the case of s = 1/2. The corresponding representation consists of two basis states withσ = ±1. In this representation

⟨1

2,−1

2

∣∣∣Sx∣∣∣12,

1

2

⟩=⟨1

2,

1

2

∣∣∣Sx∣∣∣12,−1

2

⟩=

~2⇒ Sx =

~2

(0 11 0

), (6.50)

and similarly for the other spin components. Introduce the Pauli matrices σ such that(10

S =~2σ . (6.51)

(10 Do not confuse the Pauli matrices σ with the eigenvalues of Sz operator, viz. ~σ.


From (6.50) we see that in the spinor representations they have the following form:

σx =

(0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

). (6.52)

Properties of the Pauli matrices:

1.

σ2x = σ2

y = σ2z = 1 =

(1 00 1

). (6.53)

2.

σyσz = iσx , σzσx = iσy , σxσy = iσz . (6.54)

Eq. (6.53),(6.54) can be written together in a compact form

σiσj = iεijkσk + δij 1 . (6.55)

3. Using (6.55) we derive

σi, σj = σiσj + σj σi = 2δij . (6.56)

This means that the Pauli matrices anti-commute.

4. Home assignment:

σ2 = 3 1 , (6.57)

(σ · a)(σ · b) = a · b 1 + iσ · (a× b) , (6.58)

Tr σi = 0 , (6.59)

Tr(σiσk) = 2δik , (6.60)

(a · σ)n =

an 1 n even

an−1(a · σ) , n odd(6.61)

Spin operator is a generator of rotations in the spinor space (see (2.87)):

Rnφ = eiS·nφ/~ (6.62)

Consider a rotation through 2π about the z-axis Rz2π = e2πiSz/~. The eigenvalues of this operator are

e2πiσ = cos(2πσ) + i sin(2πσ) =

1 for integer σ−1 for half-integer σ

= (−1)2s (6.63)

∗ Example. In the case s = 12 we can write (6.62) using (6.61) as

Rnφ = ei12 σ·nφ = cos

(i1

2σ · nφ

)+ i sin

(i1

2σ · nφ

)

=

∞∑

k=0

1

(2k)!(−1)k

(i1

2σ · nφ

)2k

+ i

∞∑

k=0

1

(2k + 1)!(−1)k

(i1

2σ · nφ

)2k+1

=

∞∑

k=0

1

(2k)!(−1)k

(φ

2

)2k

1 + i

∞∑

k=0

1

(2k + 1)!(−1)k

(φ

2

)2k+1

(n · σ)

= cos(φ/2) 1 + i sin(φ/2)(n · σ) . (6.64)

In particular, rotation about z-axis is given by

Rzφ = cos(φ/2) 1 + i sin(φ/2)σz =

(eiφ/2 0

0 e−iφ/2

)(6.65)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §54-56, Merzbacher,“Quantum Mechanics” 3rd edition, 16.1-16.4.


§3. Addition of angular momenta

Consider a system consisting of two parts each having angular momentum J1,2. Assume that the parts are non-

interacting so that [J1i, J2k] = 0. Then the total system can be in states with definite values of

J21 = ~2j1(j1 + 1) , and J2

2 = ~2j2(j2 + 1) (6.66)

and their z-projections

J1z = ~m1 , and J2z = ~m2 . (6.67)

Denote the corresponding state vectors by |j1m1j2m2〉 = |j1m1〉|j2m2〉. At fixed j1 and j2 there are (2j1 + 1)(2j2 + 1)eigenfunctions with different m1 and m2.

Define the operator of the total angular momentum as J = J1 + J2. It satisfies the same commutation relations asJ1 and J2:

[Ji, Jk] = i~εiklJl . (6.68)

The Casimir operator can be written as

J2 = J21 + J2

2 + 2J1 · J2 . (6.69)

Since J2 is the Casimir operator of the algebra (6.68), it commutes with all components of J , in particular, [J2, Jz] = 0.It also commutes with the Casimir operators of subsystems:

[J2, J21 ] = [J2, J

22 ] = 0 , (6.70)

because [J21 , J1k] = 0. Note, however, that [J2, J1z] 6= 0 since [J1 · J2, J1z] 6= 0.

Since Jz = J1z + J2z commutes with operators J21 , J2

2 , J1z, J2z, states |j1m1j2m2〉 are also its eigenstates corre-

sponding to the eigenvalues Jz = ~m = ~(m1 + m2). However, they are not eigenstates of the operator J2, because

J2 does not commute with J1z and J2z. Operator J1 · J2 mixes states differing by one in m1 and m2. In practice, itis advantageous to have states with definite values of J2 and Jz. Since J2

1 and J22 commute with them we can choose

as a basis set of states, eigenstates of these four operators: J2, Jz, J21 and J2

2 . I will denote such states by |j1j2jm〉.In view of completeness of both sets we can write the following unitary transformation

|j1j2jm〉 =∑

m1,m2

〈j1m1j2m2|j1j2jm〉|j1m1〉|j2m2〉 . (6.71)

The matrix elements 〈j1m1j2m2|j1j2jm〉 are called the Clebsch-Gordan coefficients. Often, when j1 and j2 are fixedthey are denoted simply by 〈m1m2|jm〉 to simplify the notation.

Properties of the Clebsch-Gordan coefficients:

1. Clebsch-Gordan coefficients are non-vanishing only if m = m1 +m2. Indeed, on the one hand

Jz|j1j2jm〉 =∑

m1,m2

〈m1m2|jm〉~(m1 +m2)|j1m1〉|j2m2〉 = ~(m1 +m2)|j1j2jm〉 . (6.72)

On the other hand, Jz|j1j2jm〉 = ~m|j1j2jm〉 implying that m1 +m2 = m.

2. At fixed j1 and j2 number j can take only the following values:

|j1 − j2| ≤ j ≤ j1 + j2 . (6.73)

A derivation of this property can be found in the book by Merzbacher, 17.5. Below I will illustrate it with anexample (Example 1).

To each value of j, at fixed j1 and j2, there correspond 2j + 1 values of m: m = −j,−j + 1, . . . , j − 1, j. Thetotal number of states with all possible values of j:

j1+j2∑

j=|j1−j2|(2j + 1) = (2j1 + 1)(2j2 + 1) . (6.74)


According to (6.71) instead of the (2j1 + 1)(2j2 + 1) basis states

|j1m1j2m2〉m1=−j1,...,j2;m2=−j2,...,j2 , (6.75)

which form a complete set in the Hilbert space, we can use another complete set

|j1j2jm〉j=|j1−j2|,...,j1+j2;m=−j,...,j (6.76)

spanning the same Hilbert space (and containing the same number (2j1 + 1)(2j2 + 1) of basis states).

3. By convention, Clebsch-Gordan coefficients are real. This implies that the inverse transformation to (6.71) is

|j1m1〉|j2m2〉 =∑

jm

〈j1j2m1m2|j1j2jm〉|j1j2jm〉 . (6.77)

4. Orthogonality and normalization∑

jm

〈j1j2m1m2|j1j2jm〉〈j1j2m′1m′2|j1j2jm〉 = δm1m′1δm2m′2

, (6.78)

∑

m1,m2

〈j1j2m1m2|j1j2j′m′〉〈j1j2m1m2|j1j2jm〉 = δjj′δmm′ , (6.79)

5. Symmetry under particle permutations:

〈j1j2m1m2|j1j2jm〉 = (−1)j1+j2−j〈j2j1m2m1|j2j1jm〉 , (6.80)

which implies that

|j1j2jm〉 = (−1)j1+j2−j |j2j1jm〉 . (6.81)

∗ Examples.

1. Consider two systems with j1 = 3 and j2 = 3/2. The total number of basis states is (3 · 2 + 1)( 32 · 2 + 1) = 28.

Let us list and count all these states in two different representations: |j1j2m1m2〉 and |j1j2jm〉. I will organizethe list starting from the largest m and decreasing it by one. Keep in mind that for a given m, m1 + m2 = mand j ≥ |m|.

|j1j2m1m2〉 |j1j2jm〉

m1 m2 # of states m j # of states

3 3/2 1 9/2 9/2 1

3 1/22 7/2

9/22

2 3/2 7/2

3 −1/23 5/2

9/232 1/2 7/2

1 3/2 5/2

3 −3/2

4 3/2

9/2

42 −1/2 7/2

1 1/2 5/2

0 3/2 3/2

2 −3/2

4 1/2

9/2

41 −1/2 7/2

0 1/2 5/2

−1 3/2 3/2

− − >

no state with j=1/2

1/2

· · · · · · · · · · · · · · · · · ·

(6.82)


The table continues symmetrically to the values m = −1/2,−3/2,−5/2,−7/2,−9/2. Going down the table Iincluded states with j = 9/2, . . . , 3/2 in order that the number of states in both sets be the same. It is seenthat for m = 1/2, the state with j = 1/2 does not exist. This agrees with the rule that 3− 3/2 ≤ j ≤ 3 + 3/2.

2. Consider the expectation value of J = J1 + J2 in a state |j1m1j2m2〉. Since J2 = J21 + J2

2 + 2J1 · J2 and〈Jx〉 = 〈Jy〉 = 0 (home assignment) we get

⟨J2⟩

= ~2j1(j1 + 1) + ~2j2(j2 + 1) + 2~2m1m2 . (6.83)

3. Using the table of the Clebsch-Gordan coefficients.

(a) j1 = 3/2, j2 = 1/2.

|m1 = 3/2,m2 = −1/2〉 =1

2|j = 2,m = 1〉+

√3

2|j = 1,m = 1〉 . (6.84)

(b) j1 = 2, j2 = 1.

|m1 = 0,m2 = 0〉 =

√3

5|j = 3,m = 0〉 −

√2

5|j = 1,m = 0〉 . (6.85)

(c) j1 = 2, j2 = 1/2.

|j = 5/2,m = 3/2〉 =1√5|m1 = 2,m2 = −1/2〉+

2√5|m1 = 1,m2 = 1/2〉 . (6.86)


§4. Matrix elements of vector operators

We have noticed in I §4 that commutation relations of vector operators with the operator of orbital angular momen-tum are similar, see (1.59) and (1.60). In II §5 we attributed this to the fact that L is proportional to the generator ofrotations: all vector quantities transform under the rotations in the same way. (This is true for any tensor quantity,but we will consider vectors for simplicity). This is why we defined a vector operator by (2.94). We can generalizethis definition as follows:

[Ji, Vj ] = i~εijkVk . (6.87)

Eq. (6.87) allows us to relate the matrix element of the operator V in the basis |jm〉 to the matrix elements of

operators J and J · V in the same basis. To derive the corresponding expression, start with the following commutator

[J2, J × V ] = 2i~J2V − (J · V )J , (6.88)

which can be proved using (6.87),(6.1) and (C7). Let us compute the matrix elements 〈jm′| . . . |jm〉 on the both sidesof (6.88). Firstly,

〈jm′|[J2, J × V ]|jm〉 = 〈jm′|J2(J × V )− (J × V )J2|jm〉 = ~2〈jm′|(J × V )j(j + 1)− j(j + 1)|jm〉 = 0 . (6.89)

Secondly, consider the following matrix element

〈jm′|(J · V )J |jm〉 = 〈jm′|(J · V )∑

j′,m′′

|j′m′′〉〈j′m′′|J |jm〉 (6.90)

It is easy to check that [J2, J · V ] = 0 and [Jz, (J · V )] = 0 (because J · V is a scalar). Therefore, state vectors |jm〉are eigenstates of operator J · V implying that

〈jm′|J · V |j′m′′〉 ∝ δm′m′′δj′j′ . (6.91)


It follows from (6.90) and (6.91) that

〈jm′|(J · V )J |jm〉 = 〈jm′|J · V |jm′〉〈jm′|J |jm〉 . (6.92)

Lastly,

〈jm′|J2V |jm〉 = ~2j(j + 1)〈jm′|V |jm〉 . (6.93)

Combining (6.88),(6.89),(6.92) and (6.93) we finally obtain

〈jm′|V |jm〉 =1

~2j(j + 1)〈jm′|J · V |jm′〉〈jm′|J |jm〉 . (6.94)

Equivalently, replace m↔ m′, take the complex conjugation on both sides and use the fact that all involved operatorsare Hermitian to write (6.94) in a slightly different way:

〈jm′|V |jm〉 =1

~2j(j + 1)〈jm|J · V |jm〉〈jm′|J |jm〉 , (6.95)

which indicates that the matrix element 〈jm|J · V |jm〉 does not depend on m at all.This is a particular case of the Wigner-Eckart theorem. The Wigner-Eckart theorem separates the geometric and

symmetry-related properties of the matrix elements (in our example 〈jm′|J |jm〉) from the other physical properties

that are contained in the reduced matrix element (in our example 〈jm|J · V |jm〉).

∗ Examples.

1. Suppose that J = J1 + J2 and let us compute the matrix elements 〈j1j2jm|J1,2|j1j2jm〉.According to (6.94)

〈jm′|J1,2|jm〉 =1

~2j(j + 1)〈j,m|J · J1,2|jm〉〈jm′|J |jm〉 . (6.96)

Decompose

J =1

2(ex − iey)J+ +

1

2(ex + iey)J− + ezJz . (6.97)

and calculate the matrix element

〈jm′|J |jm〉 =1

2~(ex − iey)

√j(j + 1)−m(m+ 1)δm′,m+1

+1

2~(ex + iey)

√j(j + 1)−m(m+ 1)δm′,m−1 + ez~mδmm′ . (6.98)

Next, using the identities

J · J1 =1

2(J2 + J2

1 − J22 ) , J · J1 =

1

2(J2 + J2

2 − J22 ) . (6.99)

we have

〈jm|J · J1|jm〉 = 〈jm|12

(J2 + J21 − J2

2 )|jm〉 =~2

2[j(j + 1) + j1(j1 + 1)− j2(j2 + 1)] . (6.100)

Notice that the reduced matrix element (6.100) does not depend on m and m′ as expected. Plugging (6.100)and (6.98) into (6.96) we derive

〈jm|J1|jm〉 = ez~m2

[1 +

j1(j1 + 1)

j(j + 1)− j2(j2 + 1)

j(j + 1)

], (6.101)

〈jm|J2|jm〉 = ez~m2

[1− j1(j1 + 1)

j(j + 1)+j2(j2 + 1)

j(j + 1)

]. (6.102)


2. Consider electron in a state with given angular orbital momentum l, and z-component m of its total angularmomentum j. The possible values of the total angular momentum are j = l+1/2 and j = l−1/2. Using (6.101)

and (6.102) with replacements J1 → L and J2 → S we have the following diagonal matrix elements

⟨l +

1

2,m; l

1

2

∣∣∣S∣∣∣l +

1

2,m; l

1

2

⟩= ez

~m2

[1− l(l + 1)

(l + 12 )(l + 3

2 )+

3

4(l + 12 )(l + 3

2 )

]= ez

~m2l + 1

, (6.103)

⟨l − 1

2,m; l

1

2

∣∣∣S∣∣∣l − 1

2,m; l

1

2

⟩= ez

~m2

[1− l(l + 1)

(l − 12 )(l + 1

2 )+

3

4(l − 12 )(l + 1

2 )

]= −ez

~m2l + 1

, (6.104)

⟨l +

1

2,m; l

1

2

∣∣∣L∣∣∣l +

1

2,m; l

1

2

⟩= ez

~m2

[1 +

l(l + 1)

(l + 12 )(l + 3

2 )− 3

4(l + 12 )(l + 3

2 )

]= ez

~m(l + 1)

2l + 1, (6.105)

⟨l − 1

2,m; l

1

2

∣∣∣L∣∣∣l − 1

2,m; l

1

2

⟩= ez

~m2

[1 +

l(l + 1)

(l − 12 )(l + 1

2 )− 3

4(l − 12 )(l + 1

2 )

]= ez

~m(l + 1)

l + 12

. (6.106)

We will see later that the magnetic dipole moment of electron is given by

µ =e

2Mc(L+ 2S) . (6.107)

Its expectation value in two states with different j are

〈µ〉j=l+ 12

=e

2Mc

⟨l +

1

2,m; l

1

2

∣∣∣(L+ 2S)∣∣∣l +

1

2,m; l

1

2

⟩= ez

e

2Mc

~m(l + 1)

l + 12

, (6.108)

〈µ〉j=l− 12

=e

2Mc

⟨l − 1

2,m; l

1

2

∣∣∣(L+ 2S)∣∣∣l − 1

2,m; l

1

2

⟩= ez

e

2Mc

~mll + 1

2

, (6.109)

• Additional reading: Landau and Lifshitz, “Quantum Mechanics: Non-relativistic theory”, §106,107, Merzbacher,“Quantum Mechanics” 3rd edition, Ch. 17.


VII.


Appendix A: Fourier analysis

Any integrable function f(x) can be expanded in a complete set of functions eiknxn in the interval −L/2 ≤ x ≤L/2, known as Fourier series, as follows

f(x) =

∞∑

n=−∞fne

iknx , kn =2πn

L. (A1)

The Fourier coefficients fn are given by

fn =1

L

∫ L/2

−L/2f(x)e−iknxdx . (A2)

In the limit L→∞ (A1),(A2) become continuous Fourier integral expansion

f(x) =

∫ +∞

−∞fke

ikx dk

2π, (A3)

fk =

∫ +∞

−∞f(x)e−ikxdx , (A4)

Three-dimensional generalization of (A3),(A4) is

f(r) =

∫ +∞

−∞fke

ik·r d3k

(2π)3, (A5)

fk =

∫ +∞

−∞f(r)e−ik·rd3r , (A6)

• Additional reading: Arfken at. al. , “Mathematical methods for physicists”, 7th edition, Ch.19, 20.1-20.4.

Appendix B: Dirac delta function

Dirac delta function δ(x) is a singular function such that

δ(x) =

0 , if x 6= 0 ,

∞ , if x = 0 ,(B1)

and∫ +∞

−∞δ(x)dx = 1 . (B2)

This definition implies the main property of the delta function: if f(x) is a smooth function and a < 0, b > 0, then

∫ b

a

f(x)δ(x)dx = f(0) . (B3)

In particular,

∫ +∞

−∞f(x)δ(x)dx = f(0) . (B4)

Shifting the origin x→ x− a we get in place of (B4)

∫ +∞

−∞f(x)δ(x− a)dx = f(a) . (B5)


Often one needs do deal with a delta function of the form δ[ϕ(x)], where ϕ is a smooth function. Suppose thatϕ(x) has one and only zero at x = x0. Then integrand in

∫ +∞

−∞f(x)δ[ϕ(x)]dx (B6)

is non-vanishing only at x = x0. Thus, we can replace ϕ with the first non-vanishing term of its Taylor expansionϕ(x) = ϕ′(x0)(x− x0). We have

∫ +∞

−∞f(x)δ[ϕ(x)]dx =

∫ +∞

−∞f(x)δ[ϕ′(x0)(x− x0)]dx. (B7)

Making change of variables y = ϕ′(x0)(x− x0) we get using (B4)

∫ +∞


1

|ϕ′(x0)|

∫ +∞

−∞f [y/ϕ′(x0) + x0]δ(y)dy =

f(x0)

|ϕ′(x0)| . (B8)

Generally, if ϕ(x) has zeros at x = xn, were n is any integer number bigger than zero, then (B8) becomes

∫ +∞


∑

n

f(xn)

|ϕ′(xn)| . (B9)

In a particular case ϕ(x) = cx, with constant c:

∫ +∞

−∞f(x)δ(cx)dx =

1

|c|f(0) . (B10)

Delta function can be represented in different forms that are often very useful in applications. As a Fourier integral:

δ(x) =

∫ +∞

−∞eikx

dk

2π. (B11)

As a limit of elementary functions:

δ(x) =1

πlimε→0

ε

x2 + ε2(B12)

=1

πlimε→0

sin(x/ε)

x(B13)

=1

2√π

limε→0

1

ε2e−x

2/4ε2 (B14)

Note that (B11) and (B13) are mutually consistent as follows:

1

2πlimε→0

∫ 1/ε

−1/ε

eikxdk =1

πlimε→0

sin(x/ε)

x(B15)

Three-dimensional delta function is defines as a product of three one-dimensional delta-functions

δ(r − r′) = δ(x− x′)δ(y − y′)δ(z − z′) . (B16)

• Additional reading: Arfken et. al. “Mathematical methods for physicists”, 7th edition, 1.11.

Appendix C: Levi-Civita symbol

Levi-Civita symbol εijk is defined as

εijk =

+1 if (i, j, k) is (1, 2, 3), (3, 1, 2), (2, 3, 1) ,

−1 if (i, j, k) is (1, 3, 2), (3, 2, 1), (2, 1, 3) ,

0 if i = j or j = k or k = i

(C1)


Levi-Civita symbol is completely antisymmetric: εikj = −εijk, εjik = −εikj etc. That is, permutation of any twonearby indices produces the minus sign. The nearby pairs of indices of εikj are i and k, k and j, j and i. With thisnotation the i’th component of vector A = B × C can be written as Ai = εijkBjCk, where summation over the

repeated indices is implied(11. The scalar product of any two vectors A and B reads A ·B = AiBi.We can use this, so called index notation to easily prove vector identities. For example,

A× (B ×C) = eiεijkAjεklmBlCm = ei(δilδjm − δimδjl)AjBlCm = B(A ·C)−C(A ·B) ,

where ei is a unit vector in i’th direction.It is helpful to remember that if matrix [S]ij is symmetric, i.e. [S]ij = [S]ji and [A]ij is antisymmetric, i.e.

[A]ij = −[A]ji, then

[S]ij [A]ij = −[S]ji[A]ji = −[S]ij [A]ij = 0 (C2)

where we first permuted the indices i→ j, j → i and than relabeled i as j and j as i.The differential operator nabla is defined as ∇ = ∂x, ∂y, ∂z, where ∂i ≡ ∂

∂xi. Some operations with nabla:

∇ ·A ≡ δij∂iAj = ∂iAi (C3)

∇φ = ei∂iφ (C4)

∇×A = εijkei∂jAk (C5)

Examples of differential identities:

∇ · (A×B) = ∂i(εijkAjBk) = εijk(∂iAj)Bk + εijk(∂iBk)Aj

= εijk(∂iAj)Bk − εikj(∂iBk)Aj = (∇×A) ·B − (∇×B) ·A

∇ · (∇×A) = ∂iεijk∂jAk = 0, (C6)

since ∂i∂j = ∂j∂i is symmetric, whereas εijk is asymmetric in any pair of its indices.A useful property of the Levi-Civita symbol:

εijkεilm = δjlδkm − δjmδkl . (C7)

It helps simplify multiple cross products.

(11 In other words, we do not write explicitly the summation sign∑

j,k. This is called the Einstein summation convention. The repeatedindices i, j are called dummy indices.


34. Clebsch-Gordan coefficients 010001-1

34. CLEBSCH-GORDAN COEFFICIENTS, SPHERICAL HARMONICS,

AND d FUNCTIONS

Note: A square-root sign is to be understood over every coefficient, e.g., for −8/15 read −√

8/15.

Y 01 =

√3

4πcos θ

Y 11 = −

√3

8πsin θ eiφ

Y 02 =

√5

4π

(3

2cos2 θ − 1

2

)

Y 12 = −

√15

8πsin θ cos θ eiφ

Y 22 =

1

4

√15

2πsin2 θ e2iφ

Y −m` = (−1)mYm∗` 〈j1j2m1m2|j1j2JM〉= (−1)J−j1−j2〈j2j1m2m1|j2j1JM〉d `m,0 =

√4π

2`+ 1Ym` e−imφ

djm′,m = (−1)m−m

′djm,m′ = d

j−m,−m′ d 1

0,0 = cos θ d1/21/2,1/2

= cosθ

2

d1/21/2,−1/2

= − sinθ

2

d 11,1 =

1 + cos θ

2

d 11,0 = − sin θ√

2

d 11,−1 =

1− cos θ

2

d3/23/2,3/2

=1 + cos θ

2cos

θ

2

d3/23/2,1/2

= −√

31 + cos θ

2sin

θ

2

d3/23/2,−1/2

=√

31− cos θ

2cos

θ

2

d3/23/2,−3/2

= −1− cos θ

2sin

θ

2

d3/21/2,1/2

=3 cos θ − 1

2cos

θ

2

d3/21/2,−1/2

= −3 cos θ + 1

2sin

θ

2

d 22,2 =

(1 + cos θ

2

)2

d 22,1 = −1 + cos θ

2sin θ

d 22,0 =

√6

4sin2 θ

d 22,−1 = −1− cos θ

2sin θ

d 22,−2 =

(1− cos θ

2

)2

d 21,1 =

1 + cos θ

2(2 cos θ − 1)

d 21,0 = −

√3

2sin θ cos θ

d 21,−1 =

1− cos θ

2(2 cos θ + 1) d 2

0,0 =(3

2cos2 θ − 1

2

)

+1

5/25/2

+3/23/2

+3/2

1/54/5

4/5−1/5

5/2

5/2−1/23/52/5

−1−2

3/2−1/22/5 5/2 3/2

−3/2−3/24/51/5 −4/5

1/5

−1/2−2 1

−5/25/2

−3/5−1/2+1/2

+1−1/2 2/5 3/5−2/5−1/2

2+2

+3/2+3/2

5/2+5/2 5/2

5/2 3/2 1/2

1/2−1/3

−1

+10

1/6

+1/2

+1/2−1/2−3/2

+1/22/5

1/15−8/15

+1/21/10

3/103/5 5/2 3/2 1/2

−1/21/6

−1/3 5/2

5/2−5/2

1

3/2−3/2

−3/52/5

−3/2

−3/2

3/52/5

1/2

−1

−1

0

−1/28/15

−1/15−2/5

−1/2−3/2

−1/23/103/5

1/10

+3/2

+3/2+1/2−1/2

+3/2+1/2

+2 +1+2+1

0+1

2/53/5

3/2

3/5−2/5

−1

+10

+3/21+1+3

+1

1

0

3

1/3

+2

2/3

2

3/23/2

1/32/3

+1/2

0−1

1/2+1/22/3

−1/3

−1/2+1/2

1

+1 1

0

1/21/2

−1/2

0

0

1/2

−1/2

1

1

−1−1/2

1

1

−1/2+1/2

+1/2 +1/2+1/2−1/2

−1/2+1/2 −1/2

−1

3/2

2/3 3/2−3/2

1

1/3

−1/2

−1/2

1/2

1/3−2/3

+1 +1/2+10

+3/2

2/3 3

3

3

3

3

1−1−2−3

2/31/3

−22

1/3−2/3

−2

0−1−2

−10

+1

−1

2/58/151/15

2−1

−1−2

−10

1/2−1/6−1/3

1−1

1/10−3/10

3/5

020

10

3/10−2/53/10

01/2

−1/2

1/5

1/53/5

+1

+1

−10 0

−1

+1

1/158/152/5

2

+2 2+1

1/21/2

1

1/2 20

1/6

1/62/3

1

1/2

−1/2

0

0 2

2−21−1−1

1−11/2

−1/2

−11/21/2

00

0−1

1/3

1/3−1/3

−1/2

+1

−1

−10

+100

+1−1

2

1

00 +1

+1+1

+11/31/6

−1/2

1+13/5

−3/101/10

−1/3−10+1

0

+2

+1

+2

3

+3/2

+1/2 +11/4 2

2

−11

2

−21

−11/4

−1/2

1/2

1/2

−1/2 −1/2+1/2−3/2

−3/2

1/2

1003/4

+1/2−1/2 −1/2

2+13/4

3/4

−3/41/4

−1/2+1/2

−1/4

1

+1/2−1/2+1/2

1

+1/2

3/5

0−1

+1/20

+1/23/2

+1/2

+5/2

+2 −1/2+1/2+2

+1 +1/2

1

2×1/2

3/2×1/2

3/2×12×1

1×1/2

1/2×1/2

1×1

Notation:J J

M M

...

. . .

.

.

.

.

.

.

m1 m2

m1 m2 Coefficients

−1/52

2/7

2/7−3/7

3

1/2

−1/2−1−2

−2−1

0 4

1/21/2

−33

1/2−1/2

−2 1

−44

−2

1/5

−27/70

+1/2

7/2+7/2 7/2

+5/23/74/7

+2+10

1

+2+1

+41

4

4+23/14

3/144/7

+21/2

−1/20

+2

−10

+1+2

+2+10

−1

3 2

4

1/14

1/14

3/73/7

+13

1/5−1/5

3/10

−3/10

+12

+2+10

−1−2

−2−10

+1+2

3/7

3/7

−1/14−1/14

+11

4 3 2

2/7

2/7

−2/71/14

1/14 4

1/14

1/143/73/7

3

3/10

−3/10

1/5−1/5

−1−2

−2−10

0−1−2

−10

+1

+10

−1−2

−12

4

3/14

3/144/7

−2 −2 −2

3/7

3/7

−1/14−1/14

−11

1/5−3/103/10

−1

1 00

1/70

1/70

8/3518/358/35

0

1/10

−1/10

2/5

−2/50

0 0

0

2/5

−2/5

−1/10

1/10

0

1/5

1/5−1/5

−1/5

1/5

−1/5

−3/103/10

+1

2/7

2/7−3/7

+31/2

+2+10

1/2

+2 +2+2+1 +2

+1+31/2

−1/20

+1+2

34

+1/2+3/2

+3/2+2 +5/24/7 7/2

+3/21/74/72/7

5/2+3/2

+2+1

−10

16/35

−18/351/35

1/3512/3518/354/35

3/2

+3/2

+3/2

−3/2−1/2+1/2

2/5−2/5 7/2

7/2

4/3518/3512/351/35

−1/25/2

27/703/35

−5/14−6/35

−1/23/2

7/2

7/2−5/24/73/7

5/2−5/23/7

−4/7

−3/2−2

2/74/71/7

5/2−3/2

−1−2

18/35−1/35

−16/35

−3/21/5

−2/52/5

−3/2−1/2

3/2−3/2

7/2

1

−7/2

−1/22/5

−1/50

0−1−2

2/5

1/2−1/21/10

3/10−1/5

−2/5−3/2−1/2+1/2

5/2 3/2 1/2+1/22/5

1/5

−3/2−1/2+1/2+3/2

−1/10

−3/10

+1/22/5

2/5

+10

−1−2

0

+33

3+2

2+21+3/2

+3/2+1/2

+1/2 1/2−1/2−1/2+1/2+3/2

1/2 3 2

30

1/20

1/20

9/209/20

2 1

3−11/5

1/53/5

2

3

3

1

−3

−21/21/2

−3/2

2

1/2−1/2−3/2

−2

−11/2

−1/2−1/2−3/2

0

1−1

3/10

3/10−2/5

−3/2−1/2

00

1/41/4

−1/4−1/4

0

9/20

9/20

+1/2−1/2−3/2

−1/20−1/20

0

1/4

1/4−1/4

−1/4−3/2−1/2+1/2

1/2

−1/20

1

3/10

3/10

−3/2−1/2+1/2+3/2

+3/2+1/2−1/2−3/2

−2/5

+1+1+11/53/51/5

1/2

+3/2+1/2−1/2

+3/2

+3/2

−1/5

+1/26/355/14

−3/35

1/5

−3/7−1/2+1/2+3/2

5/22×3/2

2×2

3/2×3/2

−3

Figure 34.1: The sign convention is that of Wigner (Group Theory, Academic Press, New York, 1959), also used by Condon and Shortley (TheTheory of Atomic Spectra, Cambridge Univ. Press, New York, 1953), Rose (Elementary Theory of Angular Momentum, Wiley, New York, 1957),and Cohen (Tables of the Clebsch-Gordan Coefficients, North American Rockwell Science Center, Thousand Oaks, Calif., 1974). The coefficientshere have been calculated using computer programs written independently by Cohen and at LBNL.

FIG. 10: Clebsch-Gordan coefficients

quantum mechanics

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