quantumtheory - university of edinburghbjp/qt/qt.pdf · and feynman, namely the the path ... these...

T H

E

U N

I V E R S

I T Y

O

F

E D I N B U

R G

H

The University of Edinburgh

School of Physics and Astronomy

Lecture Notes

Quantum Theory

September 2015

Abstract

Historically quantum mechanics began with two quite different mathematical formulations: thewave equation of Schrodinger, and the matrix algebra of Heisenberg, Born and Jordan. While theseapproaches are well suited to the time-independent non-relativistic bound state problems of atomicand molecular physics, many problems in condensed matter, statistical physics and particle physicsare better suited to a more intuitive formulation of quantum mechanics based on the ideas of Diracand Feynman, namely the the path integral or sum over histories approach. In this course we beginwith a review of the fundamental ideas of quantum mechanics; we introduce the path integral fora non-relativistic point particle, and we use it to derive time-dependent perturbation theory, theBorn series for nonrelativistic scattering, and Feynman perturbation theory. The course concludeswith an introduction to relativistic quantum mechanics and some of the basic ideas of quantum fieldtheory.

Preamble

Formulations of Quantum Mechanics

Quantum Mechanics started with two ostensibly-different formulations, both based on the Hamil-tonian:

(1925) Heisenberg, Born and Jordan: Matrix Mechanics

(1925) Schrodinger: Wave Mechanics

These were shown to be equivalent by Dirac and Schrodinger (and Born?) in 1926, and latergeneralised by Dirac’s transformation theory.

Wave Mechanics and Matrix Mechanics are well-suited to the study of non-relativistic bound statesand other time-independent systems, and also to scattering and time-dependent problems.

(1942) Feynman: Path Integral or Sum over Histories

Feynman’s formulation is based on the Lagrangian and is ideal for scattering and time-dependentproblems, and (especially) for relativistic and many-particle systems. Unfortunately, bound-stateproblems are generally harder to solve using path integrals, as we shall see.

Structure of the course:

(1) Elements of quantum mechanics;

(2) Path integrals for single particles;

(3) Perturbation theory and scattering;

(4) Relativistic quantum theory – mostly wave mechanics, plus a hint of field theory.

Acknowledgements

These notes are based on the Quantum Theory lectures written by Richard Ball and typeset byFriedmar Schutze. Many thanks to both for providing copies of their work. Any mistakes are ofcourse my responsibility - please send error reports to [email protected]

Brian Pendleton Edinburgh, September 2015

1. Quantum Kinematics 1

1. Quantum KinematicsWe begin by revisiting one of the classic experiments that led to quantum theory. The setup of thetwo-slit experiment is summarised in the diagram.

Particles (electrons, photons, . . .) are fired from a source towards a screen or “grating” whichcontains two slits, 1 and 2. They are then detected, or “observed”, on a second screen.

The classical expectation is a count profile P1 + P2, which is the sum of the count profiles P1 andP2 for the individual slits (red and blue curves).

However, the observed interference pattern in the count profile P12 of particles (purple curve) sug-gests that such particles behave like waves, so we must add amplitudes (called φ1 and φ2 in thefigure); the number of particles (intensity) at any point is given by |∑ amplitude |2. Interferencemeans that we observe no particles at all in some places - entirely contrary to classical expectation!Furthermore, there is no way to predict where any given particle will be detected - we deal onlyin probabilities. The results of this experiment (and many, many more!) are encoded in the basicaxioms of Quantum Mechanics.

1.1. Fundamental principles (or postulates, axioms)

States and linearity: In quantum mechanics, every possible physical state of a given system isin one-to-one correspondence with a one-dimensional subspace of a complex linear vector space, H,with a complex inner product - a Hilbert space.

So, if |ψ〉 ∈ H, the physical state corresponds to all vectors c |ψ〉 with c ∈ C \ 0. There is nophysical state corresponding to zero.

Since the vector space is linear, given two states |ψ1〉 , |ψ2〉 ∈ H, we can construct a third state

|ψ3〉 = c1 |ψ1〉+ c2 |ψ2〉 ∈ H, with c1, c2 ∈ C. (1.1)

This is called linear superposition of states.

Dual space: To every |ψ〉 ∈ H we associate a dual or adjoint vector 〈ψ| ∈ H⋆. The duality isantilinear:

〈ψ3| = c∗1 〈ψ1|+ c∗2 〈ψ2|We then have an inner product mapping H⊗H⋆ → C. If |ψ〉 ∈ H and 〈φ| ∈ H⋆, then

〈φ ψ〉 = 〈ψ φ〉∗ ∈ C. (1.2)

(This is like the usual dot-product a · b for real vectors a, b, but the result is in general complex.)

Since |ψ〉 and c |ψ〉 correspond to the same physical state, we usually normalise |ψ〉 to unity: wechoose:

‖ |ψ〉 ‖2 = 〈ψ ψ〉 = 1. (1.3)

This fixes |c|2 = 1, so c = eiα. The phase α is often ignored: we say “the system is in the state |ψ〉”.

2 1. Quantum Kinematics

Example: In the two-slit experiment, the Hilbert space has two states:

|1〉 (the particle goes through slit 1)

|2〉 (the particle goes through slit 2)

The state |f〉 observed in the detector will in general be some linear combination of these:

|f〉 = c1 |1〉+ c2 |2〉

Examples of two-state systems in nature: electron spin, double-well potential, neutral-kaon mixing. . .

Probability: the probability that a system observed initially in state |ψ〉 will be observed finallyin state |φ〉 is

P (ψ → φ) = | 〈φ ψ〉 |2

The complex numbers 〈φ ψ〉 are called probability amplitudes because they add and multiply justlike classical probabilities. For example:

• The amplitude for ψ → φ and then φ→ χ is 〈χ φ〉〈φ ψ〉.• The amplitude for ψ → φ or χ is 〈φ ψ〉+ 〈χ ψ〉; (indistinguishable φ, χ)

• This means that

P (ψ → φ→ χ) = |〈χ φ〉〈φ ψ〉|2 = P (ψ → φ)P (φ→ χ)

butP (ψ → φ or χ) 6= P (ψ → φ) + P (ψ → χ)

because| 〈φ ψ〉+ 〈χ ψ〉 |2 = | 〈φ ψ〉 |2 + | 〈χ ψ〉 |2 + 2Re 〈φ ψ〉∗ 〈χ ψ〉

The last term is the interference term. In particular, P (ψ → φ or χ) can be zero even if P (ψ → φ)and P (ψ → χ) are both non-zero.

In general amplitudes can interfere!

An excellent example is the two slit experiment. A theorist’s sketch of the experimental setup ofthis experiment is shown in figure 1.1.

|i〉 (source) grating |f〉 (screen)

|1〉

|2〉

Figure 1.1: Two slit experiment

The probability amplitude for a particle (electron)initially in the state |i〉 (at the source) to pass

through slit 1, and be detected in the final state|f〉 (on the second screen) is:

φ1 = 〈f 1〉〈1 i〉

Similarly, the amplitude for the electron initially inthe state |i〉 to pass through slit 2, and be detectedin the state |f〉 is:

φ2 = 〈f 2〉〈2 i〉

The probability amplitude for a particle (electron) initially in the state |i〉 (at the source) to bedetected in the final state |f〉 (on the second screen) is obtained by adding the amplitudes to pass

through through the individual slits:

〈f i〉 = φ1 + φ2 = 〈f 1〉〈1 i〉+ 〈f 2〉〈2 i〉

For the probability, we have:

P (i→ f) = | 〈f i〉 |2 = |φ1 + φ2|2

= | 〈f 1〉〈1 i〉+ 〈f 2〉〈2 i〉 |2

= | 〈f 1〉〈1 i〉 |2 + | 〈f 2〉〈2 i〉 |2 + 2Re〈f 1〉〈1 i〉〈f 2〉∗ 〈2 i〉∗

6= | 〈f 1〉〈1 i〉 |2 + | 〈f 2〉〈2 i〉 |2

(1.4)


Question: Why?

Answer: The probability P (i → f) 6= P (i → f via 1) + P (i → f via 2) because P (i → f)can vanish if the amplitudes for the different paths cancel. The last term in the second-last line ofequation (1.4) is the interference term.

Question: For a given i→ f , which way does the electron go? Through slit 1 or slit 2?

Answer: We don’t know – if we don’t look. . .

If we do look, this changes the experiment – if we find the particle goes through slit 1, then

P (i→ f) → P (i→ f via 1) = | 〈f 1〉〈1 i〉 |2

and there is no interference.

In quantum mechanics, electrons can go through slit 1, or slit 2, or “both” - provided we don’tcheck!

Distinguishability: In the two-slit experiment, we don’t know whether the particle we detectedat some point on the screen passed through slit 1 or slit 2 – the two outcomes are indistinguishable.In this case, we add amplitudes

P (ψ → ξ or χ) = |〈ξ ψ〉+ 〈χ ψ〉|2 6= P (ψ → ξ) + P (ψ → χ)

If we have two final states |ξ〉 and |χ〉, which are distinguishable, the probability for ψ → ξ or χ is

P (ψ → ξ or χ) = |〈ξ ψ〉|2 + |〈χ ψ〉|2 = P (ψ → ξ) + P (ψ → χ)

i.e. we add probabilities – just as in classical probability.

1.2. Basis states

Within our assumption for the space of states to be a Hilbert space we implicitly assumed the linearspace to be complete. This means that there exists a set of basis states |n〉 (assumed for themoment to be countable) such that for any |ψ〉 ∈ H:

|ψ〉 =∑

n

ψn |n〉 (1.5)

for some complex numbers ψn – the components of |ψ〉 in the basis |n〉. The basis vectors maybe chosen to be orthonormal:

〈m n〉 = δmn. (1.6)

Then the following equation holds:

〈m ψ〉 =∑

n

ψn 〈m n〉 = ψm. (1.7)

So ψm is the probability amplitude for |ψ〉 to go to |m〉, and we have

|ψ〉 =∑

n

|n〉〈n ψ〉 (1.8)

Since this is true for all |ψ〉 ∈ H, we often write∑

n

|n〉〈n| = 1 (1.9)

where 1 is the unit or identity operator, and the sum is over a complete set of states |n〉. This isan extremely useful result and we shall use it many, many times in what follows.

Equation (1.9) expresses the completeness of the basis states: at any moment a state must be in somelinear superposition of them. Note that completeness is necessary for the probability interpretationto work:

1 = 〈ψ ψ〉 =∑

n

〈ψ n〉〈n ψ〉 =∑

n

| 〈n ψ〉 |2 =∑

n

P (ψ → n) (1.10)

Example: In the two-slit experiment we have a two-component basis, we can represent the states|1〉 and |2〉 as

|1〉 →(10

)

and |2〉 →(01

)

,

and〈f i〉 =

∑

n=1,2

〈f n〉〈n i〉

simply expresses the fact that the electron must follow a path through either slit-1 or slit-2. Thestates |i〉 and |f〉 will in general be linear superpositions of |1〉 and |2〉.


1.3. Operators and Observables

Consider an observable which takes real values ξn in basis states |n〉. Then we can construct anoperator:

ξ ≡∑

n

ξn |n〉〈n| (1.11)

corresponding to the observable. Equation (1.11) is called the spectral representation of the operator.

It is easy to see that:

(a) ξ is a linear operator: ξ(c1 |ψ1〉+ c2 |ψ2〉) = c1ξ |ψ1〉+ c2ξ |ψ2〉(b) ξ is hermitian:

〈ψ| ξ |φ〉 =∑

n

ξn 〈ψ n〉〈n φ〉 =(∑

n

ξn 〈φ n〉〈n ψ〉)∗

(because ξ∗n = ξn)

= 〈φ| ξ |ψ〉∗ ≡ 〈ψ| ξ† |φ〉 (by definition of the Hermitian conjugate)

More succinctly, we may write simply: ξ† = ξ.

(c) |n〉 are eigenstates of ξ, with eigenvalues ξn:

ξ |n〉 =∑

m

ξm |m〉〈m n〉 = ξn |n〉 (1.12)

(d) If we have a second observable which takes values ηn on |n〉, then ξ and η commute:

ξη |n〉 = ξnηn |n〉 = ηnξn |n〉 = ηξ |n〉 ∀n, so [ξ, η] = 0.

Measurement: When we make a measurement, the state |ψ〉 of the system just before the mea-

surement collapses into some eigenstate |n〉 of ξ: the probability that we measure ξn is | 〈n ψ〉 |2.The average result over many measurements is thus:

ξ =∑

n

ξn| 〈n ψ〉 |2 =∑

n

ξn 〈ψ n〉〈n ψ〉

= 〈ψ| ξ |ψ〉

which is often called the expectation value of ξ in the state |ψ〉. We can think of the measurementprocess as projecting |ψ〉 onto |n〉:

Pn = |n〉〈n|is the appropriate projection operator, with properties P 2

n = Pn and∑

n Pn = 1.

When we make a measurement the state changes:

|ψ〉 7→ Pn |ψ〉 = |n〉〈n ψ〉

with a probability:‖Pn |ψ〉 ‖2 = 〈ψ n〉〈n n〉〈n ψ〉 = | 〈n ψ〉 |2.

Degeneracy: Often a given operator ξ will have degeneracies:

ξ |n,m〉 = ξn |n,m〉

for all m ∈Mn where Mn is a certain set. Measuring ξn then projects onto a degenerate subspace:

|ψ〉 7→ Pn |ψ〉 =∑

m

|n,m〉〈n,m ψ〉 .

To project onto a definite eigenstate requires that we measure further observables which commutewith ξ. This leads to the notion of a maximally commuting set of observables – see Quantum

Physics. (The basis states may be organised into irreducible representations of discrete or continuoussymmetries of the system: the observables then correspond to generators of these symmetries – seeSymmetries of Quantum Mechanics.)


1.4. Change of basis

Consider a change of basis |n〉 7→ |n〉. We may express the new basis states in terms of the oldones:

|n〉 =∑

m

|m〉〈m n〉 ≡∑

m

|m〉Umn (1.13)

The Umn are sometimes called transformation coefficients (Dirac). Orthonormality of the new basisthen implies that

δnn′ = 〈n n′〉 =∑

m

〈n m〉〈m n′〉 =∑

m

U †nmUmn′ (1.14)

where we used 〈n m〉 = 〈m n〉∗ = U∗mn ≡ U †

nm. So Umn is the mn element of a unitary matrix. Inoperator notation we define

|n〉 ≡ U |n〉 which gives Umn = 〈m| U |n〉 (exercise for the student)

From (1.14), U is a unitary operator: U †U = 1. Unitary operators are not in general observablesbecause they don’t have real eigenvalues.

1.5. Space as a continuum – position and wavenumber

Consider again the two slit experiment shown in figure 1.1, and let us generalise to an N -slitexperiment. Then

P (i→ f) = |〈f i〉|2 with 〈f i〉 =N∑

n=1

〈f n〉〈n i〉

Position basis: If we let N → ∞, we will need a continuous label x. We get the usual transitionfrom discrete to continuous variables

|n〉 7→ |x〉 ;∑

n

7→∫

dx ; δnm 7→ δ(x− x′).

The component of |ψ〉 in the basis |x〉 is now a function of the continuous variable x, let’s call itψ(x). We have

|ψ〉 =b∫

a

ψ(x) |x〉 dx

If we normalise our states so that〈x′ x〉 = δ(x− x′)

then

〈x′ ψ〉 =b∫

a

ψ(x) 〈x′ x〉 dx = ψ(x′)

soψ(x) = 〈x ψ〉 and ψ∗(x) = 〈ψ x〉 .

We may then write

|ψ〉 =b∫

a

dx |x〉〈x ψ〉 and thus 1 =

b∫

a

dx |x〉〈x| (1.15)

which is the completeness relation (the particle must be somewhere).

Squares of probability amplitudes are now interpreted as probability densities:

1 = 〈ψ ψ〉 =b∫

a

dx 〈ψ x〉〈x ψ〉 =b∫

a

dx ψ∗(x)ψ(x),

so |ψ(x)|2 dx is the probability that the particle is between x and x+ dx, and hence ψ(x) is theusual spatial wavefunction of wave mechanics.


In the continuous version of (1.11) we may define the position operator as

x =

b∫

a

dx x |x〉〈x| (1.16)

and verify that

x |x〉 =b∫

a

dx′ x′ |x′〉〈x′ x〉 =b∫

a

dx′ x′ |x′〉 δ(x′ − x) = x |x〉 .

From equation (1.16), the expectation value of x in the state |ψ〉 is

〈ψ| x |ψ〉 =b∫

a

dx 〈ψ x〉x 〈x ψ〉 =b∫

a

dx ψ∗(x)xψ(x) (1.17)

There is of course no need to restrict ourselves to a finite interval (a, b). We can let a → −∞ andb→ ∞ and take x ∈ (−∞,∞), in which case the transition amplitude for the continuously-infinite-slit experiment may be written

〈f i〉 =∞∫

−∞

dx 〈f x〉〈x i〉

Fourier transform basis: We do not have to stay in the position basis. For example consider theFourier transform basis:

|k〉 ≡∞∫

−∞

dx√2π

e+ikx |x〉 = ∞∫

−∞

dx |x〉〈x k〉 , (1.18)

from which we may read off:

〈x k〉 = 1√2π

eikx and hence 〈k x〉 = 1√2π

e−ikx. (1.19)

Then we obtain the following relation:

〈k′ k〉 =∞∫

−∞

dx 〈k′ x〉〈x k〉 =∞∫

−∞

dx

2πei(−k′+k)x = δ(k′ − k), (1.20)

so the transformation is unitary. We can construct a hermitian operator k such that:

k =

∫

dk k |k〉〈k| ⇒ k |k〉 = k |k〉 (exercise)

It follows that

k |x〉 =∫

dk k |k〉〈k x〉 = i ∂∂x

∫

dk |k〉〈k x〉

= i ∂∂x

|x〉 .(1.21)

where we used (1.19) to write

k 〈k x〉 = i ∂∂x

(1√2π

e−ikx) ,and hence the last expression on the first line of (1.21).

An immediate consequence of (1.21) is 〈ψ| k |x〉 = i ∂∂x 〈ψ x〉 . Taking the complex conjugate and

recalling that k is hermitian gives

〈x| k |ψ〉 = −i ∂∂xψ(x). (1.22)


Finally

〈φ| k |ψ〉 =∫

dx 〈φ x〉〈x| k |ψ〉 =∫

dxφ∗(x)

(

−i ∂∂x

)

ψ(x)

which should look familiar from wave mechanics.

Similarly, if we define

ψ(k) = 〈k ψ〉 =∫

dx 〈k x〉〈x ψ〉 =∫

dx√2π

e−ikxψ(x) (1.23)

we easily find (exercise)

〈k| k |ψ〉 = kψ(k) and 〈φ| k |ψ〉 =∫

dk φ∗(k) k ψ(k), (1.24)

while (exercise)

〈k| x |ψ〉 = i ∂∂kψ(k). (1.25)

and hence

〈φ| x |ψ〉 =∫

dk 〈φ k〉〈k| x |ψ〉 =∫

dk φ∗(k)

(i ∂∂k

)

ψ(k) .

A summary is given in table 1.1. In either basis we get the commutator:

basis x k

position: |x〉 x −i ∂∂x

wavenumber: |k〉 i ∂∂k k

Table 1.1: Operators in different bases

[

x, k]

= i (1.26)

This tells us that the two operators do not correspondto simultaneous observables (as expected). The Heisen-berg uncertainty principle can be deduced:

∆x∆k ≥ 1

2. (1.27)

Later we will show that momentum p = hk.

Historical note: The commutator (1.26) was deduced independently by Born and Dirac, and theuncertainty principle was formulated independently by each of Heisenberg, Dirac and Jordan.

Three dimensions: All of the above can be generalised to three space dimensions rather easily.For position states and operators:

〈x′ x〉 = δ(3)(x− x′) 1 =

∞∫

−∞

d3x |x〉〈x| x =

∞∫

−∞

d3x x |x〉〈x| x |x〉 = x |x〉

Similarly for the wavevector basis |k〉. The wavefunctions in the two bases are related by (exercise)

ψ(k) = 〈k ψ〉 =∫

d3x 〈k x〉〈x ψ〉 =∫

d3x

(2π)3/2e−ik·x ψ(x)


1.6. Time as a continuum

|i〉 (source) gratings |f〉 (screen)

Figure 1.2: Sketch of the slit experimentwith more gratings

We now add more gratings to the slit experiment.Let the nth grating be passed at position xn at timetn, and call this state |xn, tn〉. If we haveN gratingsthe transition amplitude becomes

〈f i〉 =∫

dx1 〈f x1, t1〉〈x1, t1 i〉

=

∫

dx1

∫

dx2

〈f x2, t2〉〈x2, t2 x1, t1〉〈x1, t1 i〉

=

∫

dx1 · · ·∫

dxN 〈f xN , tN 〉 · · · 〈x1, t1 i〉(1.28)

where t1 < t2 < . . . < tN . By adding more and more gratings the time intervals get smaller andsmaller and we fix each path between |i〉 and |f〉 more and more precisely. But we also get moreand more integrals in order to integrate over all the paths!

For simplicity (and notational convenience) we take |i〉 = |xa, ta〉 and |f〉 = |xb, tb〉 and let

tn = ta + nε ; ε =tb − taN + 1

, (1.29)

so that t0 = ta and tN+1 = tb. Then we obtain for the transition amplitude:

〈xb, tb xa, ta〉 =(

N∏

n=1

∫

dxn

)(N+1∏

n=1

〈xn, tn xn−1, tn−1〉)

. (1.30)

The expression in the first pair of parentheses tells us to “integrate over all paths”, and the expressionin the second pair is the amplitude for each path.

For N → ∞ (i.e. ε→ 0) this becomes:

〈xb, tb xa, ta〉 =xb∫

xa

Dx 〈xb, tb xa, ta〉∣∣x(t)

(1.31)

where the continuous path x(t) is such that x(ta) = xa and x(tb) = xb. So, to find the transi-tion amplitude we take the amplitude for each path x(t), and then sum (i.e. integrate) over all

paths between xa and xb. This is easy to understand physically but much harder to understandmathematically!

In fact, the limit only exists if we take care to normalise each integral carefully (see later).

What is the amplitude for each path x(t)? Intuitively, for a path which is infinitesimal (tn = tn−1+ε):

〈xn, tn xn−1, tn−1〉 ∼ exp [iεφ(xn, xn−1, tn, tn−1)] , (1.32)

where φ is real. This is because:

(a) in the limit ε → 0, the amplitude should be constant (continuity), this is just a normalisationconstant;

(b) the phase should depend only on xn, xn−1, tn, tn−1 (locality)

(c) The transition |xn−1, tn−1〉 7→ |xn, tn〉 is just a change of basis, so we expect (up to a constant)| 〈xn, tn xn−1, tn−1〉 | ∼ 1, i.e. the amplitude is just a phase (unitarity).

Using (1.30) and (1.32) we can write the transition amplitude along the path x(t) as:

〈xb, tb xa, ta〉∣∣x(t)

∼ limN→∞

N+1∏

n=1

exp iεφ(xn, xn−1, tn, tn−1)

∼ exp

i limN→∞

N+1∑

n=1

(tn − tn−1)φ(xn, xn−1, tn, tn−1)

∼ exp

i tb∫

ta

dt φ(x(t), x(t), t)

(1.33)

To say more (about φ) we need to revise and develop some classical mechanics. . .

2. Quantum Dynamics 9

2. Quantum Dynamics2.1. Classical Dynamics

In Lagrangian dynamics, the action for a path x(t) is

S[x(t)] =

tb∫

ta

L(x, x, t) dt (2.1)

where L is the Lagrangian. For a non-relativistic point particle in a potential V in one dimensionthis is:

L = T − V =1

2mx2 − V (x, t) (2.2)

Classical dynamics is based upon the principle of least action: the classical path x(t) is an extremumof the functional S. Formally

δS

δx

∣∣∣∣x=x

= 0 (2.3)

where δS/δx is the functional derivative. For a small variation of the path: x(t) 7→ x(t) + δx(t):

δS = S[x+ δx]− S[x]

=

tb∫

ta

dt (L(x+ δx, x + δx, t)− L(x, x, t))

=

tb∫

ta

dt

(∂L

∂xδx+

∂L

∂xδx

)

+O(δx2)

=

[∂L

∂xδx

]tb

ta

−tb∫

ta

dt

(d

dt

(∂L

∂x

)

− ∂L

∂x

)

δx+O(δx2)

(2.4)

where, as usual, we integrated by parts in the last step. If the end-points of the path are fixed,i.e. δx(ta) = δx(tb) = 0, then the first term in the last line of (2.4) vanishes, and we obtain Lagrange’sequation for the classical path

d

dt

(∂L

∂x

)

− ∂L

∂x= 0. (2.5)

Consider now the value of the action S on the classical path: Scl ≡ S[x(t)]. Scl will be a functionof the end points, i.e. of xa, ta, xb, and tb.

Let us now vary the endpoint (xb, tb) 7→ (xb, tb) + (δxb, δtb), but keep (xa, ta) fixed. Lagrange’sequation (2.5) holds for the classical path so, from (2.4), we get (choosing δtb = 0 for now)

δScl =

[∂L

∂xδx

]tb

ta

=[p δx

]tb

ta= p(tb) δxb − p(ta) δxa = p(tb) δxb

Hence∂Scl

∂xb= pb (2.6)

where we used the usual definition of the canonical momentum p conjugate to x: p = ∂L∂x .

Now consider dScl

dtb. From (2.1) we may write:

dScl

dtb= L(xb, xb, tb) =

∂Scl

∂tb+∂Scl

∂xbxb

This gives∂Scl

∂tb= L− pbxb = −Eb ⇒ Eb = −∂Scl

∂tb(2.7)

where Eb is the energy (or, more precisely, the Energy Function or Hamiltonian) at b.

Equations (2.6) and (2.7) are known as the Hamilton-Jacobi equations.

10 2. Quantum Dynamics

Example 1: The free particle. The Lagrangian L, and the Lagrange equation of motion (EoM) forthe classical path x(t) are

L =1

2mx2 ⇒ ¨x = 0

Integrating twice, and imposing the boundary conditions x(ta) = xa and x(tb) = xb, the solution is(exercise):

x = xa + v(t− ta) where v =xb − xatb − ta

= ˙x

so v is the (constant) velocity of the particle. The classical action is

Scl = S[x] =

∫ tb

ta

1

2m ˙x

2dt =

1

2mv2(tb − ta) =

1

2m

(xb − xa)2

(tb − ta),

and, using the Hamilton-Jacobi equations, we get

pb =∂Scl

∂xb= mv and Eb = −∂Scl

∂tb=

1

2mv2 as expected (exercise).

Example 2: The simple harmonic oscillator. The Lagrangian and Lagrange EoM are

L =1

2m(x2 − ω2x2

)⇒ ¨x+ ω2x = 0

The solution which satisfies the boundary conditions x(ta) = xa and x(tb) = xb, with T ≡ tb− ta, is

x(t) = xbsinω(t− ta)

sinωT+ xa

sinω(tb − t)

sinωT

This is important - check it carefully! Then we have, using integration by parts and the fact thatx(t) satisfies the classical equation of motion to simplify the calculation,

S[x] =

∫ tb

ta

m

2

(

˙x2 − ω2x2

)

dt =m

2

[x ˙x]tb

ta− m

2

∫ tb

ta

x(¨x+ ω2x

)dt =

m

2

[x ˙x]tb

ta+ 0

=mω

2 sinωT

[(x2a + x2b

)cosωT − 2xaxb

]

You should verify this very important result, and use the Hamilton-Jacobi equations to check that

pb = m ˙x∣∣t=tb

and Eb =m

2

(

˙x2+ ω2x2

) ∣∣t=tb

.

2.2. The Amplitude for a Path

We saw already in (1.33) that we expect the amplitude for the path x(t) to be:

〈xb, tb xa, ta〉∣∣x(t)

∼ exp

i tb∫

ta

dt φ(x, x, t)

. (2.8)

Furthermore, we expect some sort of classical ↔ quantum correspondence, and we want the classicallimit to be included in the quantum description. Classically, we have:

S[x(t)] =

tb∫

ta

dt L(x, x, t) (2.9)

So we guess that φ ∝ L. Now we need to fix up the units: we have [S] = [t][E] = [x][p]. So if weintroduce a dimensionful constant h, with units [S] (action), then we can take

∫φ dt = S/h, i.e. we

assume:〈xb, tb xa, ta〉

∣∣x(t)

= eiS[x(t)]/h. (Dirac) (2.10)

Equation (2.10) is our basic dynamical assumption, just as Schrodinger’s wave mechanics assumesthe Schrodinger equation to be the equation of motion for quantum systems. The overall (as yetundetermined) normalisation constant will be (implicitly) absorbed into Dx.Note that:

(1) If S → S + c (where c is a constant), all amplitudes change by eic/h, so physics is unchanged.

(2) If δS ∼ O(2πh) ∼ O(h), the phase changes by O(1), this sets the size of quantum fluctuations.


2.3. The Feynman Path Integral

With the assumption (2.10) the transition amplitude becomes:


xa

Dx

︸︷︷︸

sum over all paths

suitably normalised

eiS[x(t)]/h︸︷︷︸

amplitude for

each path

(2.11)

Though the notation in (2.11) is very neat, to do calculations we will have to use the limitingprocedure as given above in (1.30) and (1.33):

xb∫

xa

Dx = limN→∞

AN

N∏

n=1

∫ ∞

−∞

dxn

where AN = (ν(ε))N+1, i.e. a normalising factor ν(ε) for each discrete interval.

ta

tc

tb

Figure 2.1: Split onepath in two

Note the fundamental property: if we insert a complete set of statesinto the LHS of (2.11), we can split the path in two:

〈xb, tb xa, ta〉 =∫

dxc 〈xb, tb xc, tc〉〈xc, tc xa, ta〉

=

∫

dxc

xb∫

xc

Dxxc∫

xa

Dx exp

ih

tb∫

tc

L dt+

tc∫

ta

L dt

This is shown pictorially in figure 2.1. It is very useful for determiningthe normalisation factor ν (see below).

2.4. The classical limit (heuristic)

For ‘quantum situations’, tb & ta and xb & xa are “close”, S[x(t)] = O(h), so the phase fluctuationsare of O(1), and paths far from the classical path may be important.

For ‘classical situations’, tb, ta, xb and xa are “far apart” and S[x(t)] ≫ h in general. (Formally, theclassical limit is obtained by taking h → 0.) Now consider paths “very close” to a given path x(t).Even though δx(t) is small, δS will in general be large compared with h, because S is so large. SoeiδS/h = cos(δS/h) + i sin(δS/h) will oscillate violently and contributions from nearby paths willcancel. However, the classical path x(t) is special: δS is of order O(δx2), so nearby paths can addconstructively. So, as h → 0 the classical path gives the dominant contribution (i.e. we derive theprinciple of least action and hence classical mechanics) and therefore

〈xb, tb xa, ta〉 ∼ eiScl/h (this is the ‘semiclassical approximation’)

2.5. Momentum and Energy

Let us try to make the statements for the phase variations in the previous section a bit more rigorous.As we have seen, in classical situations Scl ≫ h and the amplitude oscillates very rapidly. To seejust how rapidly, consider a small change in the endpoint: xb → xb + δxb, keeping tb fixed. Then:

Scl → Scl +∂Scl

∂xbδxb.

So the change in phase, δS/h, is kbδxb, where the wavenumber

kb ≡1

h

∂Scl

∂xb=pbh

(by Hamilton−Jacobi.) (2.12)

Here, pb is the classical momentum at the endpoint (cf. (2.6)). So p = hk, or in operator language

p = hk, and thus from (1.26):[x, p] = ih. (2.13)

Similarly, if we change the time at the endpoint and keep xb fixed, i.e. tb → tb + δtb, we get

Scl → Scl +∂Scl

∂tbδtb.

The change in phase is now −ωbδtb, where the frequency is (cf. (2.7)):

ωb ≡ − 1

h

∂Scl

∂tb=Eb

h(by Hamilton−Jacobi) (2.14)

and we have E = hω.


2.6. The Free Particle

Let us evaluate the path integral explicitly for a free particle. The “continuum” expression for theFeynman path integral in (2.11) is elegant and succinct but we shall shall evaluate it here usinga limiting procedure. We shall make use of a range of Gaussian integrals – see separate handout.We use the following discrete approximation to the free particle Lagrangian L = T = 1/2mx2 ≈1/2m((xn − xn−1)/ε)

2, so that

〈xb, tb xa, ta〉 = limN→∞

AN

(N∏

n=1

∫

dxn

)

exp

iεh

m

2

N+1∑

n=1

(xn − xn−1

ε

)2

(2.15)

where AN is a normalisation constant to be fixed. In (2.15) we have a sequence of nested Gaussianintegrals. Doing these integrals is straightforward but tedious. Each of the integrals is of the form

I =

∞∫

−∞

ei(x−u)2/aei(u−y)2/b du

=

∞∫

−∞

exp

i(1

a+

1

b

)

u2 − 2i(xa+y

b

)

u+ i(x2a

+y2

b

)

du

=

∞∫

−∞

exp

i(1

a+

1

b

)(

u− x/a+ y/b

1/a+ 1/b

)2

exp

−i (x/a+ y/b)2

1/a+ 1/b+ i(x2

a+y2

b

)

du

(2.16)

where we completed the square in the second line. For brevity, we make the following substitutions:

α ≡ −i(1

a+

1

b

)

= −ia+ b

ab; v ≡ u+

iα

(x

a+y

b

)

(2.17)

and note that(x2

a+y2

b

)

− ab

a+ b

(x

a+y

b

)2

=1

a+ b

[

x2 +bx2

a+ y2 +

ay2

b− ab

(x

a+y

b

)2]

=1

a+ b

[

x2 +bx2

a+ y2 +

ay2

b− 2yx− bx2

a− ay2

b

]

=(x− y)2

a+ b.

(2.18)

We now plug (2.17) and (2.18) into (2.16) and use equation (1) (or (7)) from the sheet of Gaussianintegrals:

I =

∞∫

−∞

e−αv2

exp

i (x − y)2

a+ b

dv =

√π

αexp

i (x− y)2

a+ b

=

√i πaba+ b

exp

i (x− y)2

a+ b

.

(2.19)

We can now evaluate the integrals in (2.15) one at a time using (2.19), starting with the integralover x1: ∫

dx1 expi m

2εh

(

(x2 − x1)2+ (x1 − x0)

2)

This is of the form of (2.16) with a = b = 2εh/m, so ab/(a+ b) = εh/m and a+ b = 4εh/m. Hence

〈xb, tb xa, ta〉

= limN→∞

AN

(N∏

n=1

∫

dxn

)(N+1∏

n=1

expi m

2εh(xn − xn−1)

2)

= limN→∞

AN

(N∏

n=2

∫

dxn

)(N+1∏

n=3

exp

im(xn − xn−1)2

2εh

)√ iπεhm

exp

im(x2 − x0)2

4εh

= . . . = limN→∞

AN

( iπεhm

)N2

√

2 · 2N2(N + 1)

exp

im(xN+1 − x0)2

2h(N + 1)ε

= limN→∞

AN

(2πihεm

)N+1

2√

m

2πih(N + 1)εexp

im(xN+1 − x0)2

2h(N + 1)ε

Exercise: check this explicitly (a slightly laborious exercise, but worth the effort.)


Since x0 = xa, xN+1 = xb, and (N + 1)ε = tb − ta ≡ T , if we choose

AN = (ν(ε))N+1 =( m

2πihε)N+1

2

, i.e. ν(ε) =

√m

2πihεthen, since the limit N → ∞ is trivial,

〈xb, tb xa, ta〉 =√

m

2πihT exp

im2h

(xb − xa)2

T

≡ F0(T ) eiScl/h, (2.20)

since, as we showed above for the free particle,

Scl =m

2v2T =

m

2

(xb − xa)2

T.

Note that our choice for ν(ε) renders the normalisation factor or pre-factor F0(T ) independent ofxa and xb - see also below for discussion.

Notes:

(1) At large times (where Scl ≫ 1), letting x = xb +∆x, t = tb +∆t and expanding the exponent in aTaylor series we find (exercise)

〈x, t xa, ta〉 ≃ 〈xb, tb xa, ta〉 e(i/h)(pb∆x−Eb∆t),

i.e. a plane wave with momentum pb and energy Eb = p2b/2m, as expected from Hamilton-Jacobi.

(2) The free particle amplitude is a very useful object: we often call it the free-particle Green function

(see later), and write

G0(xb − xa, tb − ta) ≡ 〈xb, tb xa, ta〉 (tb > ta) (2.21)

which makes explicit its translational invariance: G0 depends only on the difference between theinitial and final positions and times. In momentum space (exercise)

G0(p, t) ≡∞∫

−∞

dx eixp/hG0(x, t) =

√m

2πiht ∞∫

−∞

dx exp

ixph

+imx22ht

= exp

(

−i p2t2mh

)

= exp

(

−iEth

)(2.22)

where E = p2/2m, i.e. a plane wave with classical energy E as expected.

(3) Normalisation: choosing ν(ε) =√

m2πihε seems strange at first, since ν(ε) → ∞ as ε → 0, so the

amplitude diverges for infinitesimal time intervals. This is not an accident: if the amplitude forfinite times is to be finite, then

limN→∞

(

ν(ε)

√

2πihεm

)

must be finite, and we must choose ν(ε) =√

m2πihε (1 + O(ε)), (so that the O(ε) term goes to zero

as ε → 0). Moreover, since 〈xb, tb xa, ta〉 is a Gaussian in (xb − xa) with width h(tb − ta)/m and(with this normalisation) unit area, as tb → ta

〈xb, tb xa, ta〉 → δ(xb − xa) = 〈xb, ta xa, ta〉

as required by orthonormality.

Finally, it is straightforward to verify that with this normalisation

〈xb, tb xa, ta〉 =∫ ∞

−∞

dx 〈xb, tb x, t〉〈x, t xa, ta〉 (2.23)

for any ta < t < tb (tutorial exercise).


(4) The result 〈xb, tb xa, ta〉 = F0(T ) eiScl/h is also not accidental.

To see this, we consider an alternative method of calculation. Write an arbitrary path x(t) asx(t) = x(t) + η(t), where x(t) is the classical path with boundary conditions x(ta) = xa andx(tb) = xb. Then η(ta) = η(tb) = 0, and η(t) describes the ‘quantum fluctuations’ about theclassical path. Furthermore, since δS

δx

∣∣x=x

= 0 there will be no terms linear in η in the action, and,since the Lagrangian is quadratic, we find S[x] = S[x+ η] = S[x] + S[η]. Explicitly

S[x] = S[x+ η] =m

2

tb∫

ta

( ˙x+ η)2 dt =m

2

tb∫

ta

( ˙x2+ η2) dt+ 2

tb∫

ta

˙xη dt

=m

2

tb∫

ta

( ˙x2+ η2) dt+ 2

[η ˙x]tb

ta− 2

tb∫

ta

η¨x dt

(2.24)

The last two terms vanish because η(ta) = η(tb) = 0, and ¨x = 0 by the equation of motion.

But S[x] = Scl, and∫ xb

xaDx =

∫ 0

0Dη (since x is fixed, hence dxn = dηn ∀n), therefore


xa

Dx eiS[x(t)]/h = eiScl/h

0∫

0

Dη eiS[η]/h ≡ F0(T ) eiScl/h, (2.25)

where T = tb − ta as before, and the path integral is over all paths η(t) with η(ta) = η(tb) = 0. Thenormalisation factor F0(T )is independent of xa and xb:

F0(T ) = 〈0, tb 0, ta〉 = 〈0, T 0, 0〉 (2.26)

by translational invariance (in time). We can compute F0(T ) by evaluating the path integral overη(t) explicitly (tutorial).

Alternatively, using (2.23)

〈0, T 0, 0〉 =∫ ∞

−∞

dx 〈0, T x, t〉〈x, t 0, 0〉

so F0(T ) =

∫ ∞

−∞

dxF0(T − t) exp

( imx22h(T − t)

)

F0(t) exp

( imx22ht

)

=

√

2πih(T − t)t

mTF0(T − t)F0(t)

(2.27)

Now let T ≫ t so that (T − t) ≃ T and F0(T − t) ≃ F0(T ), and we find

F0(t) =

√m

2πiht (2.28)

as before. Note that in this argument no discretisation of the path integral is required.

2.7. The Harmonic Oscillator

The Lagrangian is:

L =m

2x2 − m

2ω2x2. (2.29)

We use the same trick as for the free particle: we write x = x+ η, then

S[x+ η] = S[x] + S[η] (2.30)

because the cross term vanishes:

∫ tb

ta

dt(˙xη − ω2xη

)=[η ˙x]tb

ta−∫ tb

ta

dt η(¨x+ ω2x

)= 0 (2.31)

The first term on the RHS is zero because η(tb) = η(ta) = 0, and the second vanishes because xsatisfies the equation of motion. (This “trick” works for any Lagrangian quadratic in x.)


Again Dx = Dη, and hence


xa

Dx eiS[x(t)]/h = eiScl/h

0∫

0

Dη eiS[η]/h = Fω(tb − ta) eiScl/h, (2.32)

where Scl is the classical action

Scl =mω

2 sinωT

((x2a + x2b

)cosωT − 2xaxb

)

(2.33)

and the normalisation factor is

Fω(T ) =

0∫

0

Dη eiS[η]/h = 〈0, T 0, 0〉 (2.34)

which is again independent of the boundary conditions on x. As before for the free particle theexplicit evaluation of the normalisation factor is tedious. It can be computed by expanding η(t) ina Fourier series (see Feynman & Hibbs, 3.11), by matrix methods, or implicitly – using the samemethod as for the free particle:

〈0, T 0, 0〉 =∫ ∞

−∞

dx 〈0, T x, t〉〈x, t 0, 0〉

so Fω(T ) =

∫ ∞

−∞

dxFω(T − t) exp

( imωx22h

cosω(T − t)

sinω(T − t)

)

Fω(t) exp

( imωx22h

cosωt

sinωt

)

= Fω(T − t)Fω(t)

√

2πih sinω(T − t) sinωt

mω sinωT

(2.35)

As before, let T ≫ t so that Fω(T − t) ≃ Fω(T ), hence

Fω(T ) = Fω(T )Fω(t)

√

2πih sinωT sinωt

mω sinωT

(

1 +O(t/T )

)

hence Fω(t) =

√mω

2πih sinωt (2.36)

This result holds for all t, as can be checked by direct substitution into equation (2.35). Note thatas ω → 0, Fω(t) → F0(t), as it must.

2.8. The Forced Harmonic Oscillator

Consider the forced harmonic oscillator with Lagrangian

L =m

2(x2 − ω2x2) + Jx (2.37)

where the external force J(t) is non-zero but arbitrary for ta ≤ t ≤ tb. The equation of motion isnow

¨x+ ω2x =J

m. (2.38)

The action depends on both x(t) and J(t), but since it is still quadratic it again splits into twoindependent parts:

S[x, J ] = S[x+ η, J ] =m

2

∫

dt

(

( ˙x + η)2 − ω2(x+ η)2 +2J

m(x+ η)

)

= S[x, J ] + S[η, 0] +m

∫

dt

(

˙xη − ω2ηx+J

mη

)

= S[x, J ] + S[η, 0] +m[˙xη]tb

ta−m

∫

dt[η(¨x+ ω2x− (J/m))

]

= S[x, J ] + S[η, 0]

(2.39)

The fluctuation term S[η, 0] is independent of J because the coupling of J to x (and hence to η) islinear. A minor modification of the steps in equation (2.32) gives (exercise)

〈xb, tb xa, ta〉 = Fω(T ) eiScl/h (2.40)

where now Scl = S[x, J ], and the normalisation factor is the same as for the unforced harmonicoscillator.


2.9. Schrodinger’s Equation

To complete the picture, we derive the Schrodinger equation for a particle in an external potential:

L =m

2x2 − V (x, t) . (2.41)

Recalling that momentum p = hk, we introduce the momentum basis following equations (1.18)through (1.20):

|p〉 =∞∫

−∞

dx√2πh

e+ipx/h |x〉 = ∞∫

−∞

dx |x〉〈x p〉 , (2.42)

from which we may read off

〈x p〉 = 1√2πh

eipx/h ; 〈p x〉 = 1√2πh

e−ipx/h . (2.43)

The 1/√2πh is to ensure that the states |p〉 are correctly normalised:

〈p′ p〉 =∞∫

−∞

dx 〈p′ x〉〈x p〉 =∞∫

−∞

dx

2πhei(−p′+p)x/h = δ(p′ − p) and 1 =

∞∫

−∞

dp |p〉〈p| (2.44)

Now consider the infinitesimal amplitude

〈xn+1, tn+1 xn, tn〉 =√

m

2πihε expiεh

[

m

2

(xn+1 − xn

ε

)2

− V (xn, tn)

]

=

√m

2πihε exp im2hε

(xn+1 − xn)2 − iε

hV (xn, tn)

=

∞∫

−∞

dpn2πh

exp

ipnh

(xn+1 − xn)

exp

− iε2mh

p2n

exp

− iεhV (xn, tn)

,

(2.45)where the second line may be recovered from the third using equation (8) on the sheet of gaussianintegrals (exercise.)

If we introduce a basis of (time-independent) momentum eigenstates satisfying p |pn〉 = pn |pn〉, andrecall that x |xn〉 = xn |xn〉, we may use (2.43) to rewrite (2.45) as

〈xn+1, tn+1 xn, tn〉 =∞∫

−∞

dpn exp

− iεh

p2n2m

〈xn+1 pn〉 exp

− iεhV (xn, tn)

〈pn xn〉

=

∞∫

−∞

dpn 〈xn+1| exp

− iεh

p2

2m

|pn〉〈pn| exp

− iεhV (x, tn)

|xn〉

= 〈xn+1| exp

− iεh

p2

2m

exp

− iεhV (x, tn)

|xn〉 ,

(2.46)

To get the last line we used the completeness of the momentum basis∫dpn |pn〉〈pn| = 1.

Finally, we use the Baker-Campbell-Hausdorff formula

eA eB = eA+B+(1/2)[A,B]+... (2.47)

to combine the exponentials. All terms beyond the first two on the RHS are of order O(ε2), so

〈xn+1, tn+1 xn, tn〉 = 〈xn+1| exp

− iεhH(x, p, tn)

|xn〉 , (2.48)

where

H(x, p, t) =p2

2m+ V (x, t) ≡ H(t)

is the quantum-mechanical analogue of the classical Hamiltonian.


Remembering that ε = tn+1 − tn, we can rewrite (2.48) as

〈xn+1, tn+1 xn, tn〉 = 〈xn+1| exp

− i(tn+1 − tn)

hH(x, p, tn)

|xn〉 ,

which we can use to write |x, t〉 in terms of |x〉

|x, t〉 = exp

ih(t− t0)H(t0)

|x〉 (2.49)

for some t0 such that t− t0 is infinitesimal.

If we differentiate (2.49) with respect to t, and then let t− t0 → 0, we obtain

∂

∂t|x, t〉 = i

hH(t) |x, t〉 ⇒ H(t) |x, t〉 = −ih ∂

∂t|x, t〉 .

where we used H(t0) → H(t) as t− t0 → 0.

Following (1.21) we have p |x, t〉 = ih ∂∂x |x, t〉. Now define ψ(x, t) = 〈x, t ψ〉, so 〈ψ x, t〉 = ψ∗(x, t),

then (exercise)

Hψ(x, t) =

(

− h2

2m

∂2

∂x2+ V (x, t)

)

ψ(x, t) = ih ∂∂tψ(x, t) . (2.50)

Equation (2.50) is of course Schrodinger’s equation.

Notes:

1. The argument is reversible: starting from the Schrodinger equation, we can derive Feynman’s pathintegral representation. Indeed, this is the route followed by most text books.

2. We can use the expression in the third line of (2.45) to construct an alternative representation ofthe path integral, called the phase space path integral:


(N∏

n=1

∫

dxn

)(N+1∏

n=1

∫dpn2πh

)

exp

iεh

N+1∑

n=1

[

pn(xn − xn−1)

(tn − tn−1)− p2n

2m− V (xn, tn)

]

≡∫

Dx∫

Dp exp

ih

tb∫

ta

dt (px−H(x, p, t))

(2.51)

where H is the classical Hamiltonian. This has the advantage of a natural measure – there are nonormalisation factors (because of Liouville’s theorem) – and it is particularly useful in statisticalmechanics.

3. We can use (2.51) to give yet another derivation of the free particle amplitude:


(N∏

n=1

∫

dxn

)(N+1∏

n=1

∫dpn2πh

)

exp

ih

N+1∑

n=1

(

pn(xn − xn−1)−εp2n2m

)

= limN→∞

1

2πh

(N+1∏

n=1

∫

dpn

)N∏

n=1

δ(pn+1 − pn) exp

ih

(

pN+1xb − p1xa −ε

2m

N+1∑

n=1

p2n

)

=

∞∫

−∞

dp

2πhexp

ihp(xb − xa)−

i2mh

(N + 1)εp2

=

√m

2πih(tb − ta)exp

im2h

(xb − xa)2

tb − ta

where we rewrote the sum in the first lineN+1∑

n=1

pn(xn − xn−1) = pN+1xb − p1xa −N∑

n=1

xn(pn+1 − pn) .

in order to perform theN integrals over the xn. We then used the resulting delta functions to performthe integrals over the first N momenta pi, leaving just one momentum integral (over pN+1 ≡ p)which may be done using integral (8) on the sheet of Gaussian Integrals.


4. Equation (2.49) may be regarded as the solution of the Schrodinger equation for infinitesimal t− t0.If the Hamiltonian were time-independent, i.e. H(t) = H , then (2.49) would hold for all t and t0,

|x, t〉 = exp

ih(t− t0) H

|x〉 ≡ U †(t, t0) |x〉 ∀t

For a general time-dependent Hamiltonian, we must define:

|x, t〉 = U †(t, t0) |x〉 (2.52)

where, from (2.49) and the equation below it, the time evolution operator U(t, t0) satisfies theSchrodinger-like equation (exercise):ih ∂

∂tU(t, t0) = H(t) U(t, t0) (2.53)

with the boundary condition U(t0, t0) = 1.

Clearly, U must be unitary, since it changes bases |x〉 7→ |x, t〉:U−1(t, t0) = U †(t, t0) = U(t0, t).

5. The Schrodinger and Heisenberg pictures

We may writeψ(x, t) = 〈x, t ψ〉 = 〈x| U(t, t0) |ψ〉 ≡ 〈x ψ, t〉

where we have defined the time-dependent state vector |ψ, t〉 ≡ U(t, t0) |ψ〉, which from (2.53)satisfies the Schrodinger equation

H |ψ, t〉 = ih ∂∂t

|ψ, t〉 .

We say that |ψ, t〉 and |x〉 are the state vector and the position eigenstate in the Schrodinger picturefor time dependence in quantum mechanics. Similarly, |ψ〉 and |x, t〉 are the equivalent quantitiesin the Heisenberg picture. These two “pictures” for describing time dependence in the operatorformulation of quantum mechanics are of course equivalent in that they give the same physicalpredictions for all observables.

The wave-function ψ(x, t) = 〈x ψ, t〉 = 〈x, t ψ〉 is (by definition) the same in both pictures.

Clearly, Schrodinger-picture operators such as x =∫dx x |x〉〈x| are time-independent while in

the Heisenberg picture x(t) ≡∫dx x |x, t〉〈x, t| is time-dependent. So we write x |x〉 = x |x〉 and

x(t) |x, t〉 = x |x, t〉 .What is the relation between x and x(t)? Inverting (2.52) gives |x〉 = U(t, t0) |x, t〉, hence

〈x| x |x〉 = 〈x, t| U †(t, t0) x U(t, t0) |x, t〉 ≡ 〈x, t| x(t) |x, t〉This must hold ∀ |x, t〉. Hence, for the two pictures to be equivalent, we must have

x(t) = U †(t, t0) x U(t, t0) (2.54)

Similarly for other operators. Note that [x(t), x(t′)] 6= 0 unless t = t′.

However, since ih ∂∂t

〈x, t ψ〉 = 〈x, t| H |ψ〉 = 〈x| UH |ψ〉ih ∂∂t

〈x ψ, t〉 = 〈x| H |ψ, t〉 = 〈x| HU |ψ〉

we must have UH = HU ⇒ H = U †HU , so the Hamiltonian is the same in both pictures at alltimes.

Note that all Heisenberg-picture states and operators coincide with their Schrodinger-picture equiv-alents at time t = t0.

We can derive an equation of motion for the position operator in the Heisenberg picture:ih ∂∂tx(t) =

(ih ∂U †

∂t

)

xU + U †x

(ih∂U∂t

)

= −U †HxU + U †xHU

= −U †HU U †xU + U †xU U †HU

=[

x(t), H]

(2.55)

This is the Heisenberg Equation of motion for the time-evolution of the position operator in theHeisenberg picture. Since the state vector |ψ〉 is time-independent in the Heisenberg picture, itdoesn’t have an equation of motion.


The same argument can be used for any operator O(t). We say O(t) is conserved if ∂O(t)/∂t = 0,

whence[

O(t), H]

= 0. For example, momentum is conserved if and only if[

p(t), H]

= 0.

For the rest of this course we will generally adopt the Heisenberg picture.

6. The transition amplitude 〈x, t x′, t′〉 is the retarded Green function for the Schrodinger equation.For t > t′

ψ(x, t) = 〈x, t ψ〉 =∫

dx′ 〈x, t x′, t′〉〈x′, t′ ψ〉

So if we define

G(x, x′; t, t′) =

〈x, t x′, t′〉 t > t′

0 t < t′(2.56)

then θ(t− t′)ψ(x, t) =

∫

dx′ G(x, x′; t, t′)ψ(x′, t′) ∀ t, t′

Now, since ψ(x, t) satisfies the time-dependent Schrodinger equation, we find

(

− h2

2m

∂2

∂x2+ V (x, t)− ih ∂

∂t

)

θ(t− t′)ψ(x, t) = −ihδ(t− t′)ψ(x, t)

which implies

(

− h2

2m

∂2

∂x2+ V (x, t) − ih ∂

∂t

)

G(x, x′; t, t′) = −ihδ(t− t′) δ(x− x′) ∀ t, t′

To recover the first equation from the second, multiply the latter by ψ(x′, t′), integrate with respectto x′, and then compare with the last line of (2.56)

Note: We choose G(x, x′; t, t′) = 0 for t < t′ because in non-relativistic quantum mechanics we onlyconsider paths in which the particle moves forwards in time.

7. For time-independent Hamiltonians, it is useful to expand in a basis of energy eigenstates |n〉 withn = 0, 1, 2, 3, . . .

H |n〉 = En |n〉 ; un(x) ≡ 〈x n〉 . (2.57)

So for t > 0 the Green function is:

G(x, y; t) ≡ 〈x, t y, 0〉 = 〈x| e−itH/h |y〉 =∑

n

〈x| e−itH/h |n〉〈n y〉

=∑

n

e−itEn/hun(x)u∗n(y)

(2.58)

Define the Fourier transform of G(x, y; t) wrt t as G(x, y;E) =∞∫

−∞

dt eitE/hG(x, y; t) eit(iǫ)/h , soG(x, y;E) =

∑

n

∞∫

0

dt eit(E−En)/h un(x)u∗n(y) e

it(iǫ)/h = ih∑n

un(x)u∗n(y)

E − En + iǫ (2.59)

The iǫ (ǫ > 0, infinitesimal) is introduced to ensure that for positive real energies the integralconverges at the upper limit. Evidently, bound states correspond to poles in G(x, y;E) which liejust below the real axis in the complex E plane. i.e. at En − iǫ, ensuring that when we do theinverse transform we recover the retarded (or causal) Green function (i.e. G(x, y; t) = 0 for t < 0.)

[Technical note: An iǫ is also necessary for the convergence of the path integral, e.g. for the harmonicoscillator, En → En − iǫ if ω → ω − iǫ, ω > 0, i.e. ω2 → ω2 − iǫ, so

S[x] → S[x] + iǫm ∫ dt x2, and expiS/h → expiS/h exp

−ǫ(m/h)∫

dtx2

The last term is a convergence factor which damps paths with very large x2, it ensures that thepath integral gives causal propagation in non-relativistic quantum mechanics.]

Setting x = y and integrating over x in (2.58) (i.e. taking the trace) gives (for t > 0):

∞∫

−∞

dx 〈x, t x, 0〉 =∫

dx G(x, x; t) =∑

n

e−itEn/h

∫

dx |un(x)|2 =∑

n

e−itEn/h (2.60)


for orthonormal energy eigenfunctions un(x).

Hence, if we know G(x, y; t), we can use this expression to deduce the energy eigenvalues. Forexample let us consider the harmonic oscillator:

∞∫

−∞

dx 〈x, t x, 0〉 =∞∫

−∞

dx

√mω

2πih sinωt exp imω2h sinωt

2x2(cosωt− 1)︸︷︷︸iScl/h,

=1

2i 1

sin ωt2

(gaussian integral)

=e−iωt/2

1− e−iωt= e−iωt/2

∞∑

n=0

e−inωt =∑

n

e−itEn/h (using (2.60))

(2.61)

It follows that

En =

(

n+1

2

)

hω

The eigenfunctions may also be deduced in this way (tutorial.)

2.10. Single particle in an Electromagnetic Field

The Lagrangian for a particle of charge e in an electromagnetic field is (in Heaviside-Lorentz units)

L(x, x, t) =1

2m|x|2 − eφ+

e

cx ·A (2.62)

where φ(x, t) is the electric potential (also known as the “scalar” or “electrostatic” potential) andA(x, t) is the magnetic vector potential with

E = −∇φ− 1

c

∂A

∂tand B = ∇×A (2.63)

The classical Hamiltonian is

H(x, p, t) =1

2m

∣∣∣p− e

cA∣∣∣

2

+ eφ (2.64)

Exercise: verify explicitly that the Lagrangian (2.62) gives the Lorentz force F = eE + (e/c) x×Band the classical Hamiltonian above. (See Junior Honours Lagrangian Dynamics notes & tutorials.)

One can then derive the quantum Hamiltonian H(x, p, t) and the corresponding Schrodinger equa-

tion. This requires a little care because[p, A(x, t)

]6= 0 (tutorial).

Gauge invariance: Classically, the E and B fields (and thus the Lorentz force, and hence theclassical path) are unchanged under the gauge transformation

A→ A+∇χ and φ→ φ− 1

c

∂χ

∂t(exercise) (2.66)

for any function χ(x, t). However, the Lagrangian does change:

L→ L+e

cx · ∇χ+

e

c

∂χ

∂t= L+

e

c

dχ

dt(2.67)

where we used x·∇χ+ ∂χ∂t = dχ

dt , the total derivative with respect to t. But adding a total derivative tothe Lagrangian doesn’t change the classical Lagrange equations of motion (see Lagrangian Dynamics

notes), so our previous claim that classical physics is gauge invariant remains valid.

The action also changes:

S =

∫ tb

ta

dt L→ S +e

c

∫ tb

ta

dtdχ

dt= S +

e

c(χ(xb, tb)− χ(xa, ta)) (2.68)

so the quantum mechanical transition amplitude changes:


xa

Dx eiS/h →

xb∫

xa

Dx eiS/h exp

iehc

(χ(xb, tb)− χ(xa, ta))

(2.69)

From this we may deduce

|x, t〉 → exp

− iehcχ(x, t)

|x, t〉 (2.70)

which is an independent phase change locally at every point in space and time. The transitionprobability ∝ |〈xb, tb xa, ta〉|

2is of course unchanged. This is a huge symmetry of the theory, and

is known as a U(1) local gauge symmetry.


2.11. The Aharonov-Bohm effect

Consider a double slit experiment involving charged particles and a magnetic field.

1 1

22

AB

The magnetic field B points out of the page and is non-zero in the shaded area only – think of a longthin solenoid – and we assume the particles are shielded from the region of non-zero magnetic fieldperfectly. The B field corresponds to a magnetic vector potential A whose field lines form circlesaround the solenoid as shown.

Classically, we expect the field to have no effect because the particles don’t travel through the regionof non-zero magnetic field. The A field is non-zero along the paths but the vector potential has nodirect significance in classical physics.

Quantum mechanically, the dominant contributions to the transition amplitude 〈f i〉 will come frompaths close to the classical paths x1(t) and x2(t) shown in the figure. In the absence of the field, weadd the amplitudes as usual, so the contribution of these paths is

eiS[x1]/h + eiS[x2]/h = eiS[x1]/h(

1 + ei(S[x2]−S[x1])/h)

(2.71)

and we get interference from the relative phase φ = (S[x2]− S[x1])/h.

When we add a magnetic field B (assumed time-independent, so that A is also time-independent)the Lagrangian changes:

L→ L+e

cx ·A

so

S → S +e

c

∫ tb

ta

dtdx

dt· A = S +

e

c

∫ tb

ta

dx(t) · A (2.72)

Therefore, when we switch on the magnetic field, there is a change in the relative phase

δφ =e

hc

(∫ tb

ta

A · dx2(t)−∫ tb

ta

A · dx1(t))

=e

hc

∮

C

A · dx(t) ≡ e

hcΦ

(2.73)

where the curve C is the closed path x2−x1. This path clearly encircles the region of non-zeromagnetic field, but the magnetic field is zero everywhere on the closed path itself. Using Stokes’theorem we may write

Φ =

∮

C

A · dx(t) =∫

S

∇×A · dS =

∫

S

B · dS (2.74)

where S is any surface bounded by the closed curve x2−x1 in the diagram. So Φ is the total magnetic

flux passing in between the two paths, and the interference pattern shifts upwards (or downwards)due to the relative phase shift (e/hc)Φ – even though the particle hasn’t passed through any regionof non-zero magnetic field B, and has therefore felt no direct electromagnetic forces!


Notes:

(1) We get the same phase shift δφ for all paths x1(t) and x2(t) which don’t penetrate the region ofnon-zero magnetic field B, not just for the classical paths. Hence our expression (2.73) for the phaseshift holds in the full quantum theory, it’s not just a semi-classical approximation.

(2) Φ is gauge invariant. If A→ A+∇χ, then

Φ → Φ+

∮

C

∇χ · dx = Φ +

∮

C

dχ = Φ + 0 = Φ . (2.75)

(3) Only the flux passing between the two paths is included.

(4) The effect is periodic: there is no effect if δφ = 2nπ, i.e. when

Φ = 2πnhc

e= n

hc

e, n = 0, ±1, ±2, . . . (2.76)

(5) From the shift in the interference pattern, we deduce that the vector potential A(x) is not just amathematical artifice, as might be concluded from classical physics.

(6) The Aharonov-Bohm effect was first observed in 1960.

2.12. Transition Elements

Besides transition amplitudes, we are also interested in transition elements where the sum over allpaths is weighted by some function(al) of x(t). The simplest example is (ta ≤ t ≤ tb):

〈x(t)〉S ≡xb∫

xa

Dx x(t) eiS[x]/h =

∞∫

−∞

dx

xb∫

x

Dxx∫

xa

Dx exp

ih

tb∫

t

L dt

x exp

ih

t∫

ta

L dt

=

∫

dx 〈xb, tb x, t〉x 〈x, t xa, ta〉 =∫

dx 〈xb, tb| x(t) |x, t〉〈x, t xa, ta〉

= 〈xb, tb| x(t) |xa, ta〉

(2.77)

This is the matrix element of the operator x(t) between the states |xa, ta〉 and |xb, tb〉 in the Heisen-berg picture, and it’s a non-trivial function of t. It is easy to see (exercise) that for any local functionf(x(t)), we have

〈xb, tb| f(x(t)) |xa, ta〉 =xb∫

xa

Dx f(x(t)) eiS[x]/h (2.78)

Now consider correlations between x(t) and x(t′) with t 6= t′. For t > t′:

〈x(t)x(t′)〉S ≡xb∫

xa

Dx x(t)x(t′) eiS/h =

∫

dx

∫

dx′ 〈xb, tb x, t〉x 〈x, t x′, t′〉x′ 〈x′, t′ xa, ta〉

=

∫

dx

∫

dx′ 〈xb, tb| x(t) |x, t〉〈x, t| x(t′) |x′, t′〉〈x′, t′ xa, ta〉

= 〈xb, tb| x(t) x(t′) |xa, ta〉

For t < t′, we get the same expression, but with t↔ t′,

〈x(t)x(t′)〉S = 〈xb, tb| x(t′)x(t) |xa, ta〉

So, in general we have:

xb∫

xa

Dx x(t)x(t′) eiS/h = 〈xb, tb|T (x(t) x(t′)) |xa, ta〉 (2.79)

whereT (x(t)x(t′)) ≡ θ(t− t′)x(t)x(t′) + θ(t′ − t)x(t′)x(t) (2.80)

is called the time ordered product - the position operator at the earlier time appears on the rightof the one at the later time.

Note that the Heisenberg-picture operators x(t) and x(t′) do not commute (unless t = t′).


Clearly, this may be generalised to any number of local insertions:

〈f1(x(t1)) . . . fn(x(tn))〉S = 〈xb, tb|T (f1(x(t1)) . . . fn(x(tn))) |xa, ta〉

=

xb∫

xa

Dx f1(x(t1)) . . . fn(x(tn)) eiS[x(t)]/h (2.81)

Note that the quantities fi(x(ti)) in the “operator expression” on the RHS of the first line of (2.81)are non-commuting operators, whilst the quantities fi(x(ti)) in the path integral on the second lineare ordinary (commuting) functions.

The time-ordering operator orders all of the operators in the product according to time – theoperator at the earliest time is on the extreme right, . . . , and the operator at the latest time is onthe extreme left.

Between more general states, we simply insert complete sets of position eigenstates: for example

〈ψ|T (x(t1) . . . x(tn)) |φ〉 =∫

dxa

∫

dxb 〈ψ xb, tb〉〈xb, tb|T (x(t1) . . . x(tn)) |xa, ta〉〈xa, ta φ〉

=

∫

dxa

∫

dxb ψ∗(xb, tb)φ(xa, ta)︸︷︷︸

wave functions

xb∫

xa

Dx x(t1) . . . x(tn)eiS/h︸︷︷︸

transition elements

(2.82)These transition elements or matrix elements or Green functions or correlation functions (in statis-tical mechanics language) will play a central role in what follows.

We can also define transition elements with insertions of time derivatives of x(t). However, theseare a little trickier since x = p/m, and p(t) and x(t) do not commute. To understand the issuesinvolved, we consider a couple of examples.

We start by going back to the basic definition of the path integral:


AN

(N∏

n=1

∫

dxn

)

exp

iεh

N+1∑

n=1

[

m

2

(xn − xn−1

ε

)2

− V (xn, tn)

]

(2.83)

Now, since∫∞

−∞dxp

∂∂xp

f(xp) = 0 for any function f(xp) which approaches zero sufficiently quickly

as |xp| → ∞, and for any p ∈ 1, . . . , N, we must have:

0 = limN→∞

AN

(N∏

n=1

∫

dxn

)

∂

∂xp

(

F (xp) exp

iεh

N+1∑

n=1

[

m

2

(xn − xn−1

ε

)2

− V (xn, tn)

])

= limN→∞

AN

(N∏

n=1

∫

dxn

)(∂F

∂xp− iεhF (xp)

m

ε2(xp+1 − 2xp + xp−1) +

∂V

∂xp

)

exp . . .

(2.84)Technical note: The integral should be considered as the limit of the analytic continuation of anintegral with a real part in the exponential argument. So the integral is well-defined although itlooks divergent.

As ε→ 0 we can rewrite terms in (2.84):

limε→0

xp+1 − xpε

= x(t), while limε→0

xp+1 − 2xp + xp−1

ε2= lim

ε→0

1

ε

(xp+1 − xp

ε− xp − xp−1

ε

)

= x(t)

(2.85)If we now set F = 1, so ∂

∂xpF = 0, then as N → ∞ we get:

0 =

xb∫

xa

Dx(

mx(t) +∂V

∂x

)

eiS/h ⇒ 〈x〉S = − 1

m

⟨∂V

∂x

⟩

S

(2.86)

Equation (2.86) is the quantum version of the classical equation of motion and is known as Ehren-fest’s theorem.


If instead we take F = xp, so∂

∂xpF = 1, we get:

0 =

xb∫

xa

Dx eiS/h limε→0

(

1− ih

mxp

[xp+1 − xp

ε− xp − xp−1

ε

]

+ εxp∂V

∂xp

)

(2.87)

As ε → 0, we write xp → x(t) and hence xp+1 → x(t + ε). Remembering the time ordering rule,and writing mx(t) ≡ p(t), we get:

0 = limε→0

〈xb, tb|

1− ihT(x(t)p(t+ ε/2)− x(t)p(t− ε/2)

)

|xa, ta〉

= 〈xb, tb|

1− ih

(p(t)x(t)− x(t)p(t)

)

|xa, ta〉 (ta < t < tb)

(2.88)

whence the usual equal-time commutation relation

[x(t), p(t)] = ih . (2.89)

It follows that we only get the ordering of x and p correct if we are careful with the time orderingin the path integral. However, since xp+1 = xp +O(ε) (by continuity),

xp(xp − xp−1) = xp+1(xp+1 − xp) +O(ε2),

we may also write (2.87) as

0 =

xb∫

xa

Dx eiS/h limε→0

(

1− ihm

xp

(xp+1 − xp

ε

)

− xp+1

(xp+1 − xp

ε

)

+O(ε)

)

,

so: ⟨

m

(xp+1 − xp

ε

)2⟩

S

= − hiε 〈1〉S +O(1) (2.90)

But if we now take the limit ε → 0, we find that⟨x(t)2

⟩

Sand therefore

⟨p(t)2

⟩

Sare infinite! This

shows that the paths that we integrate over are not smooth. The action is finite (for paths thatcount), so from (2.83), ε(δx/ε)2 is finite, i.e. δx ∼ √

ε, δx ∼ 1/√ε. The paths are continuous but

jagged. This phenomenon is called Zitterbewegung.

There is a nice illustration on page 177 of the book by Feynman and Hibbs (and Steyer).

How then do we define the kinetic energy? The trick is to use

limε→0

1

2m〈p(t+ ε/2)p(t− ε/2)〉S (2.91)

To see that this is finite, take F = xp+1 − xp, so∂

∂xpF = −1, and so, from (2.84):

〈−1〉S =iεmh

⟨(xp+1 − xp

ε

)(xp+1 − xp

ε− xp − xp−1

ε

)

+ (xp+1 − xp)︸︷︷︸

O(ε)

∂V

∂xp

⟩

S

(2.92)

⇒ 1

2m〈p(t+ ε/2)p(t− ε/2)〉S =

m

2

⟨(xp+1 − xp

ε

)(xp − xp−1

ε

)⟩

S

+O(ε)

=

⟨

m

2

(xp+1 − xp

ε

)2⟩

S︸︷︷︸

≃− h2iε 〈1〉S+O(1) from (2.90)

+h

2iε 〈1〉S +O(ε) (2.93)

so the limit ε→ 0 exists. This provides a (trivial) example of renormalisation in quantum mechanics- using a ‘point-splitting’ regularisation.

3. Perturbation Theory 25

3. Perturbation Theory3.1. Perturbation theory from path integrals

Most dynamical systems are not exactly solvable (either classically or quantum mechanically). How-ever, we can often separate the action into a solvable part and a perturbation

S[x(t)] = S0[x(t)] + S1[x(t)]. (3.1)

For example for a particle in a slowly-varying potential:

S0[x(t)] =

tb∫

ta

dt1

2mx2 ; S1[x(t)] = −

tb∫

ta

dt V (x(t), t) (3.2)

or S0[x(t)] =

tb∫

ta

dt

(1

2mx2 − U(x)

)

; S1[x(t)] = −tb∫

ta

dt V (x(t), t) (3.3)

where V (x, t) = V (x, t) − U(x), so L0 is time-independent. Sometimes we take U = mω2x2/2, sothat L0 is the Lagrangian of a harmonic oscillator. It all depends on the particular problem we wishto solve.

Then the transition amplitude is (using (3.2)):


xa

Dx exp

ih(S0[x(t)] + S1[x(t)])

=

xb∫

xa

Dx eiS0[x(t)]/h∞∑

n=0

1

n!

( ihS1[x(t)]

)n

=

∞∑

n=0

1

n!

(

− ih

)ntb∫

ta

dt1 · · ·tb∫

ta

dtn

xb∫

xa

Dx V (x(t1), t1) · · · V (x(tn), tn) eiS0[x(t)]/h

(3.4)assuming that we can exchange the order of the infinite sum and the functional integration. Eachterm in the series is a transition element.

For ta < t1 < t2 < . . . < tn < tb, we may write:

xb∫

xa

Dx V (x(t1), t1) · · ·V (x(tn), tn) eiS0/h =

=

∫

dx1 · · ·∫

dxn 〈xb, tb xn, tn〉0 V (xn, tn) 〈xn, tn xn−1, tn−1〉0 · · ·V (x1, t1) 〈x1, t1 xa, ta〉0(3.5)

The subscript zero indicates that these are transition amplitudes evaluated with S0. All that remainsto be done is to actually evaluate the transition elements.

It is useful to represent the series in (3.4) & (3.5) pictorially. Denoting the transition amplitudes〈x, t x′, t′〉0 by straight lines and insertions of the potential by wiggles, we have:

〈xb, tb xa, ta〉 =

ta

tb

+

tat1

tb

+ta

t1 t2tb

+ . . . (3.6)

In words: the full amplitude can be written as a sum of “partial” amplitudes in which the particleis: not scattered + scattered once + scattered twice + · · · .

26 3. Perturbation Theory

The integrals over xi and ti ensure that we sum over all paths, and the 1/n! ensures that paths withdifferent time orderings are not double counted, since:

1

n!

tb∫

ta

dt1 · · ·tb∫

ta

dtn V (t1) · · ·V (tn) =

tb∫

ta

dtn

tn∫

ta

dtn−1 · · ·t3∫

ta

dt2

t2∫

ta

dt1 V (t1) · · ·V (tn) (3.7)

so ta ≤ t1 ≤ t2 ≤ · · · ≤ tn ≤ tb is the only one we need to consider.

Proof of (3.7) for n = 2. Generalisation to all n can be done by induction (tutorial). Since

tb∫

ta

dt2

tb∫

ta

dt1 V (t1)V (t2) =

tb∫

ta

dt2

t2∫

ta

dt1 V (t1)V (t2) +

tb∫

t2

dt1 V (t1)V (t2)

,

it suffices to show that:tb∫

ta

dt2

t2∫

ta

dt1 V (t1)V (t2) =

tb∫

ta

dt1

tb∫

t1

dt2 V (t1)V (t2) (swapped order of integrals)

=

tb∫

ta

dt2

tb∫

t2

dt1 V (t2)V (t1) (relabelled dummy integrands: t1 ↔ t2)

This works because the integrand is symmetric under t1 ↔ t2, i.e. V (t1)V (t2) = V (t2)V (t1).

Using the result from (3.5) and (3.7) in (3.4), we obtain the transition amplitude to all orders inperturbation theory:

〈xb, tb xa, ta〉 = 〈xb, tb xa, ta〉0 +

+

∞∑

n=1

(

− ih

)n ∫

dx1 · · ·∫

dxn

tb∫

ta

dtn

tn∫

ta

dtn−1 · · ·t2∫

ta

dt1

〈xb, tb xn, tn〉0 V (xn, tn) 〈xn, tn xn−1, tn−1〉0 · · ·V (x1, t1) 〈x1, t1 xa, ta〉0

(3.8a)

We can obtain a formal expression for the sum of this series as follows. First write (3.8a) as

〈xb, tb xa, ta〉 = 〈xb, tb xa, ta〉0 −ih

∞∫

−∞

dx

tb∫

ta

dt 〈xb, tb x, t〉0 V (x, t) 〈x, t xa, ta〉0

+

(

− ih

)2∞∫

−∞

dx

∞∫

−∞

dx′tb∫

ta

dt

t∫

ta

dt′ 〈xb, tb x, t〉0 V (x, t) 〈x, t x′, t′〉0 V (x′, t′) 〈x′, t′ xa, ta〉0 + . . .

where the dots represent all higher-order terms. Now factorise the 2nd, 3rd, . . . , terms:

〈xb, tb xa, ta〉 = 〈xb, tb xa, ta〉0 +

+

(

− ih

) ∞∫

−∞

dx

tb∫

ta

dt 〈xb, tb x, t〉0 V (x, t)

[

〈x, t xa, ta〉0 +

+

(

− ih

) ∞∫

−∞

dx′t∫

ta

dt′ 〈x, t x′, t′〉0 V (x′, t′) 〈x′, t′ xa, ta〉0 + . . .

]

The infinite sum in square brackets is just the perturbation series for 〈x, t xa, ta〉. Therefore

〈xb, tb xa, ta〉 = 〈xb, tb xa, ta〉0 −ih

∞∫

−∞

dx

tb∫

ta

dt 〈xb, tb x, t〉0 V (x, t) 〈x, t xa, ta〉 (3.8b)

We may interpret equation (3.8b) pictorially:

ta

tb

=

ta

tb

+

tat

tb

(3.9)

Full amplitude = unscattered amplitude + sum of all processes with the last scattering at time t.


We can use (3.8b) to obtain a similar equation for wavefunctions:

〈x, t ψ〉 =∞∫

−∞

dx′′ 〈x, t x′′, t0〉〈x′′, t0 ψ〉 =∞∫

−∞

dx′′ 〈x, t x′′, t0〉0 〈x′′, t0 ψ〉

− ih

∞∫

−∞

dx′′∞∫

−∞

dx′t∫

t0

dt′ 〈x, t x′, t′〉0 V (x′, t′) 〈x′, t′ x′′, t0〉〈x′′, t0 ψ〉

= 〈x, t ψ〉0 −ih

∞∫

−∞

dx′t∫

t0

dt′ 〈x, t x′, t′〉0 V (x′, t′) 〈x′, t′ ψ〉

i.e. ψ(x, t) = ψ0(x, t) −ih

∞∫

−∞

dx′t∫

t0

dt′ 〈x, t x′, t′〉0 V (x′, t′)ψ(x′, t′)

(3.10)

where ψ0(x, t) is the unperturbed wave function, which satisfies the unperturbed Schrodinger equa-tion (ih ∂

∂t− H0(x, t)

)

ψ0(x, t) = 0 .

Relation to Schrodinger’s equation: Recall that G0(x, x′; t, t′) ≡ θ(t− t′) 〈x, t x′, t′〉0 satisfies

(ih ∂∂t

− H0(x, t)

)

G0(x, x′; t, t′) = ihδ(t− t′) δ(x− x′) .

(See equations (2.56), etc.) If we rewrite the last line of equation (3.10) as

ψ(x, t) = ψ0(x, t) −ih

∞∫

−∞

dx′∞∫

t0

dt′ θ(t− t′) 〈x, t x′, t′〉0 V (x′, t′)ψ(x′, t′) ,

we find

(ih ∂∂t

− H0(x, t)

)

ψ(x, t) = 0− ihih ∞∫

−∞

dx′∞∫

t0

dt′ δ(t− t′) δ(x− x′)V (x′, t′)ψ(x′, t′) = V (x, t)ψ(x, t)

⇒ ih ∂∂tψ = (H0 + V )ψ = H ψ

The last line of (3.10) is therefore an integral equation for ψ, which is equivalent to Schrodinger’sequation.

3.2. Fixed target scattering – time-independent transitions

Consider elastic scattering of a particle of mass m in a fixed potential V (x, t). We need to find

limtb→+∞

ta→−∞

〈x b, tb x a, ta〉 = limtb→+∞

ta→−∞

〈x b| U(tb, ta) |x a〉 ≡ 〈x b| S |x a〉

where the operator S ≡ U(∞,−∞) is called the Scattering Operator or the “S-matrix”. Since U isunitary, S is also unitary: S†S = 1. (What goes in must come out!)

|x b,∞〉 is called the out state, a free particle state in the far future, and |xa,−∞〉 is the in state, afree particle state in the far past, where we have assumed that the potential V (x, t) is short-ranged:V (±∞, t) = 0.

If the interaction is time-independent, (i.e. just V (x)), then

limtb→+∞

ta→−∞

〈x b, tb x a, ta〉 = limT→∞

〈x b, T/2 xa,−T/2〉 = limT→∞

〈x b, T xa, 0〉

by time translation invariance. So, from equation (3.8a) we need to calculate

〈x b, T xa, 0〉 = 〈x b, T xa, 0〉0 −ih

∞∫

−∞

d3x

T∫

0

dt 〈x b, T x, t〉0 V (x) 〈x, t xa, 0〉0 +O(V 2)


where 〈x b, T xa, 0〉0 obviously describes the case of no scattering.

no scatteringV

O target

detector

source

(x , T)c

(x , t)

θ

(x , 0)aRp

a

r

r

a

a

b

r

pb

(x , T)b

Rb

We are treating V (x) as the perturbation, so the unperturbed Lagrangian is simply

L0 =m

2

(x21 + x22 + x23

)

In Cartesian coordinates, the path integral for the transition amplitude for a free particle in threedimensions factorises into three one-dimensional path integrals, and hence (exercise)

〈x′, t′ x, t〉0 =3∏

i=1

〈x′i, t′ xi, t〉0

The transition amplitude, A1, in first-order perturbation theory is then

A1 =−ih

∞∫

−∞

d3x

T∫

0

dt

(m

2πih(T − t)

)3/2

exp

( im|x b − x|22h(T − t)

)

V (x)( m

2πiht)3/2 exp( im|x− x a|22ht

)

Performing the integral over t using

T∫

0

dt

[(T − t)t]3/2

exp

− α

T − t− β

t

=1

T

√π

T

(1√α+

1√β

)

exp

−(√

α+√

β)2

/ T

(see separate handout) gives

A1 =−ih

( m

2πih)3 1

T

√π

T

∞∫

−∞

d3x

(−im|x a − x|22h

)−1/2

+

(−im|x b − x|22h

)−1/2

× V (x) exp

im

2hT

(

|x a − x|+ |x b − x|)2

=−ih

( m

2πih)5/2 1

T 3/2

∞∫

−∞

d3x

(1

ra+

1

rb

)

V (x) exp

im

2hT(ra + rb)

2

where we defined ra ≡ |x a − x| and rb ≡ |x b − x|.Since the potential is short-range, if we define Ra ≡ |x a|, Rb ≡ |x b|, and r ≡ |x|, then r ≪ Ra, Rb,and so

ra =(|x a − x|2

)1/2= Ra

(

1− 2xa · xR2

a

+r2

R2a

)1/2

= Ra − n a · x+O(1/Ra)

where n a = x a/Ra is a unit vector in the direction of x a. Similarly

rb = Rb−n b ·x+O(1/Rb) and (ra + rb)2 = (Ra +Rb)

2− 2(Ra+Rb) (n a+n b) ·x+O(R0a,b)

So

A1 ≃ −ih

( m

2πih)5/2 1

T 3/2

1

Ra+

1

Rb+O

(

1

R2a,b

)

exp

im

2hT(Ra +Rb)

2

×∞∫

−∞

d3xV (x) exp

− im

hT(Ra +Rb) (n a + n b) · x+O(R0

a,b)


We can measure Ra, Rb and T , and for a short-range potential (where we can treat it as a freeparticle most of the time), we deduce from the diagram on the previous page that the particle’stotal energy, initial momentum and final momentum are

E =1

2m

(Ra +Rb)2

T 2=

1

2mu2 and p

a= −p na, p

b= p n b where p = m

(Ra +Rb)

T= mu

and u is the particle’s speed. Therefore

A1 ≃ −ih

( m

2πih)5/2 1

T 3/2

(Ra +Rb

RaRb

)

expiET/h∞∫

−∞

d3xV (x) expi((p

a− p

b) · x

)

/h

︸︷︷︸

≡V (q)

where we defined the wave vector transfer q ≡ (pa− p

b)/h, so that the momentum transfer is hq.

The transition probability (per unit volume – see tutorial) is then

P (a→ b) = |A1|2 =1

h2

( m

2πh

)5 1

T 3

(Ra +Rb

RaRb

)2 ∣∣∣V (q)

∣∣∣

2

=1

h2

( m

2πh

)5 1

T

u2

R2aR

2b

|V (q)|2

Similarly P (a→ c) = |A0|2 = |〈x c, T xa, 0〉0|2 =

( m

2πh

)3 1

T 3=( m

2πh

)3 1

T

u2

(Ra +Rb)2

where x c ≡ −Rb n a, is the probability (per unit volume) for no scattering (i.e. for V (x) = 0). Theratio of the scattering and no-scattering probabilities is then:

P (a→ b)

P (a→ c)=

(m

2πh2

)2 (Ra +Rb

RaRb

)2

|V (q)|2

Following classical ideas, we describe the scattering process in terms of an effective target areaknown as the scattering cross section.

RR

c

b

a

R d2b

a bΩ

σ

dσ ba

2

R a2

(R + R ) d

θ

If particles starting from x a were to hit a small target of area dσ, at distance Ra, these par-ticles would be removed from the region around x c, where they would be spread out over area

[(Ra +Rb)/Ra]2dσ. Instead, they are scattered into solid angle dΩ and are spread out over area

R2b dΩ there. Hence

P (a→ b)R2b dΩ = P (a→ c)

(Ra + Rb)2

R2a

dσ

Rearranging gives

P (a→ b)

P (a→ c)=

(Ra +Rb

RaRb

)2dσ

dΩ

Comparing our two expressions for this ratio gives

dσ

dΩ=

(m

2πh2

)2

|V (q)|2

which is called the differential cross section in the Born approximation.


A more conventional wording of the definition of the differential cross section is (see, for example,Quantum Physics):

(Number of particles scattered into dΩ / unit time)

=

dσ × (Number of incident particles crossing scattering region / unit area / unit time)

Again, this gives

P (a → b) R2b dΩ = dσ

(

P (a→ c)

(Ra +Rb

Ra

)2)

which reproduces our previous result for dσ/dΩ.

Notes:

(1) All factors of T , Ra, Rb cancel when we construct dσ/dΩ, so we can send them to infinity withimpunity.

(2) We assumed initial and final states with definite position. However, we get the same result for anyinitial and final states, provided sufficiently localised wave functions cancel when we take the ratioP (a→ b)/P (a→ c).

(3) For a central potential V (r), r = |x|, we choose spherical polars with q parallel to the z-axis, and

the expression for V (q) simplifies to

∞∫

−∞

d3xV (r) exp(iq · x) = 2π

∞∫

0

r2dr

+1∫

−1

d(cos θ) exp(iqr cos θ)V (r) =4π

q

∞∫

0

rV (r) sin(qr) dr

Hence

dσ

dΩ=

4m2

h4q2

∣∣∣∣∣∣

∞∫

0

rV (r) sin(qr)dr

∣∣∣∣∣∣

2

with q = |pa− p

b|/h = (2p/h) sin θ/2 where θ is the scattering angle as shown in the figure.

(4) For the Coulomb potential V (r) = −e2/(4πǫ0r)∞∫

0

rV (r) sin(qr) dr =−e24πǫ0q

sodσ

dΩ=

(1

4πǫ0

)24m2e4

h4q4=

(1

4πǫ0

)2e4

16E2

1

sin4(θ/2)

(

where E =p2

2m

)

which is not only independent of h, but is the same as the classical Rutherford cross section. This“quantum cross section = classical cross section” result is unique to the 1/r potential – Rutherfordgot lucky!

Technical note (1): For the integral over the Coulomb potential to converge, we need to defineV (r) = −e2 exp(−µr)/(4πǫ0r), and take the limit µ→ 0 at the end. (See tutorial.)

Technical note (2): For compatibility with most standard texts, we have used SI units in theexpression for the Coulomb potential. In Heaviside-Lorentz units V (r) = −e2/(4πr), i.e. justremove the ǫ0.

(5) The total cross-section, σ, is

σ ≡∫

4π

dσ

dΩdΩ =

∫∫dσ

dΩsin θ dθ dφ

in spherical polar coordinates, where θ is the scattering angle shown in the figure, and φ is the anglemeasured about the the ‘beam axis’ which runs through the origin and x c

(6) These results may be obtained using time-dependent perturbation theory for plane wave states –see Quantum Physics.

(7) We may use (3.8a) to generate the O(V n) (with n > 1) contributions to the scattering amplitude.The resulting perturbation expansion is known as the Born Series.


3.3. Colliding Beams

Consider two particles of massesm1 andm2 interacting through a mutual potential. The Lagrangianis

L =1

2m1|x 1|2 +

1

2m2|x 2|2 − V (x 1 − x 2)

=1

2M |R|2 + 1

2µ|r|2 − V (r)

where

R =m1x 1 +m2x 2

m1 +m2, M = m1 +m2

︸︷︷︸

centre-of-mass motion

, r = x 1 − x 2 , µ =m1m2

m1 +m2︸︷︷︸

relative motion

O

x1

x2

r

R

In the centre-of-mass frame, R ≡ 0, so the first term drops out, and we have simply L = 12µ|r|2−V (r),

i.e. it’s the same as in fixed-particle scattering but with x→ r and m→ µ.

So if we consider a colliding beam scattering experiment,

θx /2a

−x /2

x /2

−x /2a

b

b

pb

pa

the differential cross section is simply

dσ

dΩ=

(µ

2πh2

)2

|V (q)|2 ≡ |f(θ, φ)|2

where q = (pa− p

b)/h as before, and we have defined the scattering amplitude f(θ, φ), via

dσ

dΩ≡ |f(θ, φ)|2

If m2 → ∞, then µ → m1 = m, say, and we recover the fixed target result (Born-Oppenheimerapproximation.)

However, for (non-identical) particles of the same mass m1 = m2 = m, then µ = m/2 and

dσ

dΩ=

(m

4πh2

)2

|V (q)|2

For central potentials, we have already shown that the scattering amplitude depends only on θ (seeNote (3) above), hence

dσ

dΩ= |f(θ)|2


Scattering of identical particles

If the particles are identical, we have two indistinguishable possibilities:

θ

π − θ

Direct Exchange

Classically, we add probabilities, so in a mythical world of classical “identical” particles we wouldhave

dσcldΩ

= |f(θ)|2 + |f(π − θ)|2

But in quantum mechanics, we must add amplitudes, therefore for identical bosons

dσbdΩ

= |f(θ) + f(π − θ)|2

so, for example, for the Coulomb potential,

dσbdΩ

=

(1

4πǫ0

)2e4

16E2

(1

sin2(θ/2)+

1

cos2(θ/2)

)2

=

(1

4πǫ0

)2e4

E2

1

sin4 θ

For identical spinless ‘fermions’ (i.e ignoring spin – see tutorial for spin dependence)

dσfdΩ

= |f(θ)− f(π − θ)|2 =

(1

4πǫ0

)2e4

16E2

(1

sin2(θ/2)− 1

cos2(θ/2)

)2

This gives

dσfdΩ

=

(1

4πǫ0

)2e4

E2

cos2 θ

sin4 θ

which vanishes when θ = π/2, as it must.

0 π

dσ/dΩ dσ/dΩ

00 π

Bosons

π/2 π/2

Fermions

(1/4πǫ0)2 e4/E2

So we can tell whether particles are bosons or fermions by studying the shape of the differentialcross section.


3.4. Perturbation theory in the operator formalism

Consider the transition amplitude 〈xb, tb xa, ta〉, where the position eigenstates states are in theHeisenberg picture. From (2.52) and the unitarity of U , we have

〈xb, tb xa, ta〉 =(

U †(tb, t0) |xb〉)†

U †(ta, t0) |xa〉 = 〈xb| U(tb, t0)U(t0, ta) |xa〉

= 〈xb| U(tb, ta) |xa〉(3.11)

where |xa〉 and |xb〉 are position eigenstates in the Schrodinger picture. We shall develop perturba-tion theory for U(tb, ta)

Firstly, we use (2.81) to write (3.5) as

xb∫

xa

Dx V (x(t1), t1) · · ·V (x(tn), tn) eiS0/h = 0〈xb, tb|T (V (x0(t1), t1) · · ·V (x0(tn), tn)) |xa, ta〉0

(3.12)where the states |x, t〉0 and the operators x0(t) are in the Heisenberg picture with respect to S0

rather than S,x0(t) = U †

0 (t, t0) x U0(t, t0) .

In this expression, x is the position operator in the Schrodinger picture, and the unperturbed time-evolution operator U0(t, t0) therefore satisfiesih ∂

∂tU0 = H0 U0 ,

where H0 is the unperturbed Hamiltonian. This is known as the Dirac or interaction picture fortime dependence in quantum mechanics.

Using (3.8a) the perturbative expansion (3.4) may then be written as:

〈xb, tb xa, ta〉 =∞∑

n=0

1

n!

(

− ih

)ntb∫

ta

dt1 · · ·tb∫

ta

dtn 0〈xb, tb|T (V (x0(t1), t1) · · ·V (x0(tn), tn)) |xa, ta〉0

=

∞∑

n=0

(

− ih

)ntb∫

ta

dtn

tn∫

ta

dtn−1 · · ·t2∫

ta

dt1 0〈xb, tb|V (x0(tn), tn) · · ·V (x0(t1), t1) |xa, ta〉0

(3.13)Note that

V (x0(t), t) = U †0 (t, t0)V (x, t) U0(t, t0) (3.14a)

is the potential energy operator in the interaction picture, and V (x, t) is its Schrodinger-pictureequivalent.

Since the time dependence in (3.12) and (3.13) is governed by S0, the interaction-picture states|xb, tb〉0 and |xa, ta〉0 on the RHS of (3.13) are related to the Schrodinger-picture states by

|xa, ta〉0 = U †0 (ta, t0) |xa〉 and hence 0〈xb, tb| = 〈xb| U0(tb, t0) (3.14b)

Furthermore we shall need

U0(tn, t0) U†0 (tn−1, t0) = U0(tn, t0) U0(t0, tn−1) = U0(tn, tn−1) (3.14c)

Starting from (3.13), and using (3.11) and (3.14a/b/c), we obtain another representation for equation(3.8a) in terms of the unperturbed time-evolution operators U0(tn, tn−1) and the Schrodinger-pictureoperators V (x, tn)

〈xb| U(tb, ta) |xa〉 = 〈xb| U0(tb, ta)) |xa〉

+

∞∑

n=1

(

− ih

)ntb∫

ta

dtn · · ·t2∫

ta

dt1 〈xb| U0(tb, tn)V (x, tn)U0(tn, tn−1)V (x, tn−1) · · ·

· · · V (x, t2)U0(t2, t1)V (x, t1)U0(t1, ta) |xa〉


The equation above holds for all states |xa〉 and |xb〉, so we can write it in purely operator form

U(tb, ta) = U0(tb, ta) +

∞∑

n=1

(

− ih

)ntb∫

ta

dtn · · ·t2∫

ta

dt1 U0(tb, tn)V (x, tn)U0(tn, tn−1)V (x, tn−1) · · ·

· · · V (x, t2) U0(t2, t1)V (x, t1) U0(t1, ta)(3.15a)

This is usually referred to as the Dyson series. We can sum this series formally (exercise) to obtainthe operator equivalent of (3.8b)

U(tb, ta) = U0(tb, ta)−ih

tb∫

ta

dt U0(tb, t)V (x, t) U(t, ta) (3.15b)

Equivalently, we can obtain (3.15b) by expressing (3.8b) in terms of time evolution operators andSchrodinger-picture position eigenstates (exercise). We leave it as an exercise to show that (3.15b)is the “solution” of the Schrodinger equation. Indeed, (3.15b) can be derived directly from theSchrodinger equation (tutorial).

Note that all of the potential-energy operators V (x, t), etc, in (3.15a) and (3.15b) are in theSchrodinger picture.

3.4. Time dependent transitions

A common situation is where the Lagrangian L0 (corresponding to the action S0) is time indepen-dent, whilst the Lagrangian L1 of the perturbation (corresponding to S1) is time dependent. Letus further assume that the Hamiltonian H0 of the unperturbed system has a discrete spectrum ofbound-state energies En and (Schrodinger-picture) eigenstates |n〉

H0 |n〉 = En |n〉(For the Heisenberg-picture, we simply replace |n〉 → |n, t〉.) Rather than working in the positionbasis, it is easier to use the energy eigenbasis, because then the unperturbed transition amplitudeis diagonal:

〈m, t n, t′〉0 = 〈m| U0(t, t′) |n〉 = 〈m| e−i(t−t′)H0/h |n〉 = exp

−i(t− t′)En

h

δmn (3.16)

It follows from (3.15a) that the perturbed amplitude in the energy eigenbasis is:

〈b, tb a, ta〉 = 〈b| U(tb, ta) |a〉 = 〈b| U0(tb, ta) |a〉 −ih

tb∫

ta

dt 〈b| U0(tb, t)V (x, t)U0(t, ta) |a〉+ · · ·

= 〈b| U0(tb, ta) |a〉 −ih

tb∫

ta

dt∑

m,n

〈b| U0(tb, t) |m〉〈m|V (x, t) |n〉〈n| U0(t, ta) |a〉+ · · ·

= e−i(tb−ta)Ea/hδab −ih

tb∫

ta

dt∑

m,n

e−i(tb−t)Eb/hδmbVmn(t)e−i(t−ta)Ea/hδna + · · ·

= e−i(tb−ta)Ea/hδab −ihe−i(Ebtb−Eata)/h

tb∫

ta

dt eit(Eb−Ea)/hVba(t) + · · ·

(3.17)where Vmn(t) is the matrix element of the potential

Vmn(t) ≡ 〈m|V (x, t) |n〉 =∞∫

−∞

dx 〈m|V (x, t) |x〉〈x n〉 =∞∫

−∞

dx u∗m(x)V (x, t)un(x)

The second order term in the expansion is (check this for yourself)

(

− ih

)2∑

n

tb∫

ta

dt2

t2∫

ta

dt1 e−i(tb−t2)Eb/hVbn(t2)e−i(t2−t1)En/hVna(t1)e

−i(t1−ta)Ea/h


It is easy to see how this will generalise to higher orders. Diagramatically, we have (for a 6= b):

〈b, tb a, ta〉 =

taa

b

Vba(t)

tb

+∑

n taa

nb

Vna(t1)

Vbn(t2)

tb+∑

n,m taa

nm

b

Vna(t1)

Vmn(t2)

Vbm(t3)

tb

+ · · ·

All intermediate ‘virtual’ states n,m, · · · are summed over, and the ‘interaction times’ t1 < t2 · · · < tnare integrated over. For example, at third order: a → n → m → b for all possible intermediatevalues of n and m.

If a 6= b, the first (trivial) term vanishes, and the transition probability becomes:

p(a→ b) = | 〈b, tb a, ta〉 |2 =1

h2

∣∣∣∣∣∣

tb∫

ta

dt eitωbaVba(t) +O(V 2)

∣∣∣∣∣∣

2

(3.18)

where ωba ≡ (Eb−Ea)/h is the transition frequency. This should be familiar from Quantum Physics.

Note that higher order terms interfere: from (3.18), the next correction to p(a → b) is O(V 3), notO(V 4). Writing

Vba =

tb∫

ta

dt eiωbatVba(t),

we obtain

p(a→ b) =1

h2|Vba|2 −

2

h3Re

(

V ∗ba

∑

n

VbnVna

)

. (3.19)

When the perturbation is time independent, we can do the t integrals, and the transition amplitudethrough second order becomes

e−iTEa/hδab+

(e−iTEb/h − e−iTEa/h

)

Eb − EaVba+

∑

n

(e−iTEb/h − e−iTEn/h

)(e−iTEn/h − e−iTEa/h

)

(Eb − En)(En − Ea)VbnVna

where T = tb − ta, as usual.

To first order, the transition probability (for the time-independent case with a 6= b) is simply

p(a→ b) =

∣∣e−iTEa/h

(e−iTωba − 1

)∣∣2

h2ω2ba

|Vba|2 =sin2(ωbaT/2)

h2(ωba/2)2|Vba|2 ≡ f(T, ωba)

h2|Vba|2 (3.20)

The function

f(t, ω) =sin2(ωt/2)

(ω/2)2 =

sin2(ωt/2)

(ωt/2)2 t2

consists essentially of a large peak, centred on ω = 0, with height t2 and width ≃ 2π/t, as indicatedin the figure. Thus there is only a significant transition probability to those states whose energy liesin a band of width δE ≃ 2πh/t about the initial energy, Ea.

o

t2

ω

f(t, )ω

2 / tπ


Using the standard integral

∫ +∞

−∞

sin2 x

x2dx = π we have

∫ ∞

−∞

f(t, ω) dω = 2πt.

Therefore f(t, ω) reduces to an infinite “spike” of area 2πt as t→ ∞, and we have

limt→∞

f(t, ω) ∼ 2πtδ(ω).

Now consider the case of a transition not to a single final state but to a range R of final statesaround Eb. Then we have

p(a→ R) =

∫

R

p(a→ E) (E) dE, (3.21)

where (E) dE is the number of states in the interval E → E+dE. The function (E) is called thedensity of final states.

Let us assume that the range R is small enough so that we can consider (E) and Vba to be constantin R. Since

limt→∞

sin2(ωbaT/2)/(ωba/2)2 = 2π T δ(ωb − ωa) (3.22)

then, for sufficiently large T , the only significant contributions to the integral come from the energyrange corresponding to the narrow central peak of the function f(T, ωba), so we can safely extendthe limits on the integral to infinity without affecting the result. This gives

p(a→ R) =1

h2

∞∫

−∞

(E)sin2(ωbaT/2)

(ωba/2)2 |Vba|2 dE =

|Vba|2 (Eb)

h2

∞∫

−∞

sin2(ωbaT/2)

(ωba/2)2 h dωba

=2π

h|Vba|2 (Eb)T

(3.23)

The transition rate is then

R =2π

h|Vba|2 (Eb) (3.24)

This result is known as Fermi’s Golden Rule.

3.4. Feynman Perturbation Theory

Classical mechanics: The perturbation theory considered so far is intrinsically quantum mechan-ical: the unperturbed amplitude is a Green function for the Schrodinger equation, which describesquantum propagation. We can also do perturbation theory in classical mechanics.

As an example, consider a forced anharmonic oscillator, with action S = S0[x, J ] + S1[x], andLagrangian

L = L0 + L1 =m

2x2 − m

2ω2x2 + Jx

︸︷︷︸

forced SHO

− λ

4x4

︸︷︷︸

anharmonic

(3.41)

where J = J(t) is an arbitrary function of t. The classical equation of motion is:

(∂2

∂t2+ ω2

)

x =J

m− λ

mx3 (3.42)

The homogeneous equation(∂2

∂t2+ ω2

)

x = 0

is that of a simple harmonic oscillator. Let the solution of this equation be x0(t), with boundaryconditions x0(ta) = xa and x0(tb) = xb.

To solve the full inhomogeneous equation, we construct a Green function ∆(t, t′) such that

(∂2

∂t2+ ω2

)

∆(t, t′) = −δ(t− t′)


which satisfies the boundary conditions ∆(tb, t′) = ∆(ta, t

′) = 0. It is straightforward (tutorial) toshow that

∆(t, t′) =1

ω sinωT

[

θ(t− t′) sinω(tb − t) sinω(t′ − ta)

+ θ(t′ − t) sinω(t− ta) sinω(tb − t′)

] (3.44)

where T = tb − ta as usual.

Note that this Green function is not causal, because of the boundary conditions. We have bothretarded (t′ < t) and advanced (t′ > t) pieces. We call ∆(t, t′) the Feynman Green function or theFeynman propagator. Note also that ∆(t, t′) = ∆(t′, t).

The full solution of (3.42) with the correct boundary conditions is then

x(t) = x0(t) +1

m

tb∫

ta

dt′ ∆(t, t′)(−J(t′) + λ x(t′)3

)(3.51)

The second term on the RHS is the solution of the full inhomogeneous equation with trivial (zero)boundary conditions (check this explicitly), whilst the first term “fixes up” the boundary conditions.

More precisely, equation (3.51) is an integral equation for x(t) which we can solve iterately:

x(t) = x0(t)−1

m

tb∫

ta

dt′ ∆(t, t′)J(t′) +λ

m

tb∫

ta

dt′ ∆(t, t′)

x0(t′)− 1

m

tb∫

ta

dt′′ ∆(t′, t′′)J(t′′)

3

+ · · ·

(3.52)We may simplify the problem (and hence this expression) by choosing xa = xb = 0, so x0(t) = 0.

Iterating further, we may represent (3.52) diagramatically:

x(t) = ×Jt + t t′

×

×

×

︸︷︷︸

O(λ)

+ t t′

t′′

×

×

×

×

×

︸︷︷︸

O(λ2)

+t t′

t′′t′′′

×

× ×××

×

× + t t′

t′′

t′′

×

×

×

×

×

×

×

×

︸︷︷︸

O(λ3)

+ · · ·

In these diagrams, each line corresponds to (1/m)∆(t, t′) (with appropriate times for t and t′), eachcross to −

∫dt J(t), each vertex to λ

∫dt, and the diagrams at O(λ2) (and higher) include some

combinatorial factors from the expansion of (3.51) that must be evaluated explicitly.

Notes:

(1) ∆(t, t′) propagates us from t′ to t.

(2) J(t) acts as a “source” (and a “sink”) for motion.

(3) The (λ/4)x4 anharmonic term looks like a 4-point “interaction”.

(4) We integrate over all intermediate times – time ordering is taken care of by the retarded andadvanced pieces of ∆(t, t′).

(5) Finally, we sum over all possible (tree) graphs.

This looks rather like the quantum perturbation theory we studied in the preceding sections. . .


Quantum mechanics: An interesting question is whether we can find a similar expansion for thecorresponding quantum problem, e.g. in the evaluation of transition elements.

For the forced anharmonic oscillator, we again define S[x, J ] = S0[x, J ] + S1[x], where

S0[x, J ] =

tb∫

ta

dt(m

2(x2 − ω2x2) + Jx

)

and S1[x] = −λ4

tb∫

ta

dτ x4, (3.53)

Consider the transition element

〈xb, tb|T (x(t1) · · · x(tm)) |xa, ta〉J =

xb∫

xa

Dx x(t1) · · ·x(tm) exp

ihS[x, J ]

(3.54)

where the subscript J on the LHS indicates that the transition element is evaluated in the presenceof the force or “source” J(t).

Functional differentiation: For a set of independent real variables Ji, xj ∈ R | i, j = 0 . . .N,we have the usual results for partial derivatives

∂

∂JiJj = δij and

∂

∂Ji

N∑

j=0

Jj xj = xi (3.55)

We may generalise the second result to define differentiation with respect to a function J(t) of a

continuous variable t. Define the functional derivative of the functional X [J ] =tb∫

ta

J(t)x(t) dt with

respect to the value of the function J(t) at the point t′ to be

δX [J ]

δJ(t′)≡ δ

δJ(t′)

tb∫

ta

J(t)x(t) dt = x(t′) when ta < t′ < tb, and zero otherwise. (3.56)

For this to hold for arbitrary functions x(t), we must haveδ

δJ(t′)J(t) = δ(t− t′) .

The usual rules of differentiation apply, for example

hi δ

δJ(t′)exp

ih

tb∫

ta

J(t)x(t) dt

= x(t′) exp

ih

tb∫

ta

J(t)x(t) dt

(3.57)

when ta < t′ < tb (which will always be the case in what follows), and zero otherwise.

Now use (3.53) and (3.57) to rewrite the transition element (3.54) in terms of functional derivatives

〈xb, tb|T (x(t1) · · · x(tm)) |xa, ta〉J =

xb∫

xa

Dx x(t1) · · ·x(tm) expih

S[x, 0] +

tb∫

ta

J(t)x(t) dt

=

xb∫

xa

Dx hi δ

δJ(t1)· · · hi δ

δJ(tm)exp

ih

S[x, 0] +

tb∫

ta

J(t)x(t) dt

= hmδmiδJ(t1) · · · iδJ(tm)

〈xb, tb xa, ta〉J(3.58)

In the last line, we took the functional derivatives outside the path integral to obtain an expressionfor the m-point transition element (with J 6= 0) in terms of functional derivatives of the exacttransition amplitude, also with J 6= 0.

If we set J = 0 at the end of the calculation, i.e. after performing all the functional derivatives, weget the transition element in the absence of the source J(t)

〈xb, tb|T (x(t1) · · · x(tm)) |xa, ta〉J=0 = hm[

δmiδJ(t1) · · · iδJ(tm)〈xb, tb xa, ta〉J

]

J=0

(3.59)


Perturbation theory: To obtain a perturbation expansion, we expand expiS1/h as a powerseries in λ, just as we did for ordinary time-dependent perturbation theory in (3.4):

〈xb, tb xa, ta〉J =

xb∫

xa

Dx exp

ih(S0[x, J ] + S1[x])

=∞∑

n=0

1

n!

(

− iλ4h

)ntb∫

ta

dτ1 · · ·tb∫

ta

dτn

xb∫

xa

Dx x(τ1)4 · · ·x(τn)4 expih

S0[x, 0] +

tb∫

ta

J(t)x(t) dt

(3.60)We can replace all the factors of x(τi)

4 by functional derivatives with respect to J(τi) using

x(τi)4 exp

ih

tb∫

ta

J(t)x(t) dt

=

(hi δ

δJ(τi)

)4

exp

ih

tb∫

ta

J(t)x(t) dt

(3.61)

Taking the functional derivatives outside the path integral and resumming the series gives

〈xb, tb xa, ta〉J = exp

− iλ4h

tb∫

ta

dτ

(hi δ

δJ(τ)

)4

xb∫

xa

Dx exp

ihS0[x, J ]

(3.62)

The remaining path integral in (3.62) is just the forced harmonic oscillator. From (2.40) we havexb∫

xa

Dx exp

ihS0[x, J ]

= Fω(T ) exp

ihS0[x, J ]

(2.40a)

where S0[x, J ] is the classical action for the forced harmonic oscillator, which satisfies boundaryconditions x(ta) = xa and x(tb) = xb, with T = tb − ta.

Substituting (2.40a) and (3.62) into (3.58) gives us an expression for all transition elements to all

orders in perturbation theory in terms of functional derivatives of the transition amplitude for theforced harmonic oscillator!

〈xb, tb|T (x(t1) · · · x(tm)) |xa, ta〉J

= hmδmiδJ(t1) · · · iδJ(tm)

exp

− iλ4h

tb∫

ta

dτ

(hi δ

δJ(τ)

)4

Fω(T ) exp

ihS0[x, J ]

(3.63)We evaluated S0[x, J ] for arbitrary xa and xb explicitly in tutorial-problem (2.3) (or Feynman andHibbs Problem 3-11), but things are much simpler if we choose xa = xb = 0, as we did for classicalmechanics. Let x(t) satisfy

¨x+ ω2x =J

mwith x(ta) = x(tb) = 0 . (3.64)

From our discussion of classical mechanics (3.51), this has solution

x(t) = − 1

m

tb∫

ta

dt′ ∆(t, t′)J(t′) (3.65)

Integrating the classical action S0[x, J ] by parts, and using (3.64) and (3.65), we obtain (exercise)

S0[x, J ] =

tb∫

ta

dt(m

2( ˙x

2 − ω2x2) + Jx)

=1

2

tb∫

ta

dt J(t) x(t) = − 1

2m

tb∫

ta

dt

tb∫

ta

dt′ J(t)∆(t, t′)J(t′)

(3.66)Substituting (3.66) into (3.63) gives an explicit expression for the (quantum) transition elements(with xa = xb = 0) to all orders in perturbation theory in terms of the classical Green function∆(t, t′)

1

Fω(T )〈0, tb|T (x(t1) · · · x(tm)) |0, ta〉J

=hm δmiδJ(t1) · · · iδJ(tm)

exp

− iλ4h

tb∫

ta

dτ

(h δi δJ(τ))4

exp

−i2mh

tb∫

ta

dt

tb∫

ta

dt′J(t)∆(t, t′)J(t′)

(3.67)


Harmonic oscillator: As a first example, let us evaluate the two-point correlation function forλ = 0 and set the source to zero at the end.

1

Fω(T )〈0, tb|T (x(t1) x(t2)) |0, ta〉J=0

=

h2 δ2iδJ(t1) iδJ(t2) exp

− i2mh

tb∫

ta

dt

tb∫

ta


J=0

=

h2 δ2iδJ(t1) iδJ(t2) 1− i

2mh

tb∫

ta

dt

tb∫

ta

dt′J(t)∆(t, t′)J(t′) +O(J4)

J=0

=ih2m

δ

δJ(t1)2

tb∫

ta

dt J(t)∆(t, t2) =ihm

∆(t1, t2)

t2t1

=

(3.68)

where we have again introduced an obvious diagrammatic notation. Note that we don’t need toevaluate the O(J4) term in the expansion of the exponential because its contribution vanishes whenwe set J = 0 at the end. And since ∆(t, t′) = ∆(t′, t), we also have

tb∫

ta

dt′ ∆(t2, t′)J(t′) =

tb∫

ta

dt J(t)∆(t, t2) (3.68)

Similarly, for the four-point function (exercise)

1

Fω(T )〈0, tb|T (x(t1) x(t2) x(t3) x(t4)) |0, ta〉J=0

=

h4 δ4iδJ(t1) · · · iδJ(t4) exp

− i2mh

tb∫

ta

dt

tb∫

ta


J=0

=h4 δ4iδJ(t1) · · · iδJ(t4) 1

2!

− i2mh

tb∫

ta

dt

tb∫

ta


2

=

( ihm

)2 [

∆(t1, t2)∆(t3, t4) + ∆(t1, t3)∆(t2, t4) + ∆(t1, t4)∆(t2, t3)

]

t3 t4

= +

t4t3

t1 t2

t3 t4

t1 t2t2t1

+

(3.70)

Note that we only need the contributions from the O(J4) term in the third line: the O(J2) termgives zero when we perform the four derivatives, and all terms O(Jn) with n > 4 vanish when weset J = 0 at the end.

Note that the final result is symmetric with respect to interchange of any pair of t1, · · · , t4. Ingeneral, for λ = 0

1

Fω(T )〈0, tb| x(t1) · · · x(tm) |0, ta〉J=0 =

( ihm

)m/2 ∑

all pairings

∆(ti1 , ti2)∆(ti3 , ti4) · · ·∆(tim−1, tim)

(3.71)when m is even, and zero when m is odd. The sum is over all possible pairings of times t1 to tm .This result is known as Wick’s Theorem.


Anharmonic oscillator: Now let’s evaluate the four-point function in perturbation theory tofirst-order in λ for J = 0

1

Fω(T )〈0, tb|T (x(t1) x(t2) x(t3) x(t4)) |0, ta〉J=0

=

h4 δ4iδJ(t1) · · · iδJ(t4) exp− iλ

4h

tb∫

ta

dτ

(h δiδJ(τ))4

exp

−i2mh

tb∫

ta

dt

tb∫

ta


J=0(3.72)

The O(λ) contribution may be obtained by expanding the first exponential to first order, giving atotal of eight functional derivatives. So we need only the O(J8) term from the second exponential

h4 δ4iδJ(t1) · · · iδJ(t4) − iλ4h

tb∫

ta

dτ

(h δiδJ(τ))4

1

4!

− i2mh

tb∫

ta

dt

tb∫

ta


4

(3.73)

Clearly, this will result in many different terms. Let us first perform the derivatives with respectto J(τ), keeping only those terms in which each derivative acts on a different

∫∫J∆J term in

∫∫J∆J

4, which gives

h4 δ4iδJ(t1) · · · iδJ(t4) (− iλ4h

) (hi )4

8 · 6 · 4 · 24!

tb∫

ta

dτ

− i2mh

tb∫

ta

dt′ ∆(τ, t′)J(t′)

4

= −i3!λh

tb∫

ta

dt

4∏

i=1

( ihm

∆(t, ti)

)

= t

t2

t4t3

t1

(3.74)

where we changed the dummy integration variable from τ to t in the last expression. All the otherterms give rise to disconnected diagrams, which we ignore - see below.

To keep track of the form of each term, diagrams are very useful. The Feynman Rules for drawingFeynman diagrams (or graphs) are collected in table 3.1. Note that in the diagrams the binomialcoefficients are omitted.

vertices × t −i 3!λh tb∫

ta

dt

propagator t t′ ihm∆(t, t′)

external field t × ih

tb∫

ta

dtJ(t)

Table 3.1: Feynman rules

hω

a†

a

Figure 3.2: Creation andannihilation operators

Interpretation: Remember, that for a harmonic oscillator we maywrite (in the Heisenberg picture)

x(t) =

√

h

2mω

(a†(t) + a(t)

)

=

√

h

2mω

(a†(0) e−iωt + a(0) eiωt

)(3.83)


Think of the insertions x(t) as creating or destroying a quantum at time t. a†(t) creates a modewith energy E > 0 at time t, a(t) destroys it. The transition element in (3.74) describes how twoquanta interact and scatter off each other due to the anharmonic perturbation, which gives us thebasis of a multi-particle quantum theory.

For coupled oscillators, the anharmonic term describes phonon-phonon scattering in simple modelsof crystals.

This brings us to the end of the examinable material for the non-relativistic part of the course.What follows should be useful for Modern Quantum Field Theory.

End of examinable material on non-relativistic quantum theory

Loops and legs: We can represent the perturbation expansion of equation (3.67) for 〈0, tb 0, ta〉Jby setting the index m = 0 and using Feynman diagrams. For J = 0, we find, through O(λ2) (notproven here)

〈0, tb 0, ta〉0 = Fω(T )

(

1 + · · ·)

(3.75)

+ ++ +

Curved lines in the diagrams above represent propagators ih/m∆(t, t′) as before. Curved lineswhich begin and end at the same point represent propagators ih/m∆(t, t), i.e with t = t′.

If J(t) 6= 0, we must perform the derivatives without expanding the exponential of the∫∫J∆J term.

We find, for example,

〈0, tb 0, ta〉J = Fω(T )eiS0[x,J]/h

(

1 + · · ·)

(3.76)+ + +

If instead we consider a ratio, the factors Fω(T ) eiS0[x,J]/h cancel not proven here – see MQFT). For

example, expanding up to order O(λ2), we get

〈0, tb| x(t) |0, ta〉J〈0, tb 0, ta〉J

=

classical × +

×

×

×

+ ×

×

×

×

×

+ · · ·

O(h) +

×

+ ×

×

×

+

×

×

×

+

×

×

×

+ · · ·

O(h2) + × +

×

+ × + · · ·

Notes:

(1) We have only drawn the “connected” diagrams because all the “disconnected” ones cancel betweennumerator and denominator. The proof, however, needs more machinery.

(2) The new graphs which we didn’t encounter in the diagrammatic version of (3.52) for the classicalsolution x(t) contain propagator “loops”, and these give the quantum corrections to the classicalresult. To show this, we need to count the powers of h. Each propagator brings an h, each source


brings a h−1, while each vertex gives us a factor of h−1. Hence, for a graph of order hn, the numberP of propagators, the number V of vertices, and the number S of sources are related by

n = P − V − S. (3.77)

But, from the diagrams above, this is also equal to the number of loops (the proof is by induction).Thus in the classical limit all loop graphs are suppressed as h → 0 and the tree graphs sum to theclassical solution as in (3.52).

(3) To calculate scattering amplitudes, we need to take ta → −∞ and tb → ∞. We also require to pickout definite “in” and “out” states in this limit, but ∆(t, t′) oscillates wildly for large times. Thetrick is to take ω′2 ≡ ω2 − iε, with infinitesimal (but non-zero) ε > 0, in the action, so

S′ = S + iε ∞∫

−∞

x2 dt ⇒ eiS′/h = eiS/h−ε/h∫

x2 dt. (3.78)

So states with x 6= 0 as tb,a → ±∞ are exponentially suppressed. We have

ω′ =√

ω2 − iε = ω

(

1− iεω2

) 12

≃ ω

(

1− iε2ω2

)

=

(

ω − iε2ω

)

= (ω − iε′). (3.79)

where ε′ = ε/(2ω) > 0.

In the limit tb,a → ±∞ the first term in ∆(t, t′) in (3.44) becomes

limtb→∞

limta→−∞

sinω′(tb − t) sinω′(t′ − ta)

sinω′T= lim

tb→∞lim

ta→−∞

12ieiω′(tb−t) 1

2ieiω′(t′−ta)

12ieiω′T

=1

2ie−iω′(t−t′)

(3.80)

This, together with a similar analysis for the second term in (3.44), gives

limtb→∞

limta→−∞

∆(t, t′) =1

2iω′

(e−iω′(t−t′)θ(t− t′) + eiω′(t−t′)θ(t′ − t)

)≡ ∆F (t− t′) (3.81)

which is the “standard” Feynman propagator. We can now drop the prime on ω′ (until we need itagain.) Adding a small imaginary term to the action is called the i-epsilon prescription.

Since “in” and “out” states with x(t) 6= 0 are suppressed, the in and out states are just groundor “vacuum” states of the SHO. (This argument can be refined.) To calculate the “scatteringamplitude”, we have to find the vacuum expectation value with J = 0. For m = 4, we find

〈0|T (x(t1) · · · x(t4)) |0〉〈0 0〉 = + + + · · · (3.82)

The first “tree level” diagram is O(λ) whilst the next two (and all permutations thereof) are O(λ2)and carry one more power of h associated with the loop.

hω

a†

a

Figure 3.2: Creation andannihilation operators

Interpretation: Remember, that for a harmonic oscillator we maywrite (in the Heisenberg picture)

x(t) =

√

h

2mω

(a†(t) + a(t)

)

=

√

h

2mω

(a†(0) e−iωt + a(0) eiωt

)(3.83)

Think of the insertions x(t) as creating or destroying a quantum out of the vacuum at time t. a†(t)creates a mode with energy E > 0 at time t, a(t) destroys it. The scattering amplitude in (3.82)


describes how two quanta interact and scatter off each other due to the anharmonic perturbation,which gives us the basis of a multi-particle quantum theory.

For coupled oscillators, the anharmonic term describes phonon-phonon scattering in simple modelsof crystals.

Note that the Feynman propagator ∆F (t − t′) takes modes with E > 0 forwards in time, t > t′:“retarded”; whilst it takes modes with E < 0 backwards in time, t < t′: “advanced”. The E < 0modes are like E > 0 modes moving backwards in time - but everything adds up to give a causalanswer!

Quantum field theory is constructed by replacing the position operators x(t) by scalar field operators

φ(x, t). These create and destroy field quanta, which we now interpret as particles, and the Feynmanpropagator ∆F (t − t′) gets replaced by its four-dimensional equivalent. Then (3.82) gives us theamplitude for scattering of pairs of Higgs bosons! (See MQFT or Standard Model courses.)

quantumtheory - university of edinburghbjp/qt/qt.pdf · and feynman, namely the the path ... these...

Documents