quantum chemistry 1 (chem 565) fall...

Quantum Chemistry 1 (CHEM 565)Fall 2018

Gerald KniziaDepartment of ChemistryThe Pennsylvania State University

Quantum Chemistry 1 (CHEM 565), Fall 2018

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Grand Topic #3

3. Simple Model Problems

Free particles & Wave packets (Plane wave solutions; Superposition principle; Gaussian

wave packets)

Numerical solution of 1-particle quantum problems (Dimensionless EOM; Real-space

grid discretization; Time evolution; Quantum phenomena (eigenstates, tunneling, Husimi-

repr.); Pseudo-spectral methods; Basis function decomposition/1)

Particle in a box & co (1D Case: Steps, barriers, wells; Transmission/reflection,

resonance, tunneling; 3D Case: Separation of variables)

Vibrations & Harmonic oscillator (H.-o. (1D,3D), Polyatomic case, Non-harmonic

oscillator (1D) & QM approximation methods (subspace projections, perturbation expansion,

variational method)), outlook: field quantization

Spherical potentials & Angular momenta (Separation of diatomic nuclear Schrödinger

equation; Rotational spectra; Ro-vibrational spectra )

Spin & Two-level systems (Single spin ½; Coupled spins; Combining orbital & spin-

degrees of freedom)


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3.2 Particles & boxes

Simple Model Problems—Particle in a box & co.

We will consider some piece-wise constant potentials („step potentials“). Why?

While not terribly realistic, these are analytically comparatively simple tohandle,

...and already show a variety of important quantum phenomena(in particular, transmission, reflection, and tunneling of wave packets)

Additionally, their eigenstates (plane waves and exponentials) are highlyrelevant as basis functions for expanding approximate wave functions inrealistic model systems (e.g., uniform electron gas, and structures with periodic

boundary conditions (polymer/slab/solids))

...and this is where the quantum version of the ideal gas lives( ensemble states ).↘ ρ̂


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Piecewise constant potentials—General solution approach for eigenstates

Divide real-space into regions of constant potential

Constant potential ansatz or , individually for eachsub-region

Stick sub-potential-solutions together by demanding that values of wavefunctions and derivatives of wave functions agree at boundaries.

(...unless boundary goes to a region, in which case at the boundary and no

condition on the derivative arises)

⇒ ψ( ) = Cx⃗ eı xki ψ( ) = Cx⃗ e xρi

i

V (x) = ∞ ψ(x) = 0


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Consider the following potential:

V (x) =⎧⎩⎨⎪⎪

∞ for x ≤ 00 for x ∈ (0,L)∞ for x ≥ L


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Particle in a box—Standard boundary conditions

Time-independent Schrödinger equation for 1 particle in positionrepresentation (where ):

Particle cannot be in (non-zero measure) region with infinite potential

The wave function needs to be continuous (earlier discussion of particle flux!). Inner function with ( ) has boundary conditions and

.

= −ıℏP̂ ∂x

(− + V (x))ψ(x) = Eψ(x)ℏ2

2md2

dx2

⇒ ψ(x)ψ(x)

==

0 for x ≤ 00 for x ≥ L

⇒ ψ(x) x ∈ [0,L] ψ(0) = 0ψ(L) = 0


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Schrödinger equation for inner part of wave function:

with linear homogenous ordinary differential equation (ODE) with

constant coefficients

ψ(x) = Eψ(x)−ℏ2

2md2

dx2

ψ(x) + ψ(x) = 0d2

dx2

2mE

ℏ2

ψ“(x) + q ⋅ ψ(x) = 0

q = 2mE

ℏ2 →

ψ“(x) + qψ(x) = 0 (with q = )2mE

ℏ2


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Ansatz: . Inserting this into the differential equation:

General solution: with

Almost identical to previously discussed free particle case, but...

ψ(x) = eıkx

ψ“(x) + qψ(x)(ık + q)2eıkx eıxk

((ık + q))2 eıkx

k2

⇒ k

====

=

000 (note: exp(x) is never 0)q

± = ±q√2mE

ℏ

− −−−−√⇒ ψ(x) = +C1e

ıkx C2e−ıkx k = 2mE

ℏ

− −−−√


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This provides an infinity of different solutions ( and are free). However,not all of them fit to the boundary conditions and .

To see which ones do, it is helpful to reformulate the exponentials intotrigonometric functions with

E C1

ψ(0) = 0 ψ(L) = 0

exp(ıkx) = cos(kx) + ı sin(kx) (Euler formula).


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Inserting this into the general solution for :

with , two new arbitrary constants (instead of and ). This used:

x ∈ [0,L]

ψ(x) =====:

+C1eıkx C2e

−ıkx

(cos(kx) + ı sin(kx)) + (cos(−kx) + ı sin(−kx))C1 C2

(cos(kx) + ı sin(kx)) + (cos(kx) − ı sin(kx))C1 C2

( + ) cos(kx) + ı( − ) sin(kx)C1 C2 C1 C2

A cos(kx) + B sin(kx)

A B C1 C2

cos(x) = cos(−x) and sin(−x) = − sin(x).


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Since , we can instantly see that :

Thus, only functions of the form can potentially becompatible with .

ψ(x) = A cos(kx) + B sin(kx)

ψ(0) = 0 A = 0

0 = ψ(0) = A + B = Acos(0) =1

sin(0) =0

ψ(x) = B sin(kx)ψ(0) = 0


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For to hold, for must be located at one of the zeros of theSinus function, which lie at ( ). Therefore:

produces (not normalizable). Furthermore, solutionsfor and are identical. So only need be considered.

ψ(L) = 0 kx x = L

n ⋅ π n ∈ Z

sin(kL) = 0⇒ kL = πn

⇒ L = nπ2mE

ℏ2

− −−−−√⇒ E = = (n ∈ Z)

n2π2ℏ2

2mL2

n2h2

8mL2

n = 0 ψ(x) = B sin(0) = 0n −n n ≥ 1


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The free constant in can be fixed by demanding that(i.e., that the wave function is normalized).

The Schrödinger equation itself has an infinite, continuous range ofsolutions for any energy.

However, of those only a discrete subset is compatible with theboundary conditions of wave function continuity[*] and normalizability.

([*]: At points where is finite, the wave function also must be differentiable)

E = = (n = 1, 2, 3, …)n2π2ℏ2

2mL2

n2h2

8mL2

B ψ(x) = B sin(x)∫ |ψ(x) dx = 1|2

x V (x)


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Comment 2: This is a general feature of bound states: Since

is a 2nd order ODE, once you know and at any point , you cancompute for all other , for any constant choice of

( initial value problem, Picard–Lindelöf theorem).

Quantization of the energy thus comes from the conditions on thephysicality of the so-computed wave functions (i.e., boundary conditions and

normalizability).

It is not a consequence of the Schrödinger equation.

(− + V (x) − E)ψ(x) = 0ℏ2

2md2

dx2

ψ(x) (x)ψ′ x

ψ(x) x E

→


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Note that the lowest possible energy (the ground state energy) is not zero

Comment 3: Absolute energies are not observable, but differences are.

In many cases the form of the potential changes during a process, andthen differences in zero-point energy between the initial and final states canmatter.

(e.g., consider the vibrational ground states for a chemical reaction for

diatomics. The ground state energies of the inter-nuclear potentials , ,

, are not observable, but they add a generally non-zero contribution to the

reaction energy. This is experimentally well confirmed.)

E = = (n = 1, 2, 3, …)n2π2ℏ2

2mL2

n2h2

8mL2

V (x)

AB + CD → AC + BD

( )VAB RAB ( )VCD RCD

( )VAC RAC ( )VBD RBD


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Comment 4: Note that the wave function lives entirely in a region ofvanishing potential (i.e., at all points where ). That meansthat all of the (non-zero) energy of the states is kinetic energy.

Furthermore, the kinetic energy scales as

That is, it becomes larger the more a particle is confined.

E = = (n = 1, 2, 3, …)n2π2ℏ2

2mL2

n2h2

8mL2

ψ(x)V (x) = 0 |ψ(x) ≠ 0|2

∝ .Ekin1L2


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Comment 4: (cont'd) (...kinetic energy becomes larger the more a particle is confined)

This, too, is a general feature of quantum systems. It can be considered asa consequence of the Heisenberg uncertainty relation : reducingspace (reducing ) leads to an increase in momentum fluctuations ( ), and

thus kinetic energy ( ).

Lowering kinetic energy of electrons by spreading them over more space isa central mechanism in the formation of chemical bonds.

E = = (n = 1, 2, 3, …)n2π2ℏ2

2mL2

n2h2

8mL2

δx ⋅ δp ≥ ℏ/2δx δp

∝⟨ ⟩P 2

2m


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General step potentials

Schrödinger equation in region with constant potential:

Case 1, (i.e., classically allowed region):

Introduce positive constant such that

General solution to Schrödinger equation in this region:

where , are complex constants (fixed by boundary conditions)

( + V)ψ(x) = Eψ(x) ⇔ ψ(x) + ψ(x) = 0−ℏ2

2md2

dx2

d2

dx2

2m(E − V )

ℏ2

E > V

k

=k2 2m(E − V )

ℏ2

ψ(x) = +C1eıkx C2e

−ıkx

C1 C2


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Schrödinger equation in region with constant potential:

Case 2, (i.e., classically forbidden region):

Introduce positive constant such that

General solution to Schrödinger equation in this region:

where , are complex constants (fixed by boundary conditions)

( + V)ψ(x) = Eψ(x) ⇔ ψ(x) + ψ(x) = 0−ℏ2

2md2

dx2

d2

dx2

2m(E − V )

ℏ2

E < V

ρ

=ρ2 2m(V − E)

ℏ2

ψ(x) = +C1eρx C2e

−ρx

C1 C2


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In general: Write wave function like above in specific regions, and thenstitch them together using the boundary conditions at the region boundaries.

Let us consider some specific cases:

Potential Steps (reflection)

Potential Barriers (transmission and tunneling)

Potential Wells


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3.2 Particles & boxes: Potential Steps

Case 1: (partial reflection). Then:E > V0

k1

k2

=

=

2mE

ℏ2

− −−−−√2m(E − )V0

ℏ2

− −−−−−−−−−√(x) = +ψ1 A1e

ı xk1 A~

1e−ı xk1

(x) = +ψ2 A2eı xk2 A

~2e

−ı xk2


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Transmission and reflection at potential steps [Case 1: . (cont'd)]

Now fix constants via boundary conditions. Since the Schrödinger equationis homogenous, we can only determine the three ratios , ,

.

We have three constants, but only two conditions: and. We shall thus only consider the case , which

amounts to an incident particle coming from .(why? we have seen in HW6 that a wave function carries a probability density flux in

direction)

E > V0

/A~

1 A1 /A2 A1

/A~

2 A1

(0) = (0)ψ1 ψ2

(0) = (0)ddxψ1

ddxψ2 = 0A

~2

x = −∞e+ı ⋅k⃗

x ⃗ +k⃗


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Transmission and reflection at potential steps [Case 1: . (cont'd)]

The conditions and then give:

: Superposition of two waves with wave vector . One is associatedwith the incident wave, the other with the reflected wave.

: Transmitted wave. (Note also the case.)

E > V0

k1

k2

=

=

2mE/ℏ2− −−−−−−√2m(E − )/V0 ℏ2− −−−−−−−−−−−√


~1e

−ı xk1

(x) =ψ2 A2eı xk2

(0) = (0)ψ1 ψ2 (0) = (0)ψ′1 ψ′

2

= =A~

1

A1

−k1 k2

+k1 k2

A2

A1

2k1

+k1 k2

ψ1 ±k1

ψ2 → 0V0


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Transmission and reflection at potential steps

One can then define transmission ( ) and reflection ( ) coefficients. (Theseare consistent with the ratio of probability flux on both sides):

Note that

Unlike for a classical particle, which just reduces its momentum whenpassing a potential step, in QM a particle has a non-zero probability ofturning back.

T R

J

R

T

=

=

= … = 1 −∣

∣∣A~

1

A1

∣

∣∣2

4k1k2

( +k1 k2)2

= … =k2

k1

∣∣∣A2

A1

∣∣∣2 4k1k2

( +k1 k2)2

T + R = 1


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Case 2: (total reflection). Then:E ≤ V0

k1

ρ2

=

=

2mE

ℏ2

− −−−−√2m( − E)V0

ℏ2

− −−−−−−−−−√(x) = +ψ1 A1e

ı xk1 A~

1e−ı xk1

(x) = +ψ2 B2exρ2 B

~2e

− xρ2


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Total reflection at potential steps [Case 2: . (cont'd)]

For the solution to be normalizable, we must have (prefactor of .Otherwise as )

The conditions and then give:

The reflection coefficients becomes:

As in classical mechanics, the particle is reflected. But it does have a non-zero, exponentially decaying probability of being found in the classicallyforbidden region at .

E ≤ V0

= 0B2 e xρ2

| (x) → ∞ψ2 |2 x → ∞

(0) = (0)ψ1 ψ2 (0) = (0)ψ′1 ψ′

2

= =A~

1

A1

− ık1 ρ2

+ ık1 ρ2

B2

A1

2k1

+ ık1 ρ2

R = = = 1∣

∣∣A~

1

A1

∣

∣∣2

∣∣∣

− ık1 ρ2

+ ık1 ρ2

∣∣∣2

x > 0


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Note: There is only one total WF. It is:

k1

ρ2

=

=

2mE

ℏ2

− −−−−√2m( − E)V0

ℏ2

− −−−−−−−−−√(x) = +ψ1 A1e

ı xk1 A~

1e−ı xk1

(x) = +ψ2 B2exρ2 B

~2e

− xρ2

ψ(x) = { (x) for x < 0ψ1

(x) for x ≥ 0ψ2


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3.2 Particles & boxes: Potential Barriers

Case 1: (resonances). Then:E > V0

k1

k2

k3

=

=

=

2mE/ℏ2− −−−−−−√2m(E − )/V0 ℏ2− −−−−−−−−−−−√

k1


~1e

−ı xk1


~2e

−ı xk2


~3e

−ı xk1


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Resonance trapping in potential barriers

Let us choose, as before, (i.e., particle incident from )

The matching conditions at :

then give and in terms of , and the matching conditions at :

give and in terms of and (and thus ).

= 0A~

3 x = −∞

x = l

(l) = (l) (l) = (l)ψ2 ψ3d

dxψ2

ddx

ψ3

A2 A~

2 A3 x = 0

(0) = (0) (0) = (0)ψ1 ψ2d

dxψ1

ddx

ψ2

A1 A~

1 A2 A~

2 A3


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(After some calculation) we find:

and...

A1

A~

1

=

=

(cos( l) − ı sin( l))k2+k2

1 k22

2k1k2k2 eı lk1 A3

ı sin( l)−k2

2 k21

2k1k2k2 eı lk1 A3


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With and we can compute the reflection coefficient and thetransmission coefficient :

Inserting and this gives:

/A~

1 A1 /A3 A1 R

T

R

T

=

=

=∣

∣∣A~

1

A1

∣

∣∣2 ( − sin( lk2

1 k22)2 k2 )2

4 + ( − sin( lk21k

22 k2

1 k22)2 k2 )2

= = 1 − R∣

∣∣A~

3

A1

∣

∣∣2 4k2

1k22

4 + ( − sin( lk21k

22 k2

1 k22)2 k2 )2

k1 k2

T =4E(E − )V0

4E(E − ) + sinV0 V 20 (l )2m(E − )/V0 ℏ2− −−−−−−−−−−−√ 2


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Transmission coefficient varies periodically with barrier length :

Maximum: whenever ( ) (i.e., whenever is aninteger multiple of the half wavelength )

Minimum whenever

T =4E(E − )V0

4E(E − ) + sinV0 V 20 (l )2m(E − )/V0 ℏ2− −−−−−−−−−−−√ 2

l

T = 1 l = nπk2 n ∈ Z l

π/k2

T = (1 + )V 20

4E(E− )V0

−1

l = (n + )πk212


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What is happening here? At resonance, the partial waves reflected at and are in constructive interference. Standing waves in region II!

Wave packet analysis would show that near resonance, wave packetsspends a long time in region II (resonance scattering).

x = 0x = l ⇒


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Case 1: (tunneling). Then:E ≤ V0

k1

ρ2

k3

=

=

=

2mE/ℏ2− −−−−−−√−2m(E − )/V0 ℏ2− −−−−−−−−−−−−−√

k1


~1e

−ı xk1

(x) = +ψ2 A2exρ2 A

~2e

− xρ2


~3e

−ı xk1


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Tunneling through potential barriers

(same calc as before... (or replace by )):

If , then , and we get:

The transmission coefficient is not zero, but decays exponentially with .

k2 −iρ2

T = =∣

∣∣A~

3

A1

∣

∣∣2

4E( − E)V0

4E(E − ) + sinhV0 V 20 (l )2m( − E)/V0 ℏ2− −−−−−−−−−−−√ 2

z = l ≫ 1ρ2 sinh(z) = ( − ) ≈12 ez e−z 1

2 ez

T ≈16E( − E)V0

V 20

e−2 lρ2

l


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What does that mean in practice? E.g., consider an electron. With theelectron mass, we get as decay length :

[ —the electron volt, is a microscopic unit of energy; it is the energygained by an electron when accelerated over a potential difference of .Note: . The unit , the Ångström, is exactly

(or ]

If an electron with then hits an energy barrier with height and length , we get a (large!) transmission probability of

1/ρ2

= ≈1ρ2

ℏ

2m( − E)V0− −−−−−−−−−√

1.96Å

( − E)/eVV0− −−−−−−−−−√

eV1V

1eV ≈ 1.602 ⋅ J ≈ 96.49kJ/mol10−19 Åm10−10 0.1nm

E = 1eV = 2eVV0

1Å

T ≈ 0.78 = 78%.


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For a proton, with a times larger mass than an electron, the decaylength decreases to

If a proton with then hits the same energy barrier (with height and length ), its transmission coefficient is only

That is quite a difference... the power of exponential scaling.

For macroscopic objects—with much larger masses—tunneling is thuspractically impossible.

≈ 1840

= ≈ .1ρ2

ℏ

2m( − E)V0− −−−−−−−−−√

4.6 ⋅ Å10−2

( − E)/eVV0− −−−−−−−−−√

E = 1eV= 2eVV0 1Å

T ≈ 4 ⋅ .10−19


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3.2 Particles & boxes: Potential Wells

Bound states: Let (other case: handled before w/ )− ≤ E ≤ 0V0 ≤ 0V0

ρ

k

=

=

−2mE/ℏ2− −−−−−−−√2m(E + )/V0 ℏ2− −−−−−−−−−−−√

(x) = +ψ1 A1eρx A

~1e

−ρx

(x) = +ψ2 A2eıkx A

~2e

−ıkx

(x) = +ψ3 A3eρx A

~3e

−ρx


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Bound states in potential wells

Since must be bounded in region I, we need . The matchingconditions at then give:

and the conditions at :

ψ = 0A~

1

x = −a/2

A2

A~

2

=

=

e(−ρ+ık)a/2 ρ + ık

2ıkA1

− e−(ρ+ık)a/2 ρ − ık

2ıkA1

x = +a/2

A3

A1

A~

3

A1

=

=

((ρ + ık − (ρ − ık )e−ρa

4ıkρ)2eıka )2e−ıka

sin(ka)+ρ2 k2

2kρ


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Bound states in potential wells

On the other hand, we also need , since otherwise is unboundedon the right hand side.

This leads to condition:

Since and depend on , this means that, unlike in the previous cases,only discrete values of lead to feasible (normalizable) wave functions. Thebound state energy is quantized.

(in general: bound state WFs are normalizable and quantized, unbound state WFs are not

normalizable (and thus not by themselves capable of representing physical states) and have

continuous spectra)

= 0A3 ψ(x)

= ⇔ = ±( )ρ − ık

ρ + ık

2

e2ıka ρ − ık

ρ + ıkeıka

ρ k E

E


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Summary

Important classes of quantum states (scattering states, bound states) andquantum behavior (transmission, reflection, tunneling, and resonant trapping of wave

packets) already emerge in simple piecewise constant potentials. These aresusceptible to analytical treatment.

Quantization and / coefficients arise from the Schrödinger inconjunction with boundary conditions on the wave function. (continuity of

, continuity of (unless ), normalizability (if bound state))

Treating these phenomena is helpful for making sense of quantum behaviorin complex problems

R T

ψ(x) ψ(x)∇⃗ V ( ) = ∞x⃗


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Grand Topic #3



wave packets)












degrees of freedom)


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3.2 Particles & boxes: 3D box

The three-dimensional box

Let us consider the simplest case in 3 spatial dimensions: The 3D boxpotential with infinite potential walls (box sizes: ):

(One can similarly consider 3D boxes with periodic boundary conditions HW)

a × b × c

W(x, y, z) = { 0 if x ∈ [0, a] ∧ y ∈ [0, b] ∧ z ∈ [0, c]+∞ otherwise

→


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The three-dimensional box

The wave function is zero outside of the box (since there).

Within the box, the WF is obtained from the Schrödinger equation for 1particle in position representation:

where

Note that nothing couples the space dimensions , , .

|ψ⟩( )r ⃗ a × b × c W = ∞

( − + )ψ(x, y, z) = Eψ(x, y, z)Δℏ2

2mW(x, y, z)

=0 inside

Δ = = + + (Laplace operator).∇⃗ 2 ∂2

∂x2

∂2

∂y2

∂2

∂z2

x y z


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The three-dimensional box—Separation Ansatz

Ansatz for wave function: Product of functions in individual space directions

An ansatz like this is called separation ansatz (because individual variablesare treated with separate uncoupeld functions.)

Let us compute the Laplacian for this:

ψ(x, y, z) = f(x)g(y)h(z)

Δψ

ψ(x, y, z)∂2

∂x2

ψ(x, y, z)∂2

∂y2

ψ(x, y, z)∂2

∂z2

=

=

=

f“(x)g(y)h(z)

f(x)g“(y)h(z)

f(x)g(y)h“(z)


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Inserting this into the Schrödinger equation , and diving by

we get:

Or, for example,

⇒ Δψ(x, y, z) = f“(x)g(y)h(z) + f(x)g“(y)h(z) + f(x)g(y)h“(z)

( Δ − E)ψ = 0−ℏ2

2m

ψ

− E = 0−ℏ2

2mf“(x)g(y)h(z) + f(x)g“(y)h(z) + f(x)g(y)h“(z)

f(x)g(y)h(z)

f(x)g(y)h(z)

f(x)g(y)h(z)

⇔ − − − − E = 0ℏ2

2mf“(x)

f(x)ℏ2

2mg“(y)

g(y)ℏ2

2mh“(z)

h(z)

− = + + E =:ℏ2

2mf“(x)

f(x)ℏ2

2mg“(y)

g(y)ℏ2

2mh“(z)

h(z)Ex


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Note that the lhs depends on , and rhs on and . Neverthess, theexpressions are supposed to be equal for all , , .

That can only work if both sides evaluate to constants which do not dependon at all!

The other two functions can be similarly arranged to give:

− = + + E =:ℏ2

2mf“(x)

f(x)ℏ2

2mg“(y)

g(y)ℏ2

2mh“(z)

h(z)Ex

x y z

x y z

x, y, z

− = − = − =ℏ2

2mf“(x)

f(x)Ex

ℏ2

2mg“(y)

g(y)Ey

ℏ2

2mh“(z)

h(z)Ez


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So with this separation of variables, we can decompose the one 3DSchrödinger equation into 3 one-dimensional Schrödinger equations, for the

directions separately:

If , , fulfill those equations, then the total fulfills

i.e., the 3D Schrödinger equation for

.

x, y, z

− f“(x) = f(x) − g“(x) = g(x) − h“(z) = h(z)ℏ2

2mEx

ℏ2

2mEy

ℏ2

2mEz

f g h ψ(x, y, z) = f(x)g(y)h(z)

( ( + + ) − ( + + )) f(x)g(y)h(z) = 0−ℏ2

2m∂2x ∂2

y ∂2z Ex Ey Ez

( Δ − E)ψ(x, y, z) = 0−ℏ2

2m

E = + +Ex Ey Ez


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We have already dealt with the 1D-equations. The boundary conditions areas before (e.g., ).

The wave functions and energies are:

where , , . Any combination ofthose quantum numbers yields a feasible solution for the 3D box.

f(0) = f(a) = 0

f(x)

g(y)

h(z)

=

=

=

sin( )2a

−−√ πxnx

a

sin( )2b

−−√ πyny

b

sin( )2c

−−√ πznz

c

=Ex

n2xh

2

8ma2

=Ey

n2yh

2

8mb2

=Ez

n2zh

2

8mc2

∈ {1, 2, 3, …}nx ∈ {1, 2, 3, …}ny ∈ {1, 2, 3, …}nz

, ,nx ny nz


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Re-combining the solutions in the individual directions:

Note that the wave functions are automagically normalized:

ψ(x, y, z)

E

=

=

sin( ) sin( ) sin( )8abc

− −−−√ πxnx

a

πyny

b

πznz

c

( + + )h2

8mn2x

a2

n2y

b2

n2z

c2

∭ |ψ(x, y, z) dx dy dz = |f(x) dx ⋅ |g(y) dy |h(z) dz = 1|2 ∫ a

0|2 ∫ b

0|2 ∫ c

0|2


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Degeneracy of states—Cubic box

Let us consider the case of a cube, where . Then the possibleenergy eigenvalues are:

in any combination of , , .

Unlike in the 1D case, we here can get degeneracy: Multiple non-equivalentwave functions with the same energy (e.g., and

and all have the same energy)

The number of different orthogonal states with the same energy is calleddegeneracy of an energy level.

a = b = c

E = ( + + ) = ( + + )h2

8mn2x

a2

n2y

b2

n2z

c2

h2

8ma2n2x n2

y n2z

∈ {1, 2, 3, …}nx ∈ {1, 2, 3, …}ny ∈ {1, 2, 3, …}nz

= 2, = 1, = 1nx ny nz

= 1, = 2, = 1nx ny nz = 1, = 1, = 2nx ny nz


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Degeneracy of states—Cubic box

Sometimes people distinguish between systematic degeneracy andaccidental degeneracy

Systematic degeneracy is degeneracy resulting from a symmetry of theproblem. For example, the states and

can be switched to one another by exchanging the and axes. A transformation under which the Hamiltonian is invariant.

Accidental degeneracy arises when two energy levels are degenerate buthave qualitatively different wave functions not related by simple symmetries.For example, the states and happen to have the same energy, but they qualitatively different wavefunctions not obtained from each other via symmetry transforms.

= 2, = 1, = 1nx ny nz

= 1, = 2, = 1nx ny nz x

y

= 7, = 4, = 1nx ny nz = 8, = 1, = 1nx ny nz


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Notes on the separation ansatz

Note 1: Separation ansatz for partial differential equation (PDE) only worksin case of certain potentials and boundary shapes, such that thecoordinates can be uncoupled. It is not possible to solve all PDEs like that.But in problems with high symmetry (e.g., spherical/elliptical potentials,rectangular wells, etc.) it is very useful.

Note 2: How do we know that we get all possible solutions of the differentialequations like this? And not just some solutions which happen to have theform ?

Normally we can't—some solutions cannot be written like this. But due tolinearity, all of the 3D solutions can be decomposed into a linearcombination of such „axis-aligned“ functions

ψ(x, y, z) = f(x)g(y)h(z)

⇒


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Grand Topic #3



wave packets)












degrees of freedom)


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quantum chemistry 1 (chem 565) fall...

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