This content has been downloaded from IOPscience. Please scroll down to see the full text.

Download details:

IP Address: 54.39.106.173

This content was downloaded on 27/02/2020 at 12:27

Please note that terms and conditions apply.

You may also be interested in:

Quantum Mechanics: Lecture notes: Introduction

K K Likharev

An extended Konno-Wadati Formalism and Some New Backlund Transformations

A Roy Chowdhury

Nonlinear quantum fluid equations for a finite temperature Fermi plasma

Bengt Eliasson and Padma K Shukla

Magnetoelastic anomalies due to spin fluctuations in weakly itinerant ferromagnetic systems

P Mohn, D Wagner and E P Wohlfarth

Binding energy of exciton in a nanowire superlattice in magnetic and electric fields

J E Galván-Moya, W Gutiérrez and C Moscoso

Relativistic Coulomb-distorted plane waves

Ch Hofmann, J Augustin, J Reinhardt et al.

Higher spin realization of the DS/CFT correspondence

Dionysios Anninos, Thomas Hartman and Andrew Strominger

Gain of information in a quantum measurement

R Balian

Analytical model of non-Markovian decoherence in donor-based charge quantum bits

F Lastra, S A Reyes and S Wallentowitz

IOP Publishing

Quantum MechanicsProblems with solutions

Konstantin K Likharev

Chapter 1

Introduction

Problem 1.1. The actual postulate made by N Bohr in his original 1913 paper wasnot directly Eq. (1.8) of the lecture notes, but an assumption that at quantum leapsbetween adjacent large (quasiclassical) orbits with n ≫ 1, the hydrogen atom eitheremits or absorbs energy ΔE = ℏω, where ω is its classical radiation frequency—according to classical electrodynamics, equal to the angular velocity of electron’srotation1. Prove that this postulate is indeed compatible with Eqs. (1.7) and (1.8).

Solution: Solving the classical Eq. (1.9),

vπε

=mr

er4

,e

2 2

02

together with Eq. (1.8),

v = ℏm r n,e

for the electron’s velocity v and its orbit’s radius r, we get

vπε πε

=ℏ

= ℏen

rnm

e4

1,

4.

2

0

2 2

e

2

0

Hence the angular velocity of electron’s rotation is

⎛⎝⎜

⎞⎠⎟

vωπε

= =ℏ

*r

m en41

. ( )e3

2

0

2

3

On the other hand, for n ≫ 1, the energy difference between the adjacent energylevels (Δn = ±1) may be calculated from Eqs. (1.12) and (1.13)—which also followfrom Eqs. (1.8) and (1.9):

1 See, e.g. Part EM section 8.2.

doi:10.1088/2053-2563/aaf3a6ch1 1-1 ª Konstantin K Likharev 2019

⎛⎝⎜

⎞⎠⎟πε

Δ ≡ − ≈ = =ℏ

**+E E EdEdn

En

m en41

. ( )n nn

1H3

e2

2

0

2

3

Comparing Eqs. (*) and (**) we see that ΔE = ℏω, i.e. that the actual Bohr’spostulate is indeed compatible with Eqs. (1.7) and (1.8).

Problem 1.2. Use Eq. (1.53) of the lecture notes to prove that the linear operators ofquantum mechanics are commutative: ˆ + ˆ = ˆ + ˆA A A A2 1 1 2, and associative:

ˆ + ˆ + ˆ = ˆ + ˆ + ˆA A A A A A( ) ( )1 2 3 1 2 3 .

Solution: These relations look obvious, but the reader should remember that in theoperators we face a different mathematical entity, and cannot take for granted anyproperties that have not been postulated. For example, for any two usual functionsΨ1 and Ψ2 (which are c-numbers at each point of their arguments), we may alwayswrite Ψ1Ψ2 = Ψ2Ψ1, but for operators’ ‘products’, such commutation is generallyinvalid—see, e.g. Eq. (2.14). This is why we should be careful.

First, let us use Eq. (1.53), with the index swap 1↔2, to write

ˆ + ˆ Ψ = ˆ Ψ + ˆ ΨA A A A( ) .2 1 2 1

The operands on the right-hand side of this equation are just functions (notoperators!), and obey the rules of the ‘usual’ algebra. In particular, these termsare commutative, so that the RHS is equal to that of Eq. (1.53). Hence, the left-handsides of these relations have to be equal as well:

ˆ + ˆ Ψ = ˆ + ˆ ΨA A A A( ) ( ) .2 1 1 2

Since this relation is valid for an arbitrary function Ψ, it gives the required proof thatthe operators are commutative as well.

Similarly, we may use Eq. (1.53) twice to write

ˆ + ˆ + ˆ Ψ = ˆ + ˆ Ψ + ˆ Ψ = ˆ Ψ + ˆ Ψ + ˆ ΨA A A A A A A A A[( ) ] ( ) .1 2 3 1 2 3 1 2 3

Again, the operands on the right-hand side of this equation are just (complex)functions, and may be regrouped as

ˆ Ψ + ˆ Ψ + ˆ Ψ = ˆ Ψ + ˆ Ψ + ˆ ΨA A A A A A( ).1 2 3 1 2 3

Now we may apply Eq. (1.53) twice to the right-hand side of the above relation, towrite

ˆ Ψ + ˆ Ψ + ˆ Ψ = ˆ + ˆ + ˆ ΨA A A A A A( ) [ ( )] .1 2 3 1 2 3

Comparing the initial and final expressions of our calculation, we get

ˆ + ˆ + ˆ Ψ = ˆ + ˆ + ˆ ΨA A A A A A[( ) ] [ ( )] .1 2 3 1 2 3

Since this equality is valid for any Ψ, we may conclude that the linear operators areindeed associative.

Quantum Mechanics: Problems with solutions

1-2

Problem 1.3. Prove that for any time-independent Hamiltonian operator H and twoarbitrary complex functions f(r) and g(r),

∫ ∫ˆ = ˆf Hg d r Hf g d rr r r r( ) ( ) ( ) ( ) .3 3

Solution: Using the fact (discussed in section 1.5 of the lecture notes) that the set ofeigenfunctions ψn of the given Hamiltonian operator (i.e. the set of stationary statesof the system) is full, we may expand the function g(r) and the complex conjugate ofthe function f(r) in series over the set, just as was done with the function Ψ(r, 0) inEq. (1.67):

⎡⎣⎢⎢

⎤⎦⎥⎥

∑ ∑

∑ ∑

ψ ψ

ψ ψ

= =

≡ = ≡

*

* **

* *

g g r f f r

f f f r f

r r

r r r

( ) ( ), ( ) ( ), so that

( ) [ ( )] ( ) ( ),

n n

n n

n n n n

n n n n

where fn and gn are some (generally, complex) coefficients. Plugging these expres-sions (with one of the summation indices n denoted as n′) into each side of theequality to be proved, and taking the constant coefficients out of the spatialintegrals, we may transform them as

∫ ∫

∫ ∫

∑

∑

ψ ψ

ψ ψ

ˆ = ˆ

ˆ = ˆ′

′

*′

*′

*′

*′

f Hg d r f g H d r

Hf g d r f g H d r

r r r r

r r r r

( ) ( ) ( ) ( ) ,

( ) ( ) ( ) ( ) ,

n n

n n

,

,

n n n n

n n n n

3 3

3 3

Now using Eq. (1.60) with n → n′, ψ ψˆ =′ ′ ′H En n n , in the first expression, and itscomplex conjugate, ψ ψˆ =* *H En n n ,

2 in the second one, and then employing theorthonormality condition (1.66), we get

∫ ∫

∫ ∫

∑ ∑ ∑

∑ ∑ ∑

ψ ψ δ

ψ ψ δ

ˆ = = =

ˆ = = =

′ ′

′ ′

*′ ′ *

′*

′ ′ ′ *

*′

*′

*′ ′ *

f Hg d r f g E d r f g E f g E

Hf g d r f g E d r f g E f g E

r r r r

r r r r

( ) ( ) ( ) ( ) ,

( ) ( ) ( ) ( ) ,

n n n n n

n n n n n

, ,

, ,

n n n n n n n n n n n n n

n n n n n n n n n n n n n

3 3,

3 3,

so that the left-hand and right-hand sides of the relation in question are indeedequal.

2 The eigenenergies En are real numbers, so they do not change at the complex conjugation, and neither are theHamiltonians of the type (1.41). (In chapter 4 we will see that the statement we are proving is valid for anyHermitian operator, in particular any Hamiltonian.)

Quantum Mechanics: Problems with solutions

1-3

Problem 1.4. Prove that the Schrödinger equation (1.25), with the Hamiltonianoperator given by Eq. (1.41), is Galilean form-invariant, provided that the wave-function is transformed as

⎧⎨⎩⎫⎬⎭

vΨ′ ′ ′ = Ψ − ⋅ℏ

+ℏ

t t im

im t

r rv r

( , ) ( , ) exp2

,2

where the prime sign denotes the variables measured in the reference frame 0′ thatmoves, without rotation, with a constant velocity v relatively to the ‘lab’ frame 0.Give a physical interpretation of this transformation.

Solution: The non-relativistic (‘Galilean’) transform of the coordinates and time atthe transfer between the reference frames is expressed by the following relations3:

′ = − ′ = *t t tr r v , . ( )

The Galilean form-invariance means that the wavefunctions Ψ′(r′, t′) and Ψ(r, t),related as specified in the assignment, satisfy similar Schrödinger equation in thereference frames—respectively, 0 and 0′:

ℏ ∂Ψ′∂ ′

= − ℏ ∇′ Ψ′ + ′ ′ ′ Ψ′

ℏ∂Ψ∂

= − ℏ ∇ Ψ + Ψ**

it m

U t

it m

U t

r

r

2( , ) , and

2( , ) .

( )

22

22

For the proof, let us note that the functions U′(r′, t′) and U(r, t) describe the samepotential energy of the particle, i.e. must give the same value at the same space–timepoint:

′ ′ ′ ≡ ′ − =U t U t t U tr r v r( , ) ( , ) ( , ).

(Note also that the wavefunction transform, suggested in the assignment, gives asimilar relation for the probability density to find the particle at the same space–timepoint:

′ ′ ′ ≡ Ψ′ ′ ′ = Ψ ≡w t t t w tr r r r( , ) ( , ) ( , ) ( , ),2 2

just as it should.)Next, considering t′, at fixed r′, as a function of arguments r(t) ≡ {r1(t), r2(t), r3(t)}

and t, we may use the general rule of differentiation of a function of severalvariables4, and then Eq. (*) in the form r = r′ + vt′, to write5

∑ ∇∂∂ ′

= ∂∂

+ ∂∂

∂∂ ′

≡ ∂∂

+ ⋅=t t r

r

t tv ,

j 1

3

j

j

3 See, e.g. Part EM section 9.1, in particular figure 9.1 and Eq. (9.2).4 See, e.g. Eq. (A.23).5 This expression is essentially the convective derivative, which was discussed several times in this lectureseries—see especially Part CM section 8.3.

Quantum Mechanics: Problems with solutions

1-4

while at fixed t′, Eq. (*) yields ∇′ = ∇, so that ∇′2 = ∇2. With these relations, astraightforward differentiation of the left-hand side of the transformation inquestion, plugged into the first of Eqs. (**), immediately yields the second of theseequations, i.e. proves the Galilean form-invariance of the Schrödinger equation.

For the interpretation of the wavefunction’s transform, let us apply it to thesimplest case of a monochromatic, plane de Broglie wave given by Eqs. (1.29) of thelecture notes, describing a free particle, whose momentum p = ℏk and (kinetic)energy E = ℏω are c-numbers, i.e. have definite values:

Ψ = ⋅ℏ

−ℏ{ }t a i iEt

rp r

( , ) exp .

The proved transform shows that in the moving reference frame the wavefunction isa similar plane wave:

⎧⎨⎩⎫⎬⎭

vΨ′ ′ ′ = ⋅ℏ

−ℏ

− ⋅ℏ

+ℏ

≡ ′ ⋅ ′ℏ

− ′ℏ{ }

t a i iEt

im

im t

a i iE t

rp r v r

p r

( , ) exp2

exp ,

2

where

v v′ ≡ − ′ ≡ − ′ ⋅ − = − ⋅ +m E Em

Em

p p v p v p v, and2 2

.2 2

These are exactly the Galilean transform expressions for the momentum and thekinetic energy of the particle, given by the non-relativistic classical mechanics.Indeed, expressing the particle’s momentum via its velocity u (in the lab frame) asp = mu, so that E = mu2/2, we get

v′ = − = ′ ′ = − ⋅ + = ′m m m E

mum

m mup u v u u v, and

2 2 2,

2 2 2

where u′ = u − v is the particle’s velocity as observed from the moving referenceframe.

Problem 1.5.* Prove the so-called Hellmann–Feynman theorem6:

λ λ∂∂

= ∂∂

E H,n

n

6Despite the theorem’s common name, H Hellmann (in 1937) and R Feynman (in 1939) were not the first onesin the long list of physicists who have (apparently, independently) discovered this fact. Indeed, it may be tracedback at least to a 1922 paper by W Pauli, and was carefully proved by P Güttinger in 1931.

Quantum Mechanics: Problems with solutions

1-5

where λ is some c-number parameter, on which the time-independent HamiltonianH , and hence its eigenenergies En, depend.

Solution: Multiplying both parts of the fundamental Eq. (1.60) of the lecture notesby ψn*, and integrating the result over space, we get

∫ ∫ψ ψ ψ ψˆ =* *H d r E d rr r r r( ) ( ) ( ) ( ) .n n n n n3 3

On the right-hand side of this relation, we may take the constant En out of theintegral, and then use the orthonormality condition (1.66) to get the followingexpression for the eigenenergy7:

∫ ψ ψ= ˆ **E H d rr r( ) ( ) . ( )n n n3

Let us differentiate both parts of this relation over the parameter λ, taking intoaccount that not only H and En, but also the eigenfunctions ψn may depend on it:

⎡⎣ ⎤⎦⎡⎣⎢⎢

⎤⎦⎥⎥

∫

∫λ λ

ψ ψ

ψλ

ψ ψλ

ψ ψψ

λ

∂∂

= ∂∂

ˆ

≡∂

∂ˆ + ∂ ˆ

∂+ ˆ ∂

∂

**

*

** *

EH d r

HH

H d r

r r

rr r r r

r

( ) ( )

( )( ) ( ) ( ) ( )

( ).

( )

nn n

nn n n n

n

3

3

Next, let us spell out the general equality whose proof was the task of problem 1.3,for the particular case when ψ= *f r r( ) ( )n , while g(r) = ∂ψn(r)/∂λ:

∫ ∫ψψ

λψ

ψλ

ˆ ∂∂

= ˆ ∂∂

* *H d r H d rrr

rr

( )( )

( )( )

.nn

nn3 3

Applying this equality into the last term of Eq. (**), and then using Eq. (1.60),ψ ψˆ =H En n n, in the first term, and its complex conjugate, ψ ψˆ =* *H En n n ,

8 in the lastterm, we get

⎡⎣⎢⎢

⎤⎦⎥⎥∫ ∫λ

ψλ

ψψ

λψ ψ

ψλ

∂∂

= ∂ ˆ∂

+∂

∂+

∂∂

*****

*E Hd r E d rr r

rr r

r( ) ( )

( )( ) ( )

( ). ( )n

n n nn

n nn3 3

Now let us stop for a minute, and differentiate over λ the wavefunctions’orthonormality condition (1.66), written for the particular case n′ = n, at the secondstep using the normalization condition (1.22c):

⎡⎣⎢⎢

⎤⎦⎥⎥∫ ∫λ

ψ ψψ

λψ ψ

ψλ λ

∂∂

≡∂

∂+

∂∂

= ∂∂

≡**

*d r d rr rr

r rr

( ) ( )( )

( ) ( )( )

1 0.n nn

n nn3 3

7Note that according to Eq. (1.64) of the lecture notes, Eq. (*) means that the Hamiltonian is nothing morethan the operator corresponding to a very special observable, the system’s energy—the fact which was alreadymentioned at its introduction in section 1.2.8 See the footnote to the model solution of problem 1.3.

Quantum Mechanics: Problems with solutions

1-6

But this means that the integral of the last term in Eq. (***) equals zero, and thatequality is reduced to

∫λψ

λψ

λ∂∂

= ∂ ˆ∂

≡ ∂∂

*E Hd r

Hr r( ) ( ) ,n

n nn

3

thus proving the Hellmann–Feynman theorem.

Problem 1.6.* Use Eqs. (1.73) and (1.74) of the lecture notes to analyze the effect ofphase locking of Josephson oscillations on the dc current flowing through a weaklink between two superconductors (frequently called the Josephson junction),assuming that an external microwave source applies to the junction a sinusoidalac voltage with frequency ω and amplitude A.

Solution: Let us assume that the phase locking has happened, so that our dc biaspoint is already on nth current step (1.76); then for the total voltage across thejunction we may write

ω ω= ℏ +V t ne

A t( )2

cos .

Then Eq. (1.73) yields the following differential equation for the Josephson phaseevolution,

φτ

τ τ ωω

= + ≡ ≡ℏ

dd

n a t aeA

cos , with and2

,

which may be readily integrated:

φ τ τ φ= + +a nsin ,0

where φ0 is a (so far, arbitrary) integration constant.As a result, the Josephson supercurrent (1.74) is equal to

τ τ φ

τ τ φ τ τ φ

= + +

≡ + + +

I I a n

I a n a n

sin ( sin )

[sin ( sin ) cos cos ( sin ) sin ].

c 0

c 0 0

Calculating its time-average (i.e. the dc component),

τ τ φ τ τ φ¯ = + + +I I a n a n[sin ( sin ) cos cos ( sin ) sin ],c 0 0

for example as

∫πτ τ¯ =

π

π

−

+f f d

12

( ) ,

we see that the first term in the square brackets vanishes due to the asymmetry of thefunction under the integral, while for the calculation of the second term we may use

Quantum Mechanics: Problems with solutions

1-7

the well-known integral representation of the Bessel functions of the first kind of aninteger order m:9

τ τ= −J a a m( ) cos( sin ),m

where m is an arbitrary integer. Taking m = −n, we see that the dc current on the nthstep is

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

φω

ω

¯ = ≡ ≡ℏ

≡ℏ

*− −I I I I J a I J

eA

I I JeA

sin , where ( )2

, so that

2.

( )n n n n

n n

0 c c

c

Let us assume that the external circuit fixes the dc current through the junction; thenthe phase shift φ0 may self-adjust to fit the external current only if it is in the range10

− ⩽ ¯ ⩽ +I I I .n n

Hence the full size of nth current step is twice the ∣In∣ given by Eq. (*); a look at theplot of the Bessel functions11 shows that it oscillates as a function of the ac voltageamplitude A, gradually diminishing at eA ≫ nℏω. Exactly this behavior (predictedby B Josephson in his Nobel-prize-winning 1962 paper12) was very soon observedexperimentally by S Shapiro13; as a result, one can frequently meet the term Shapiro(or ‘Josephson–Shapiro’) steps.

Problem 1.7. Calculate ⟨x⟩, ⟨px⟩, δx, and δpx for the eigenstate {nx, ny, nz} of a particleplaced inside a rectangular, hard-wall box described by Eq. (1.77) of the lecture notes,and compare the product δxδpx with the Heisenberg’s uncertainty relation.

Solution: Since the spatial factors X, Y and Z of wavefunctions given by Eq. (1.84) ofthe lecture notes, and by similar relations for Y and Z, are already normalized andreal, we may use Eq. (1.23) to write

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

∫ ∫

∫ ∫π π

= =

= ≡ −

*x X x xX x dx X x xdx

an xa

xdxa

n xa

xdx

( ) ( ) ( )

2sin

11 cos

2.

a a

x

ax

x x

ax

x

0 0

2

0

2

0

x x

x x

9 See, e.g. Eq. (A.42a), taking into account that the imaginary part of the function under the integral on itsright-hand side is odd and 2π-periodic.10 For a quantitative analysis of stability of such phase locking we would need to take into account other,dissipative component(s) of the current through the junction. However, as the analysis of the simple, standardmodel of phase-locking (see, e.g. Part CM section 5.4) shows, one of two physically different values of thephase difference φ0, at which Eq. (*) is satisfied (say, φ0 = sin−1(I/In) and φ0′ = π − φ0), is always stable.11 See, e.g. Part EM figure 2.16.12 [1]13 [2]

Quantum Mechanics: Problems with solutions

1-8

Integrating the second term in the parentheses by parts, we get

⎛⎝⎜

⎞⎠⎟∫ ∫π

ππ

π

ππ

= − = +

= + =

=

=

=

=

xa

a an

xdn xa

an

n xa

dx

a an

n xa

a

12 2

sin2

21

2sin

2

2 (2 )cos

22

.

x

x x

x x

x ax

x

x

x

ax

x

x x

x

x

x x

x ax

2

0 0

20

x x

x

This simple result is hardly surprising, because the wavefunctions X(x) are eithersymmetric or antisymmetric about the central point ax/2—see figure 1.8 of thelecture notes. Acting absolutely similarly, but repeating the integration by partstwice, we get

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟∫ π

π= = −x

an xa

x dx an

2sin

13

12

,x

ax

xx

x

2

0

22 2

2 2

x

so that, according to Eqs. (1.33) and (1.34),

⎛⎝⎜

⎞⎠⎟δ

π= − = −( )x x x a

n1

121

2.x

x

2 2 1/2

2 2

1/2

Notice that neither ⟨x⟩ nor δx depend on other quantum numbers (ny and nz), andthat the uncertainty of the coordinate is smallest for nx = 1 (in particular for theground state), with δxmin ≈ 0.18 ax, and increases with nx, approaching the limitδxmax = ax/√12 ≈ 0.29 ax at nx → ∞.

For the particle’s momentum, the calculations are simpler:

⎜ ⎟⎛⎝

⎞⎠∫ ∫

∫

∫

π π

π π π

π π π

= ˆ = − ℏ ∂∂

= − ℏ

= − ℏ = − ℏ =

*

=

=

p X x p X x dxa

n xa

ix

n xa

dx

in

an xa

n xa

dx

in

an xa

dx ia

n xa

( ) ( )2

sin sin

2sin cos

sin2 1

2cos

20.

x

a

xx

ax

x

x

x

x

x

ax

x

x

x

x

x

ax

x x

x

x x

x a

0 0

20

20

0

x x

x

xx

This result could be also predicted in advance, because, as was discussed insection 1.7 of the lecture notes, the standing wave X(x) may be represented as a sumof two traveling waves with equal amplitudes, and equal and opposite momenta px =±ℏkx = ±ℏπnx/ax, so that the average momentum vanishes. This reasoning impliesthat ⟨px

2⟩ may be calculated from Eq. (1.37), with two possible states having equalprobabilities: W+ = W− = ½:

⎛⎝⎜

⎞⎠⎟

π= + = + = ℏ+ + − − + −( )p p W p W p p

na

12

.xx

x

2 2 2 2 22

As a sanity check, this calculation may be confirmed directly from Eqs. (1.33)and (1.34):

Quantum Mechanics: Problems with solutions

1-9

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

∫ ∫

∫

∫

π π

π π

π π

π π π π

= ˆ = −ℏ ∂∂

= ℏ

= ℏ −

= ℏ + ℏ = ℏ

*

=

=

p X x p X x dxa

n xa x

n xa

dx

ana

n xa

dx

ana

n xa

dx

na

na

n xa

na

( ) ( )2

sin sin

2sin

11 cos

2

2sin

2.

x

a

xx

ax

x

x

x

x

x

x

ax

x

x

x

x

ax

x

x

x

x

x

x

xx

x a

x

x

2

0

2

0

22

2

22

0

2

22

0

22

2

0

2

x x

x

x

x

Now we can calculate the momentum’s uncertainty,

δ π= − = ℏ( )p p pn

a,x x x

x

x

2 2 1/2

and the uncertainty product:

⎛⎝⎜

⎞⎠⎟δ δ

π= ℏ −x pn

1212

.xx

2 2 1/2

This expression shows that for the lowest quantum number, nx = 1, theuncertainty product, (δxδpx)min ≈ 0.568 ℏ, is just slightly (about 12%) larger thanthe Heisenberg’s minimum (0.5 ℏ). On the other hand, at nx → ∞ the uncertaintyproduct grows as (π/√12)nxℏ ≈ 0.907nxℏ.

Problem 1.8. Looking at the lower (red) line in figure 1.8 of the lecture notes, itseems plausible that the 1D ground-state function (1.84) of the simple potential well(1.77) may be well approximated with an inverted parabola:

= −X x C x a x( ) ( ),xtrial

where C is the normalization constant. Explore how good this approximation is.

Solution: A convenient ‘global’ measure of the approximation quality is theproximity of the expectation value (1.23) of the system’s Hamiltonian, given bythe guessed approximation (in the variational method, to be discussed in section 2.9of the lecture notes, called the trial function):

∫ ψ ψ= ˆ **H H d rr r( ) ( ) , ( )trial trial trial3

where ψtrial r( ) is properly normalized,

∫ ψ ψ =* d rr r( ) ( ) 1,trial trial3

Quantum Mechanics: Problems with solutions

1-10

to the genuine ground state energy Eg, which, according to Eq. (1.60), satisfies asimilar relation, but with the genuine ground-state wavefunction ψg r( ):

∫ ∫ ∫ψ ψ ψ ψ ψ ψ= ˆ = = =* * *H H d r E d r E d r Er r r r r r( ) ( ) ( ) ( ) ( ) ( ) .g gg g g3

g g3

g g3

g

In our 1D case, the normalization condition is

∫ ∫≡ − =*X x X x dx C x a x dx( ) ( ) ( ) 1,a a

0trial trial

2

0

2 2

where, for the notation simplicity, a ≡ ax. Working out this simple integral, we get

∫

∫

= −

= − + = − + ≡

−C x a x dx

a x ax x dx aa

aa a a

( )

( 2 )3

24 5 30

.

a

a

2

0

2 2

0

2 2 3 4 23 4 5 5

Now using the fact that inside our simple quantum-well U(x) = 0, so thatˆ = −ℏH m d dx( /2 ) /2 2 2 in all the region where Xtrial ≠ 0, we get

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

∫ ∫

∫

∫ ∫

= ˆ = − ℏ

= − − ℏ −

= ℏ − = ℏ −

= ℏ − ≡ ℏ

* *H X x HX x dx Xm

ddx

X dx

C x a xm

dx

a mx a x dx

a max x dx

a ma

a ama

( ) ( )2

( )2

( 2)

30( )

30( )

302 3

5 .

a a

a

a a

trial0

trial trial0

trial

2 2

2 trial

2

0

2

5

2

0 5

2

0

2

5

2 2 3 2

2

Comparing this result with the exact ground state energy given by Eq. (1.85) withnx = 1 and ax = a,

π= ℏ ≈ ℏE

ma ma24.935 ,g

2 2

2

2

2

we see that the approximation given by this simple trial function is indeed prettygood, giving a ∼1% accuracy, even in the absence of adjustable parameters that areused in the genuine variational method.

Problem 1.9. A particle placed into a hard-wall, rectangular box with sides ax, ay,and az, is in its ground state. Calculate the average force acting on each face of thebox. Can the forces be characterized by certain pressure?

Solution:Directing the coordinates axes along the corresponding sides of the box, wemay describe the situation by the boundary problem described by Eq. (1.78b) of the

Quantum Mechanics: Problems with solutions

1-11

lecture notes, so that the ground state energy Eg of the particle is expressed byEq. (1.86) with the lowest possible values of the quantum numbers, nx = ny = nz = 1:

⎛⎝⎜⎜

⎞⎠⎟⎟π≡ = ℏ + +E E

m a a a21 1 1

.x y z

g 1,1,1

2 2

2 2 2

Since this energy (though kinetic by its origin) is a function of the box size only, itmay be considered as a contribution to the effective potential energy of the box, andhence the force acting on any of two faces normal to axis x may be calculated as

π= −∂∂

= ℏF

E

a ma.x

x x

g2 2

3

Since the area of this face is Ax = ayaz, the force-to-area ratio is

P π≡ = ℏFA a a a

.xx

x y z

2 2

3

Since the calculations for two other face pairs may be done absolutely similarly,and give similar results (with the proper index replacements), this expression showsthat generally

P P P≠ ≠ ,x y z

and hence the exerted forces cannot be characterized by a unique pressureP , whichby definition14 should be isotropic. Only in the particular case when the box is cubic,with sides ax = ay = az ≡ a and volume V = a3, then we may speak of pressure:

P P P P π π= = ≡ = ℏ = ℏma mV

.x y z

2 2

5

2 2

5/3

Note that the resulting ‘equation of state’,P =V const5/3 , differs from that of theideal classical gas (P =V const). As will be discussed in chapter 8, such ‘quantumequations of state’ remain the same even if the cubic box is filled by an arbitrarynumber N of non-interacting particles—either bosons or fermions (though thedependence of the pressure on N is different for these two cases)15.

Problem 1.10. A 1D quantum particle was initially in the ground state of a verydeep, rectangular potential well of width a:

⎧⎨⎩= − < < ++∞

U xa x a

( )0, for /2 /2,

, otherwise.

14 See, e.g. Part CM sections 7.2 and 8.1.15As statistical mechanics shows (see, e.g. Part SM chapter 3), at sufficiently high temperature the pressurebecomes isotropic and classical (with P =V const), regardless of the shape of the box, the number of theparticles, and their quantum properties.

Quantum Mechanics: Problems with solutions

1-12

At some instant, the well’s width is abruptly increased to a new value a′ > a, leavingthe potential symmetric with respect to the point x = 0, and then left constant.Calculate the probability that after the change, the particle is still in the ground stateof the system.

Solution: According to Eqs. (1.69) and (1.84) of the lecture notes (with theappropriate shift of the origin), the normalized initial state of the system (beforethe change of the well’s width) is

⎛⎝⎜

⎞⎠⎟

⎧⎨⎩⎫⎬⎭

⎧⎨⎪⎩⎪

πΨ = −

ℏ×

<*x

aiE

txa

xa

( , 0)2

expcos , for

2,

0, otherwise,( )

1/2g

with the ground-state energy Eg given by Eq. (1.85) with ax = a and nx = 1:

π=Ema2

.g

2

2

This ‘old’ state serves as the initial condition for the final state of the system,

∑ ψΨ = −ℏ=

∞

{ }x t c x iE

t( , ) ( )exp ,n 1

n nn

where ψn(x) are the normalized eigenstates of the new (expanded) well. In particular,according to the same Eq. (1.84), the new ground state is

⎛⎝⎜

⎞⎠⎟ψ π=

′ ′x

ax

a( )

2cos .1

1/2

The constant coefficient c1, which in particular determines the probabilityW1 = ∣c1∣2

of particle’s remaining in the ground state, may be found from the 1D versionof Eq. (1.68):

∫ ψ= Ψ−∞

+∞*c x x dx( ) ( , 0) ,1 1

giving

⎡⎣⎢

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎤⎦⎥

⎡⎣⎢⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎤⎦⎥

∫

∫

π π

π π

ππ π

ππ

=′ ′

=′

+′

+ −′

=′

+′

+′

+ −′

−′

≡ ′′ − ′

−

+

+

− −

caa

xa

xa

dx

aax

a ax

a adx

aa a aa

a a a aa

a a

a aa a

aa

2( )

cos cos

2( )

cos1 1

cos1 1

2( )

1 1sin

21 1 1 1

sin2

1 1

4( )

cos2

,

a

a

a

1 1/2 /2

/2

1/2 0

/2

1/2

1 1

1/2 3/2

2 2

Quantum Mechanics: Problems with solutions

1-13

so that16

ππ= = ′

′ − ′W c

aaa a

aa

16( )

cos2

.1 12

2

3

2 2 22

As a sanity check, if the well is virtually unchanged, a′= a+ ε→ a, then cos(πa/2a′)→πε/2a, (a′2 − a2) → 2aε, so that c1 → 1, and W1 → 1, as it should be. On the otherhand, if the final well is much wider than the initial one, a≪ a′, then cos(πa/2a′) ≈ 1,andW1 ≈ (16/π2) a/a′≪ 1. This is also reasonable, because the relatively sharp initialdistribution gives a contribution to many final eigenfunctions, with a smallprobability for the particle to be in any particular one of them.

(Additional question for the reader: could a similar problem be rationallyformulated for a′ < a, i.e. a sudden well’s shrinkage rather than its extension?)

Problem 1.11. At t = 0, a 1D particle of mass m is placed into a hard-wall, flat-bottom potential well

⎧⎨⎩= < <+∞

U xx a

( )0, for 0 ,

, otherwise,

in a 50/50 linear superposition of the lowest (ground) state and the first excited state.Calculate:

(i) the normalized wavefunction Ψ(x, t) for an arbitrary time t ⩾ 0, and(ii) the time evolution of the expectation value ⟨x⟩ of the particle’s coordinate.

Solutions:

(i) Our linear superposition is described with the wavefunction

ψ ψΨ = +x C x x( , 0) [ ( ) ( )],1 2

where ψ1 and ψ2 are the two lowest-energy eigenfunctions of this problem, whichwere by-products of the 3D calculation in section 1.7 of the lecture notes—see Eqs.(1.84) and (1.85):

⎛⎝⎜

⎞⎠⎟ψ π π= = ℏ = …x

anxa

En

mn( )

2sin ,

2, with 1, 2, .n n

1/2 2 2 2

The coefficient C (or rather its modulus) may be readily calculated from thenormalization requirement:

16Note that this result would not be affected by adding an arbitrary phase to the wavefunction (*), because itwould just shift the phase of the complex coefficient c1.

Quantum Mechanics: Problems with solutions

1-14

∫∫ ψ ψ ψ ψ

≡ Ψ Ψ

≡ + + =*

*

*

W x x dx

C x x x x dx

( , 0) ( , 0)

[ ( ) ( )] [ ( ) ( )] 1.( )

a

a0

2

01 2 1 2

Since the wavefunctions ψ1,2 are orthonormal,

∫ ∫ψ ψ ψ ψ= =* *x x dx x x dx( ) ( ) 1, ( ) ( ) 0,a a

01,2 1,2

01,2 2,1

Eq. (*) yields ∣C∣2 = 2, i.e. ∣C∣ = 1/√2. Note that the normalization condition leavesthe phase of C arbitrary.

According to Eq. (1.69) of the lecture notes, the wavefunction at arbitrarytime t is

⎡⎣⎢

⎤⎦⎥

π ω π ω

ω ω

Ψ = − + −

=ℏ

=ℏ

=ℏ

**x t

axa

i tx

ai t

E E E

( , )1

sin exp{ } sin2

exp{ } ,

with , 4 .

( )1/2 1 2

11

22 1

(ii) Now we may use this wavefunction and the basic Eq. (1.23) to calculate theexpectation value of the particle’s coordinate:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

∫

∫

∫

π ω π ω

π ω π ω

π π π π ω

ω ω ω

= Ψ ˆ Ψ

= +

× − + −

= + +

≡ −

*x x t x x t dx

axa

i tx

ai t

xxa

i tx

ai t dx

axa

xa

xa

xa

t xdx

( , ) ( , )

1sin exp{ } sin

2exp{ }

sin exp{ } sin2

exp{ }

1sin sin

22 sin sin

2cos ,

with .

a

a

a

0

01 2

1 1

0

2 2

2 1

Transforming the product of two sine functions into the difference of cos functionsof combinational arguments17, and working the resulting four integrals by parts18,we finally get

πω= − ***x a a t

12

169

cos . ( )2

Evidently, this formula describes sinusoidal oscillations of the particle, with theamplitude (16/9π2)a ≈ 0.18a, around the middle of the well (x0 = a/2).

17 See, e.g. Eq. (A.18c).18 See, e.g. Eq. (A.25).

Quantum Mechanics: Problems with solutions

1-15

At least three comments are due here. First, this problem is a good reminder thatthe quantum-mechanical averaging ⟨…⟩ is by no means equivalent to the averagingover time, and its result may still be a function of time—as in this case.

Second, recall that ⟨x⟩ does not oscillate if the system is in either of the involvedstationary states, so that the oscillations (***) are the result of the states’interference. The frequency ω of the oscillations is proportional to the differencebetween the energies of the involved stationary states, in our case

ω ω ω π π πℏ ≡ ℏ − = − = − =E Ema ma ma

( )4

2 23

2,2 1 2 1

2

2

2

2

2

2

i.e. to the frequency of the potential radiation at the transition between thecorresponding energy levels—see Eq. (1.7) of the lecture notes.

Finally, the above result is exactly valid only if Eq. (**) is taken in its displayedform, which follows from Eq. (1.69) with real coefficients cn—in our case c1 and c2.Generally, these coefficients may have different phases φ1 ≡ arg(c1) and φ2 ≡ arg(c2).Repeating the above calculations for this case, we may readily find that the time-dependent factor cosωt in Eq. (***) becomes cos(ωt − φ), where φ ≡ φ2 − φ1. So,while the common phase of a wavefunction drops out of all expectation values, themutual phase shift(s) between is its components in a linear superposition do affectthe expectation values—in our particular case, of the coordinate.

Problem 1.12. Calculate the potential profiles U(x) for that the followingwavefunctions,

(i) Ψ = − −c ax ibtexp{ }2 , and(ii) Ψ = − ∣ ∣ −c a x ibtexp{ },

(with real coefficients a > 0 and b), satisfy the 1D Schrödinger equation for a particlewith mass m. For each case, calculate ⟨x⟩, ⟨px⟩, δx, and δpx, and compare theproduct δxδpx with the Heisenberg’s uncertainty relation.

Solutions: Both these wavefunctions may be represented as the product ψn(x)exp{−iEnt/ℏ}, with En = ℏb, so that in accordance with the discussion in sections1.5–1.6 of the lecture notes, we only need to calculate the corresponding functionsU(x) from the stationary Schrödinger equation (1.65), which may be rewritten as

ψψ

= + ℏU x E

m

d

d x( )

12

nn

n2 2

2

with the given eigenfunctions ψn.

(i) In this case, ψn = c exp{−ax2}, so that a direct differentiation yields

= + ℏ −U x Em

a x a( )2

(4 2 ).n

22 2

Quantum Mechanics: Problems with solutions

1-16

Now notice that if we introduce, instead of a, the constant

ω ≡ ℏ *am

2, ( )0

the above expression may be rewritten as

⎜ ⎟⎛⎝

⎞⎠

ω ω= + − ℏU x

m xE( )

2 2,n

02 2

0

while the corresponding wavefunction becomes

⎧⎨⎩⎫⎬⎭ψ

ω= − ≡ −ℏ

c ax cm x

exp{ } exp2

.n2 0

2

Since, according to the stationary Schrödinger equation, the origins of En andUmaybe shifted (simultaneously) by an arbitrary constant, we may select this constant sothat

ω= ℏE

2,n

0

and U(x) becomes the well-known expression for potential energy of a harmonicoscillator of frequency ω0, and mass m:

ω=U xm x

( )2

.02 2

Hence, ‘by chance’ (actually, not quite :-), we have found one of the eigenstates ofthis very important 1D system. Later in the course, we will see that this is actually itsmost important, lowest-energy (ground) state, usually marked with quantumnumber n = 0.

Now, after finding the constant c (or rather its modulus) from the normalizationcondition

⎜ ⎟⎛⎝

⎞⎠∫ ∫ψ ψ π= ≡ − ≡

−∞

+∞*

−∞

+∞dx c ax dx

ac1 exp{ 2 }

2,n n

2 21/2

2

(at the last step using the well-known, and very important, Gaussian integral19), wecan use Eq. (1.23) of the lecture notes to calculate the expectation values20

= = = = ℏx p xa

p a0, 0,14

, ,x x2 2 2

so that

δ δ= = ℏxa

p a1

2, .x1/2

1/2

19 See, e.g. Eq. (A.36b).20 The calculation of two last averages requires one more Gaussian integral, given by Eq. (A.36c).

Quantum Mechanics: Problems with solutions

1-17

So, the product δxδpx equals ℏ/2, i.e. has the smallest value allowed by theuncertainty relation (1.35).21 In the notation (*), customary for the harmonicoscillator description, the above results for the coordinate and momentum variancesread

ωω= ℏ = ℏ

xm

pm

2,

2.x

2

0

2 0

Notice that the averages of the kinetic and potential energies of the oscillator areequal to each other:

ω ω= = ℏp

mm x

2 2 4,x

202 2

0

just as at classical oscillations of the system.

(ii) In this case, ψn = c exp{−a∣x∣}, so that a similar calculation of U(x) gives

= ℏ + ℏ −U x b em

ddx

e( )2

.a x a x2 2

2

At x ≠ 0, this expression gives a constant (equal to ℏb + ℏ2a2/2m), but the point x = 0requires a special calculation, because here the wavefunction has a ‘cusp’, and is notanalytically differentiable. However, using the notions of the sign function sgn(x)and the Dirac’s delta-function δ(x),22 we can still write formulas valid for all x:

δ= − = − **− − − −ddx

e a x eddx

e a a x esgn( ) , [ 2 ( )] . ( )a x a x a x a x2

2

so that, finally, we get a potential U(x) describing (besides an inconsequentialconstant U0) an ultimately narrow 1D potential well:

Wδ δ= ℏ + ℏ − ≡ −U x bm

a a x U x( )2

[ 2 ( )] ( ),2

20

where

W≡ ℏ + ℏ ≡ ℏ >U bam

am2

, 0.0

2 2 2

In this notation, the eigenfunction and the eigenenergy become

W

W

ψ = − ≡ −ℏ

= ℏ = − ℏ ≡ −ℏ

{ }c a x cm

x

E b Uam

Um

exp{ } exp ,

2 2.

2

0

2 2

0

2

2

21Note that this relation also holds for more general Gaussian wave packets, to be discussed in section 2.2.22 If you need a reminder, see, e.g. section A.14 of appendix A.

Quantum Mechanics: Problems with solutions

1-18

In chapter 2 of the lecture notes, we will see that these results describe the onlylocalized eigenstate of such a well; they will be broadly used in this course as thebasis for discussion of more complex problems.

Now after the wavefunction’s normalization (cc* = 1), Eq. (1.23) of the lecturenotes, after a straightforward integration, yields23

= =x xa

0,1

2.2

2

Calculating the expectation values of px and px2, we should be careful not to loose the

functions sgn(x) and δ(x)—see Eq. (**):

ψ

ψ δ

ˆ = − ℏ = −

ˆ = − ℏ = −ℏ −

− −

− −

p iddx

ce ica x e

pddx

ce ca a x e

( ) sgn( ) ,

( ) [ 2 ( )] .

xa x a x

xa x a x2 2

2

22

Now the integration (1.23) yields

δ δ= = ℏ = ℏp p a x p0, , so that

2.x x x

2 2 2

We see that for this non-Gaussian eigenfunction, the uncertainty product issubstantially (by ∼ 40%) larger than the minimum possible value ℏ/2.

Problem 1.13. A 1D particle of mass m, moving in the field of a stationary potentialU(x), has the following eigenfunction

ψκ

=xC

x( )

cosh,

where C is the normalization constant, and κ is a real constant. Calculate thefunction U(x) and the state’s eigenenergy E.

Solution: After calculating the second derivative of the eigenfunction,

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

ψ κ κκ

ψ κ κκ

κκ

κκ

= − = − = −ddx

C xx

ddx

ddx

C xx

Cx

xx

sinhcosh

,sinh

cosh cosh2 sinhcosh

1 ,2

2

2 2

2 2

2

we may plug it into the 1D version of the stationary Schrödinger equation (1.65):

ψ ψ ψ− ℏ + =m

ddx

U x E2

( ) ,2 2

2

23 For the second integration, we may use the table integral given by Eq. (A.34d ) for n = 2.

Quantum Mechanics: Problems with solutions

1-19

getting

⎛⎝⎜

⎞⎠⎟

κκ

κκ κ κ

− ℏ − + =m

Cx

xx

U xC

xE

Cx2 cosh

2 sinhcosh

1 ( )cosh cosh

.2 2 2

2

Canceling the common multiplier C/cosh κx ≠ 0, we may rewrite this relation asfollows

⎛⎝⎜

⎞⎠⎟

κ κκ

− = ℏ − *U x Em

xx

( )2

2 sinhcosh

1 . ( )2 2 2

2

The function U(x) and the eigenenergy E are defined to an arbitrary constant(essentially the energy reference level), provided that their difference is definite—asspecified by Eq. (*). It is convenient to select this constant so that U(x) → 0 at x →±∞. Since in these limits the expression in the parentheses of Eq. (*) tends to 1, weshould associate this constant level with (−E), so that

κ= − ℏE

m2.

2 2

Now plugging this value back to Eq. (*), we get a result that may be recast into avery simple form:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

κ κκ

κ κκ

κ

κ κ κκ

κκ

= ℏ − + = ℏ − − ℏ

≡ − ℏ − ≡ − ℏ

U xm

xx

Em

xx m

mx x

x m x

( )2

2 sinhcosh

12

2 sinhcosh

12

22(cosh sinh )

cosh1

cosh.

2 2 2

2

2 2 2

2

2 2

2 2 2 2

2

2 2

2

A plot of this function is shown with the black line in figure below, together withthe calculated eigenenergy (dashed horizontal line), both in the units of ℏ2κ2/m, andthe eigenfunction ψ(x) (red line). Due to the simple eigenfunction describing thelocalized state of the particle (which may be shown to be its ground state), thispotential is a convenient and popular model for description of ‘soft’ confinement inone dimension.

Quantum Mechanics: Problems with solutions

1-20

Problem 1.14. Calculate the density dN/dE of the traveling-wave states in largerectangular potential wells of various dimensions: d = 1, 2, and 3.

Solution: Let us calculate the number N3 of 3D states with kinetic energy ℏ2k2/2mbelow a certain fixed value E. For that, we may integrate Eq. (1.90) of the lecturenotes over the region of k-space satisfying the requirement

<ℏ

kmE2

,22

i.e. over a sphere with the so-called Fermi radius kF ≡ (2mE)1/2/ℏ:

∫π ππ

ππ= = ≡

ℏ<N

Vd k

Vk

V mE

(2 ) (2 )43 (2 )

43

(2 ).

k k3 3

33 F

33

3/2

33/2

F

From here, the 3D density of states is

ππ

π=

ℏ=

ℏdNdE

V mE V

mE

(2 )43

(2 ) 32

(2 )4

.33

3/2

31/2

3/2

2 31/2

Note that the density grows with energy.The absolutely similar calculation for the rectangular 2D well, based on

Eq. (1.99), yields

∫π ππ

ππ= = ≡

ℏ<N

Ad k

Ak

A mE

(2 ) (2 ) (2 )2

,k k

2 22

2 F2

2 2F

so that the 2D density of states does not depend on energy:

ππ

π=

ℏ=

ℏdNdE

A mA

m(2 )

22

.22 2 2

Finally, for 1D particles, Eq. (1.100) yields

∫π π π= = ≡

ℏ<N

ldk

lk

l mE

2 22

22

(2 ),

k kF1

1/21/2

F

so that the 1D density of states,

π π=

ℏ=

ℏ− −dN

dEl m

E lm

E2

2(2 ) 1

2(2 )

2,1

1/21/2

1/21/2

decreases with energy.

Problem 1.15.* Use the finite difference method with steps a/2 and a/3 to find as manyeigenenergies as possible for a 1D particle in the infinitely deep, hard-wall potentialwell of width a. Compare the results with each other, and with the exact formula24.

24You may like to start from reading about the finite difference method—see, e.g. Part CM section 8.5 or PartEM section 2.11.

Quantum Mechanics: Problems with solutions

1-21

Solution: The eigenproblem is described by the ordinary differential equation (1.83),which includes the second derivative of the wavefunction X(x). In the finitedifference method, we are approximating the derivative with the following finitedifference25:

≈ − + + −d Xdx

X x h X x h X xh

( ) ( ) 2 ( ),

2

2 2

where h (not to be confused with either ℏ or ℏ/2π!) is the selected step along axis x.For h = a/2, the only reasonable choice is to select the point x in the middle of the

potential well (in the notation of figure 1.8, at x = a/2), so that the points (x − h) and(x + h) are on the well’s walls, where X = 0. Thus Eq. (1.83) turns into a very simplerelation

+ − + =Xh

k X0 0 2

0,2

2

where X ≡ X(a/2) and k ≡ kx. This homogeneous equation cannot be used to get X,but assuming that X ≠ 0 (i.e. that the wavefunction is nonvanishing), it gives simpleresults for the eigenvalue of the standing wave’s vector k and hence for theeigenenergy E (≡ Ex) = (ℏ2/2m)k2:

= ≡ ≈ = ℏ ≡ ℏk

h a aE

mh ma2 2 2 2.83

, 22

4 .2

2

2

2

These values should be compared with the exact analytical results (1.84)–(1.85)for the lowest (ground) eigenstate (nx = 1):

π π= ≈ = ℏ ≈ ℏk

a aE

ma ma3.14

,2

4.93 .1 12

2

2

2

2

So, this large step (in the numerical-method lingo, ‘coarse mesh’) has made the mathvery simple, but has allowed the calculation of only one, ground eigenstate, and witha relatively poor accuracy: ∼10% for k and ∼20% for eigenenergy. This could beexpected, because such coarse mesh corresponds to the approximation of thegenuine sinusoidal solutions (1.84) with a single quadratic parabola.

So it is only natural to explore a slightly finer mesh with h = a/3, making a similarapproximation for two interleaved segments of the same length 2h = 2a/3: x ∈ [0,2a/3] and x ∈ [a/3, a]. Applying the finite-difference version of Eq. (1.83),

− + + − + =X x h X x h X xh

k X( ) ( ) 2 ( )

0,2

2

to the central points x− ≡ h = a/3 and x+ ≡ 2h = 2a/3 of these two segments, we gettwo equations for the corresponding wavefunction’s values X− and X+:

25 See, e.g. Part CM Eq. (8.65) or Part EM Eq. (2.220).

Quantum Mechanics: Problems with solutions

1-22

+ − + = + − + =+ −−

− ++

X Xh

k XX X

hk X

0 20,

0 20.

22

22

This system of two linear, homogeneous equation is consistent if its determinantequals zero:

− +

− +=h

kh

h hk

2 1

1 20.

22

2

2 22

This quadratic equation for k2 has two solutions:

= ± = ±kh h a2 1

(18 9)1

,22 2 2

giving the following two eigenvalue sets:

= − = = ℏ

= + ≈ = ℏ

− −

+ +

ka a

Ema

ka a

Ema

(18 9)1 3

, 4.5 ;

(18 9)1 5.20

, 13.5 .

1/22

2

1/22

2

The first of them is just a better approximation for the ground state, with a ∼5%accuracy for k and a ∼10% accuracy for energy. The second result is a much cruderdescription of the next (first excited) state, whose exact parameters are given by thesame Eqs. (1.84) and (1.85) with nx = 2:

π π= ≈ = ℏ ≈ ℏk

a aE

ma ma2 6.27

, 42

19.7 .2 12

2

2

2

2

Evidently, an even finer mesh, with a smaller h, would allow a more precisedescription of more eigenstates, for the price of solving a larger system ofhomogeneous linear equations26. For this particular problem, which has a simpleanalytical solution, this numerical method makes sense only as a demonstration, butfor eigenstates of particles moving in more complex potential profiles U(x), this isone of very few possible approaches. (A different, frequently more efficientnumerical approach will be discussed in section 6.1 of the lecture notes—seeEq. (6.7) and its discussion.)

References[1] Josephson B 1962 Phys. Lett. 1 251[2] Shapiro S 1963 Phys. Rev. Lett. 11 80

26All popular public-domain and commercial software packages, including those listed in section A.16(iv) ofappendix A, have efficient standard routines for such solutions.

Quantum Mechanics: Problems with solutions

1-23