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Solutions to Quantum Mechanics by Claude

Cohen-Tannoudji, Bernard Diu, and Franck Laloe

Benjamin D. Suh

September 21, 2020

4 CONTENTS

3.12 Infinite one-dimensional well................................ 54 3.13 Infinite two-dimensional well............................... 55 3.14 Matrices........................................... 56

Chapter 1

Waves and Particles. Introduction

to the Fundamental Ideas of

Quantum Mechanics

1.1 Multiple-Slit Experiment

A beam of neutrons of constant velocity, mass Mn(Mn ≈ 1. 67 x 10 −^27 kg) and energy E, is incident on a linear chain of atomic nuclei, arranged in a regular fashion as shown in the figure (these nuclei could be, for example, those of a long, linear molecule). We call l the distance between two consecutive nuclei, and d, their size (d  l). A neutron detector D is placed far away, in a direction which makes an angle of θ with the direction of the incident neutrons.

Figure 1.1: Multiple-slit Experiment

1.1.a Describe qualitatively the phenomena ob-

served at D when the energy E of the incident neu-

trons is varied.

From Complement DI , the fringe separation of the double-slit experi- ment is given by

λD l

. Using the Planck-Einstein relations(A-1), we know that as energy increases, the wavelength decreases. Thus, the distance between successive peaks decreases as well, so we end up seeing more peaks in a given area.

1.2. BOUND STATE OF A PARTICLE IN A ”DELTA FUNCTION POTENTIAL” 7

1.2 Bound State of a Particle in a ”Delta Function Poten-

tial”

Consider a particle whose Hamiltonian H [operator defined by formula (D-10) is:

H = −

ℏ^2

2 m

d^2 dx^2

− αδ(x)

where α is a positive constant whose dimensions are to be found.

1.2.a Integrate the eigenvalue equation of H between − and +. Letting

 approach 0, show that the derivative of the eigenfunction φ(x)

presents a discontinuity at x = 0 and determine it in terms of α, m,

and φ(0).

Nothing doing, let’s do as the question asks,

∫ (^) 

−

H |φ〉 dx =

−

E |φ〉 dx

−

ℏ^2

2 m

d^2 dx^2

φ(x) dx −

−

αδ(x)φ(x) dx = E

−

φ(x) dx

On the right, the integral disappears as  → 0. The delta function only turns on if the integral includes x = 0, and it picks out that value in that integral,

ℏ^2

2 m

dφ(x) dx

x=

x=−

− αφ(0) = 0

dφ(x) dx

x=

dφ(x) dx

x=−

2 mαφ(0) ℏ^2

As  → 0, the left side would disappear unless there were a discontinuity. Since the right side is not equal to zero, there must be a discontinuity in the derivative of φ(x).

1.2.b Assume that the energy E of the particle is negative(bound state).

φ(x) can then be written:

x < 0 φ(x) = A 1 exp(ρx) + A′ 1 exp(−ρx) x > 0 φ(x) = A 2 exp(ρx) + A′ 2 exp(−ρx)

8 CHAPTER 1. WAVES AND PARTICLES. INTRODUCTION TO THE FUNDAMENTAL IDEAS OF QUANTUM

Express the constant ρ in terms of E and m. Using the results of the preceding question, calculate the matrix M defined by:

( A 2 A′ 2

= M

A 1

A′ 1

Then, using the condition that φ(x) must be square-integrable, find the possible values of the energy. Calculate the corresponding normalized wave functions. We can find ρ by using the Schrodinger equation(B-8),

ℏ^2

2 m

d^2 dx^2

φ(x) = Eφ(x)

We only need to look at φ(x) at a single time, so lets look at the case x < 0,

ℏ^2

2 m

(A 1 ρ^2 exp(ρx) + A′ 1 ρ^2 exp(−ρx)) = E(A 1 exp(ρx) + A′ 1 exp(−ρx))

ℏ^2 ρ^2 2 m

= E

ρ^2 = −

2 mE ℏ^2 To calculate the required matrix, we use the discontinuity in the derivative,

dφ(x) dx

x=

dφ(x) dx

x=−

2 mαφ(0) ℏ^2

The first term we will need to use the x > 0 case and x < 0 for the second,

ρA 2 exp(ρ) − A′ 2 ρ exp(−ρ) − A 1 ρ exp(ρ) + A′ 1 ρ exp(−ρ) = −

2 mα ℏ^2

(A 1 + A′ 1 )

Note that we chose an arbitrary case for x = 0. Because φ(x) must be continuous, A 1 + A′ 1 = A 2 + A′ 2. Letting  → 0, we have,

A 2 − A′ 2 − A 1 + A′ 1 = −

2 mα ρℏ^2

(A 1 + A′ 1 )

A 2 = −

2 mα ρℏ^2

(A 1 + A′ 1 ) + A 1 − A′ 1 + A′ 2

Using the continuity of the wavefunction,

2 A 2 = A 1

2 mα ρℏ^2

2 mα ρℏ^2

A′ 1

10 CHAPTER 1. WAVES AND PARTICLES. INTRODUCTION TO THE FUNDAMENTAL IDEAS OF QUANTUM

1.3 Transmission of a ”delta function” potential barrier”

Consider a particle placed in the same potential as in the preceding exercise. The particle is now propagating from left to right along the x axis, with a positive energy E.

1.3.a Show that a stationary state of the particle can be written:

x < 0 φ(x) = exp(ikx) + A exp(−ikx) x > 0 φ(x) = B exp(ikx)

where k, A, and B are constants which are to be calculated in terms of the energy E,

of m and of α(watch out for the discontinuity in dφ dx

at x = 0. By observation, the given wavefunction is not dependant on time, and it follows the form given by (D-7), so it is a stationary state. To determine the constants, we can look at the Schrodinger equation(B-8),

−

ℏ^2

2 m

d^2 dx^2 φ(x) − αδ(x)φ(x) = Eφ(x)

This holds true for all values of x, so let’s look at x > 0. The delta function term dies,

−

ℏ^2

2 m

(−Bk^2 exp(ikx)) = EB exp(ikx)

ℏ^2 k^2 2 m

= E

k^2 =

2 mE ℏ^2 At x = 0, we use the wavefunction continuity, B = 1 + A as well as the discontinuity in the first derivative, dφ(x) dx

x=

dφ(x) dx

x=−

2 mαφ(0) ℏ^2

ikB − ik(1 − A) = −

2 mαB ℏ^2

B(ikℏ^2 + 2mα) = ikℏ^2 (1 − A)

B =

ikℏ^2 ikℏ^2 + mα

A = −

mα ikℏ^2 + mα

1.3. TRANSMISSION OF A ”DELTA FUNCTION” POTENTIAL BARRIER” 11

1.3.b Set −EL = −mα^2 / 2 ℏ^2 (bound state energy of the particle). Calcu-

late, in terms of the dimensionless parameter E/EL, the reflec-

tion coefficient R and the transmission coefficient T of the bar-

rier. Study their variations with respect to E; what happens when

E → ∞? How can this be interpreted? Show that, if the expression

of T is extended for negative values of E, it diverges when E → EL,

and discuss this result.

To find the reflection coefficient, we look at the parameters of incoming and outgoing particles. A particle moving to the right has magnitude 1, while a particle moving to the left has magnitude A,

R = |A|^2 =

m^2 α^2 k^2 ℏ^4 + m^2 α^2

m^2 α^2 m^2 α^2 (1 + E/EL)

R =

1 + E/EL

Similarly, a particle moving to the right after passing through the barrier has magnitude B,

T = |B|^2 =

k^2 ℏ^4 k^2 ℏ^4 + m^2 α^2

T =

E/EL

1 + E/EL

We see that T + R = 1. As E → ∞, T → 1 and R → 0, which means the wavefunction has more and more energy to overcome the barrier. As energy increases, the probability that the particle will go through the barrier increases. At E = −EL, it has the same energy as the barrier, which is why we see that discontinuity. Classically, think of this as a ball hitting the top edge of a wall. Small variations around this point determine if the ball is reflected or transmitted.

1.4. DELTA POTENTIAL, FOURIER TRANSFORM 13

We can reverse our Fourier transform to get φ(x),

φ(x) =

αφ(0) 2 πℏ

−∞

exp

ipx ℏ

(p^2 / 2 m − E) dp

φ(x) =

αφ(0)m ℏ

− 2 mE

exp

− 2 mE ℏ

x

To determine the allowed energy, let’s set x = 0 and match sides,

φ(0) =

αmφ(0) ℏ

− 2 mE

In order for this to hold true, αm ℏ

− 2 mE

E = −

α^2 m 2 ℏ^2

1.4.b The average kinetic energy of the particle can be written (cf. chap.

III):

Ek =

2 m

−∞

p^2 | φ¯(p)|^2 dp

Show that, when φ¯(p) is a ”sufficiently smooth” function, we also have:

Ek = −

ℏ^2

2 m

−∞

φ∗(x)

d^2 φ dx^2

dx

These formulas enable us to obtain, in two different ways, the energy Ek for a particle in the bound state calculated in (a). What result is obtained? Note that, in this case φ(x) is not ”regular” at x = 0, where its derivative is discontinuous. It is then necessary to differentiate φ(x) in the sense of distributions, which introduces a contribution of the point x = 0 to the average value we are looking for. Interpret this contribution physically: consider a square well, centered at x = 0, whose width a approaches 0 and whose depth V 0 approaches infinity (so that aV 0 = α), and study the behaviour of the wave function in this well. Starting with the second equation and substituting in the definitions from Appendix I,

φ∗(x)

d^2 φ dx^2

2 πℏ

−∞

exp

ipx ℏ

φ^ ¯(p)

ipx ℏ

exp

ipx ℏ

dp

14 CHAPTER 1. WAVES AND PARTICLES. INTRODUCTION TO THE FUNDAMENTAL IDEAS OF QUANTUM

Ek =

4 mπℏ

−∞

−∞

p^2 | φ¯(p)|^2 dp dx

The constants in front disappear when we integrate over all x,

Ek =

2 m

−∞

p^2 | φ¯(p)|^2 dp

16 CHAPTER 1. WAVES AND PARTICLES. INTRODUCTION TO THE FUNDAMENTAL IDEAS OF QUANTUM

The bound states are given by

exp(−ρl) = ±

2 ρ μ

(i)Ground State. Show that this state is even (invariant with respect to reflection about the point x = l/ 2 ), and that its energy Es is less than the energy −EL introduced in problem 3. Interpret this result physically. Represent graphically the corresponding wave function.

1.6. SQUARE WELL POTENTIAL 17

1.6 Square Well Potential

Consider a square well potential of width a and depth V 0 (in this exercise, we shall use systematically the notation of 2-c-α of complement HI ). We intend to study the properties of the bound state of a particle in a well when its width a approaches zero.

1.6.a Show that there indeed exists only one bound state and calculate

its energy E

we find E ≈ −

mV 0 a^2

2 ℏ^2

, that is, an energy which varies

with the square of the area aV 0 of the well

Chapter 2

The Mathematical Tools of

Quantum Mechanics

2.1 Hermitian Operator

|φn〉 are the eigenstates of a Hermitian operator H (H is, for example, the Hamilto- nian of an arbitrary physical system). Assume that the states |φn〉 form a discrete orthonormal basis. The operator U (m, n) is defined by:

U (m, n) = |φm〉 〈φn|

2.1.a Calculate the adjoint U †(m, n) of U (m, n)

Using the definition of the adjoint,

U †(m, n) = |φn〉 〈φm|

2.1.b Calculate the commutator [H, U (m, n)]

Let’s act the commutator on a vector (looking ahead, we’ll set the vector as |φn〉,

[H, U ] |φn〉 = HU |φn〉 − U H |φn〉

= H |φm〉 〈φn|φn〉 − |φm〉 〈φn|H|φn〉

Since H has eigenkets φn and φm, i.e., { H |φm〉 = m |φm〉 H |φn〉 = n |φn〉

[H, U (m, n)] = m − n

20 CHAPTER 2. THE MATHEMATICAL TOOLS OF QUANTUM MECHANICS

2.1.c Prove the relation

U (m, n)U †(p, q) = δnq U (m, p)

Writing this out,

U (m, n)U †(p, q) = |φm〉 〈φn|φq 〉 〈φp|

The middle section dies unless n = q, leaving us the delta function,

= δnq |φm〉 〈φp|

U (m, n)U †(p, q) = δnq U (m, p)

2.1.d Calculate Tr{U (m, n)}, the trace of the operator U (m, n)

By definition, the trace is given by

Tr U =

α

〈α|U |α〉

where α are the basis states.

=

α

〈α|φm〉 〈φn|α〉

Since |φn〉 form a basis, and since we sum over the basis states, at least one part dies unless m = n,

Tr U = δmn

2.1.e Let A be an operator, with matrix elements Amn = 〈φm|A|φn〉. Prove

the relation:

A =

m,n

AmnU (m, n)

Let’s start from the right side. Acting it on |φn〉, we only need to sum over m since we can use orthonormality for n, ∑

m,n

AmnU (m, n) |φn〉 =

m

〈φm|A|φn〉 |φm〉 〈φn|φn〉

Amn is a scalar, so we can move that around for free,

=

m

|φm〉 〈φm|A|φn〉

Performing the sum, the first part becomes identity, so we can remove it, ∑

m,n

AmnU (m, n) |φn〉 = A |φn〉