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Columbia University Department of Physics QUALIFYING EXAMINATION Wednesday, January 10, 2018 10:00AM to 12:00PM Modern Physics Section 3. Quantum Mechanics Two hours are permitted for the completion of this section of the examination. Choose 4 problems out of the 5 included in this section. (You will not earn extra credit by doing an additional problem). Apportion your time carefully. Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s) which question you are answering (e.g., Section 3 (QM), Question 2, etc.). Do NOT write your name on your answer booklets. Instead, clearly indicate your Exam Letter Code. You may refer to the single handwritten note sheet on 8 1 2 × 11” paper (double-sided) you have prepared on Modern Physics. The note sheet cannot leave the exam room once the exam has begun. This note sheet must be handed in at the end of today’s exam. Please include your Exam Letter Code on your note sheet. No other extraneous papers or books are permitted. Simple calculators are permitted. However, the use of calculators for storing and/or recov- ering formulae or constants is NOT permitted. Questions should be directed to the proctor. Good Luck! Section 3 Page 1 of 6

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Columbia UniversityDepartment of Physics

QUALIFYING EXAMINATION

Wednesday, January 10, 201810:00AM to 12:00PM

Modern PhysicsSection 3. Quantum Mechanics

Two hours are permitted for the completion of this section of the examination. Choose4 problems out of the 5 included in this section. (You will not earn extra credit by doing anadditional problem). Apportion your time carefully.

Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 3 (QM), Question 2, etc.).

Do NOT write your name on your answer booklets. Instead, clearly indicate your ExamLetter Code.

You may refer to the single handwritten note sheet on 812” × 11” paper (double-sided) you

have prepared on Modern Physics. The note sheet cannot leave the exam room once theexam has begun. This note sheet must be handed in at the end of today’s exam. Pleaseinclude your Exam Letter Code on your note sheet. No other extraneous papers or booksare permitted.

Simple calculators are permitted. However, the use of calculators for storing and/or recov-ering formulae or constants is NOT permitted.

Questions should be directed to the proctor.

Good Luck!

Section 3 Page 1 of 6

1. Let |0〉 be the ground state of the harmonic oscillator with angular frequency ω, withHamiltonian H = 1

2mp2 + 1

2mω2x2. The ground state wave function in position repre-

sentation is ψ0(x) = 〈x|0〉 = Ae−x2/2`2 with ` =

√hmω

and A = `−12π−

14

(a) Express the operator x in the momentum representation and verify the canonicalcommutation relation.

(b) Using the above, express the creation and annihilation operators a and a† in themomentum representation, and verify that the canonical commutation relation be-tween them holds.

(c) Using (b), calculate the ground state wave function ψ0(p) = 〈p|0〉 in the momentumrepresentation.

(d) Check your answer by direction calculation starting from the position space repre-sentation ψ0(x).

Section 3 Page 2 of 6

2. For the infinite square well with walls located at x = a and x = −a, the ground stateenergy is E1 = π2h2/8ma2 and the ground state wavefunction is ψ1(x) = 1√

acos(πx/2a).

The position momentum uncertainty relationship for this state is ∆x∆p = κh/2. Findκ.

(A potentially useful formula is∫dx x2 cos2(bx) = x3

6+(x2

4b− 1

8b3

)sin(2bx) + x cos(2bx)

4b2.)

Section 3 Page 3 of 6

3. Consider the two potential energy functions:

V1(x) =mω2

2x2, x > 0

= ∞, x ≤ 0

V2(x) =mω2

2x2, −∞ ≤ x ≤ ∞

A particle with mass m, subject to the potential V1, is initially in its quantum groundstate.

(a) Write the normalized wave function, φ(x), of this state. (Hint: This wave functioncan be determined easily from simple harmonic oscillator solutions you alreadyknow.)

(b) At a certain time (say, t = 0), the impenetrable barrier at x = 0 is very suddenlyremoved so that for all t > 0 the same system is subject to the potential V2 insteadof V1. If the energy is then measured, what is the probability of finding the systemin its new ground state, ψ0(x)?

Section 3 Page 4 of 6

4. In this problem we consider a one-dimensional quantum harmonic oscillator with fre-quency ω and mass m. The Hamiltonian of the system is

H = hω(n+ 1/2).

The orthonormal eigenstates of the Hamiltonian are denoted by |n〉 and have the propertythat n|n〉 = n|n〉 where n = a†a.

The eigenstates of the operator a are called quasi-classical states for reasons that weexamine in this problem. Consider an arbitrary complex number α.

(a) Show that the state

|α〉 = e−|α|2/2∑n

αn√n!|n〉

is an eigenstate of a with eigenvalue α, that is a|α〉 = α|α〉. Also show that it isproperly normalized.

(b) Calculate the expectation value of energy 〈H〉 = 〈α|H|α〉 for a quasi-classical state|α〉.

(c) Also calculate the expectation values of position 〈x〉 and momentum 〈p〉. Recallthat

x =

√h

2mω(a+ a†) and p = −i

√mhω

2(a− a†).

(d) Calculate the root mean square deviations ∆x and ∆p for the position and themomentum in this state. Show that ∆x∆p = h/2.

(e) Now, suppose that at time t = 0 the oscillator is in a quasi-classical state |α〉 withα = |α|eiφ, where |α| is a real positive number. Show that at any later time tthe oscillator is also in a quasi-classical state that can be written as e−iωt/2|α(t)〉.Determine the value of α(t) in terms of |α|, φ, ω, and t.

(f) Evaluate 〈x〉(t) and 〈p〉(t). Justify briefly why these states are called quasi-classical.

Section 3 Page 5 of 6

5. Consider a particle in one dimension subject to the double-delta-function potential en-ergy

V (x) = −g δ(x− a)− g δ(x+ a)

where g is a positive constant.

(a) Find an equation from which the energy of the lowest bound state can be deter-mined.

(b) Using this equation, find approximate expressions for the ground state energy inthe limits where a is very large and where a approaches zero.

Section 3 Page 6 of 6

1 Harmonic oscillator in momentum representation (Greene)

1.1 Problem

Let |0〉 be the ground state of the harmonic oscillator with angular frequency ω, with Hamil-

tonian H = 12mp2 + 1

2mω2x2. The ground state wave function in position representation is

ψ0(x) = 〈x|0〉 = Ae−x2/2`2 with ` =

√hmω

and A = `−12π−

14

(a) Express the operator x in the momentum representation and verify the canonical com-

mutation relation.

(b) Using the above, express the creation and annihilation operators a and a† in the mo-

mentum representation, and verify that the canonical commutation relation between

them holds.

(c) Using (b), calculate the ground state wave function ψ0(p) = 〈p|0〉 in the momentum

representation.

(d) Check your answer by direction calculation starting from the position space represen-

tation ψ0(x).

1.2 solution

(a) On wave functions ψ(p), x acts as x = ih∂p. The canonical commutation relations are

[x, p] = [ih∂p, p] = ih.

(b) a = 1√2

(1`x+ i `

hp), a† = 1√

2

(1`x− i `

hp), [a, a†] = 1

2(1 + 1) = 1.

(c) a|0〉 = 0 in momentum representation becomes (∂p + `2

h2p)ψ0(p) = 0, which is solved by

ψ0(p) = B e−`2p2/2h2

, B = h−12 `+ 1

2π−14 .

(d) ψ0(p) = 〈p|0〉 =∫dx 〈p|x〉〈x|0〉 =

∫dx 1√

2πhe−ipx/hAe−x

2/2`2 = 1√2πh

A√

2π` e−`2p2/2h2

=

B e−`2p2/2h2

.

Section 3 - Problem 1Greene

2 Uncertainty in potential well (Hughes)

2.1 Problem

For the infinite square well with walls located at x = a and x = �a, the ground state

energy is E1

= ⇡2h2/8ma2 and the ground state wavefunction is 1

(x) = 1pa

cos(⇡x/2a).

The position momentum uncertainty relationship for this state is �x�p = h/2. Find .

(A potentially useful formula isRdx x2 cos2(bx) = x

3

6

+�x

2

4b

� 1

8b

3

�sin(2bx) + x cos(2bx)

4b

2 .)

2.2 Solution

The uncertainties are �x =phx2i � hxi2, �p =

php2i � hpi2. By symmetry, it is clear that

hxi = 0 and hpi = 0. Moreover since H = p

2

2m

, we immediately get hp2i = 2mhHi = 2mE1

=

⇡2h2/4a2. Finally hx2i = 1

a

Ra

�a

dx x2 cos2(⇡x/2a), which with the help of the useful formula

is readily computed to give hx2i = a2(13

� 2

2 ). Thus �x�p = h/2 with =q

2

3

� 2.

(Note that > 1, consistent with the general Heisenberg uncertainty relation �x�p � h/2.)

Page 2

Section 3 - Problem 2 Hughes

3 Potential jump (Humensky)

3.1 Problem

Consider the two potential energy functions:

V1(x) =mω2

2x2, x > 0

= ∞, x ≤ 0

V2(x) =mω2

2x2, −∞ ≤ x ≤ ∞

A particle with mass m, subject to the potential V1, is initially in its quantum ground state.

(a) Write the normalized wave function, φ(x), of this state.

(b) At a certain time (say, t = 0), the impenetrable barrier at x = 0 is very suddenly

removed so that for all t > 0 the same system is subject to the potential V2 instead of

V1. If the energy is then measured, what is the probability of finding the system in its

new ground state, ψ0(x)?

3.2 Solution

(a) Find φ(x) for V1(t = 0).

H|E〉 = E|E〉

x > 0 :

(p2

2m+

1

2mω2x2

)|E〉 = E|E〉

For x > 0, the differential equation is identical to that of the harmonic oscillator, so

for x > 0 the wavefunctions take the form of harmonic oscillator energy eigenstate

wavefunctions, i.e. Hermite polynomials times a Gaussian. Due to the infinite potential

for x ≤ 0, we must in addition require φ(0) = 0. This is the case for the eigenfunctions

antisymmetric under x → −x. The lowest energy state satisfying this is the n = 1

excited state of the harmonic oscillator, which takes the form

φ(x) = Ax e−ax2/2 , a =

h,

where A is fixed by requiring 1 =∫∞

0dx φ(x)2 = 1

2A2∫∞−∞ dx x

2e−ax2. The integral

can be evaluated by observing it equals (−∂a)∫dx e−ax

2= (−∂a)

√π√a

=√π

2 a3/2 . Thus

A = 2(mωh

)3/4π−1/4.

Page 3

Section 3 - Problem 3Humensky

(b) Probability to be in ground state after barrier removed: now the ground state is the

usual ground state for a harmonic oscillator:

ψ0(x) =(mωπh

)1/4

e−(

mωx2

2h

)

The probability to be in this state for t > 0 is time-independent and given by

P (ψ0) = |〈ψ0|φ〉|2

where

〈ψ0|φ〉 =2√π

(mωh

)∫ ∞0

x e−mωx2

h dx

To solve this integral, change variables to y = (mω/h)x2; then dy = 2 (mω/h)xdx,

〈ψ0|φ1〉 =1√π

∫ ∞0

e−ydy =1√π,

and

P (ψ0) = |〈ψ0|φ〉|2 =1

π.

Page 4

Section 3 - Problem 3Humensky

4 Quasi-classical states (Will)

4.1 Problem

In this problem we consider a one-dimensional quantum harmonic oscillator with frequency

ω and mass m. The Hamiltonian of the system is

H = hω(n+ 1/2).

The orthonormal eigenstates of the Hamiltonian are denoted by |n〉 and have the property

that n|n〉 = n|n〉 where n = a†a.

The eigenstates of the operator a are called quasi-classical states for reasons that we

examine in this problem. Consider an arbitrary complex number α.

(a) Show that the state

|α〉 = e−|α|2/2∑n

αn√n!|n〉

is an eigenstate of a with eigenvalue α, that is a|α〉 = α|α〉. Also show that it is properly

normalized.

(b) Calculate the expectation value of energy 〈H〉 = 〈α|H|α〉 for a quasi-classical state |α〉.

(c) Also calculate the expectation values of position 〈x〉 and momentum 〈p〉. Recall that

x =

√h

2mω(a+ a†) and p = −i

√mhω

2(a− a†).

(d) Calculate the root mean square deviations ∆x and ∆p for the position and the momen-

tum in this state. Show that ∆x∆p = h/2.

(e) Now, suppose that at time t = 0 the oscillator is in a quasi-classical state |α〉 with

α = |α|eiφ, where |α| is a real positive number. Show that at any later time t the

oscillator is also in a quasi-classical state that can be written as e−iωt/2|α(t)〉. Determine

the value of α(t) in terms of |α|, φ, ω, and t.

(f) Evaluate 〈x〉(t) and 〈p〉(t). Justify briefly why these states are called quasi-classical.

4.2 Solution

Page 5

Section 3 - Problem 4Will

Section 3 - Problem 4Will

Section 3 - Problem 4Will

Section 3 - Problem 4Will

5 Ground state of double delta potential (Weinberg)

5.1 Problem

Consider a particle in one dimension subject to the double-delta-function potential energy

V (x) = −g δ(x− a)− g δ(x+ a)

where g is a positive constant.

(a) Find an equation from which the energy of the lowest bound state can be determined.

(b) Using this equation, find approximate expressions for the ground state energy in the

limits where a is very large and where a approaches zero.

5.2 Solution

Page 9

Section 3 - Problem 5Weinberg

Section 3 - Problem 5Weinberg