580_11_final_2

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580 FINAL EXAM December 13, 2011 All problems have equal weight. Try to make your solutions legible! A possibly useful integral: Z 0 exp(-αx + iβx)= 1 α - Problem 1. Consider a system which is (i) periodic in x, and (ii) allows the coordinate x to only take discrete values, x =0, a, 2a, 3a,..., (N - 1)a. The state vector of the system is |Ψ >, and the amplitude for the particle to found at x is <x|Ψ >. The state vector is normalized, x < Ψ|x >< x|Ψ >=1. The Hamiltonian is defined by <x|H |Ψ >= - 1 a 2 (<x + a|Ψ > -2 <x|Ψ > + <x - a|Ψ >) (a) Take the case N =4, and using the definition above, write out the matrix <x|H |x 0 >. This will be a 4 × 4 matrix, with rows and columns labeled by x/a =0, 1, 2, 3. Use the periodicity of the system, so e.g. < 4a|Ψ >=< 0|Ψ >, < -a|Ψ >=< 3a|Ψ >, etc (b) Show that the vectors |Ψ 1 > and |Ψ 2 > given by <x|Ψ 1 >= s 1 2 1 0 -1 0 <x|Ψ 2 >= s 1 2 0 1 0 -1 are eigenvectors of the matrix <x|H |x 0 >, with the same eigenvalue. Do this and find the eigenvalue E by writing out the equations X x 0 <x|H |x 0 >< x 0 |Ψ j >= E<x|Ψ j >, for j =1, 2 (c) A perturbing Hamiltonian V acts on the system where the matrix elements of V are 0 iV 0 0 -iV 0 0 0 0 0 0 0 iV 0 0 0 -iV 0 0 The common eigenvalue E will be split into two eigenvalues by the perturbing Hamiltonian. To first order in V 0 , calculate the shifts in the eigenvalue.

description

2nd quantization

Transcript of 580_11_final_2

Page 1: 580_11_final_2

580 FINAL EXAM December 13, 2011

All problems have equal weight. Try to make your solutions legible!A possibly useful integral:

∫ ∞

0exp(−αx + iβx) =

1

α− iβ

Problem 1. Consider a system which is (i) periodic in x, and (ii) allows the coordinate

x to only take discrete values, x = 0, a, 2a, 3a, . . . , (N − 1)a. The state vector of the

system is |Ψ >, and the amplitude for the particle to found at x is < x|Ψ > . The

state vector is normalized,∑

x < Ψ|x >< x|Ψ >= 1. The Hamiltonian is defined by

< x|H|Ψ >= − 1

a2(< x + a|Ψ > −2 < x|Ψ > + < x− a|Ψ >)

(a) Take the case N = 4, and using the definition above, write out the matrix< x|H|x′ > . This will be a 4 × 4 matrix, with rows and columns labeled byx/a = 0, 1, 2, 3. Use the periodicity of the system, so e.g.< 4a|Ψ >=< 0|Ψ >,< −a|Ψ >=< 3a|Ψ >, etc

(b) Show that the vectors |Ψ1 > and |Ψ2 > given by

< x|Ψ1 >=

√1

2

10−10

< x|Ψ2 >=

√1

2

010−1

are eigenvectors of the matrix < x|H|x′ >, with the same eigenvalue. Do thisand find the eigenvalue E by writing out the equations

x′< x|H|x′ >< x′|Ψj >= E < x|Ψj >,

for j = 1, 2

(c) A perturbing Hamiltonian V acts on the system where the matrix elements ofV are

0 iV0 0−iV0 0 0 0

0 0 0 iV0

0 0 −iV0 0

The common eigenvalue E will be split into two eigenvalues by the perturbingHamiltonian. To first order in V0, calculate the shifts in the eigenvalue.

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Problem 2. A radial wave function Ψl(r) satisfies the radial Schrodinger equation

(1

r2∂rr

2∂r − l(l + 1)

r2− U(r) + k2

)Ψl(r) = 0

The radial current is defined by

Jr(r) =1

2i(Ψ∗

l (r)∂rΨl − (∂rΨl(r))∗Ψl(r))

(a) Use the Schrodinger equation to show that Jr satisfies the current conservation

equation

1

r2

∂r(r2Jr) = 0

(b) At large r, Ψl takes the form

Ψl(r) → 1

r(Aeikr + Be−ikr)

Calculate Jr for large r. What condition does current conservation place on the

constants A and B?

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Problem 3. A particle is moving in one dimension with coordinate x ranging −∞ to

+∞. The scaled Hamiltonian is given by

H ′ =2m

h2 H = − d2

dx2+ U(x),

where U(x) = −α exp(−λ|x|) with both α and λ positive. A normalized trial wave

function for a possible bound state is given by

Ψt(x) =

√β

2exp(−β

2|x|)

(a) Calculate the expected value of H ′ in the state Ψt.

(b) Sketch your results as a function of β for fixed α, λ. Use your result in (a)

to derive a range of β values for which the expected value of H ′ is negative.

Derive a second condition for the value of β which gives the lowest value of the

expectation of H ′. (NOTE Both of these conditions may be left as equations.

It is not necessary to solve them.)

(c) Answer the following questions yes or no, no explanation needed.

(1) If for a certain value of β, the expectation of H ′ is positive, does that mean

a bound state does not exist?

(2) If for a certain value of β, the expectation of H ′ is negative, does that

mean a bound state does exist?

(3) If for a certain value of β, the expectation of H ′ is negative, is the eigen-

value of H ′ greater than this expected value?

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Problem 4. A standard harmonic oscillator has particle mass m and frequency ω. The

oscillator is acted upon by an external force given by F (t) = λδ(t). For t > 0, the

oscillator’s interaction picture state vector is a coherent state given by

|ΨI(t) >= |α(t) >, where α(t) =i

hλx0

so for this external force, α is constant in time.

Recall the following results from class:

XI = x0(ae−iωt + a†eiωt) where x0 =

√h

2mω,

PI = −ip0(ae−iωt − a†eiωt) where p0 =

√hmω

2,

and

< OI(t) >=< ΨI(t)|OI(t)|ΨI(t) >

(a) Compute < XI(t) >, and < PI(t) > for t > 0.

(b) Compute < XI(t)XI(t) >, and < PI(t)PI(t) > for t > 0.

(c) Compute the expected value of the energy of the oscillator for t > 0.

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Problem 5. A particle of mass m is moving in one dimension in a static potential given

by

V0(x) = − h2λ

2mδ(x)

In this potential, there is a bound state with energy and wave function given by

Eb = − h2λ2

8m, Ψb(x) =

√λ

2exp(−λ

2|x|)

In adddition, there is an external time dependent potential V1(x, t) acting. This is

given by

V1(x, t) = W exp(−iω0t + ik0x) + W ∗ exp(+iω0t− ik0x),

where W,ω0, k0 are all constants, with ω0 > 0, k0 > 0.

(a) Find the range of values of ω0 for which it is possible to knock the particle out

of the bound state and into a plane wave state.

(b) Assuming ω0 is in the allowed range, calculate the transition rate for the particle

to be emitted in a plane wave state with wave vector k > 0.

(c) Do the same for the particle to be emitted into a plane wave state with wave

vector k < 0. Take the ratio of the rate for k > 0 to that for k < 0. Comment

briefly on the magnitude of this ratio for the case |k| = k0.

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Problem 6. This problem concerns adding two angular momentum 1 states. This could

occur if two electrons both had l = 1. The allowed values of total angular mo-

mentum for this situation are 2, 1, 0. The angular momentum operators for angular

momentum 1 are: (h = 1 here)

Lx + iLy =√

2

0 1 0

0 0 1

0 0 0

Lx − iLy =√

2

0 0 0

1 0 0

0 1 0

Lz =

1 0 0

0 0 0

0 0 −1

(a) Construct the state with total angular momentum j = 2, and m = 0.

(b) Construct the state with total angular momentum j = 1, and m = 0.

(c) Construct the state with total angular momentum j = 0, and m = 0.

EXPLICIT CALCULATIONS REQUIRED. NO CREDIT FOR QUOT-

ING FROM TABLES