580 FINAL EXAM SOLUTIONS December 13, 2011

Problem 1. (a) From the given information, we have

< x|H|x′ >= − 1

a2

−2 1 0 11 −2 1 00 1 −2 11 0 1 −2

(b)

H|Ψ1 >= − 1

a2

−2 1 0 11 −2 1 00 1 −2 11 0 1 −2

√

1

2

10−10

=2

a2

√1

2

10−10

H|Ψ2 >= − 1

a2

−2 1 0 11 −2 1 00 1 −2 11 0 1 −2

√

1

2

010−1

=2

a2

√1

2

010−1

E =

2

a2

(c)

V |Ψ1 >=

0 iV0 0−iV0 0 0 0

0 0 0 iV00 0 −iV0 0

√

1

2

10−10

= −iV0

√1

2

010−1

so

V |Ψ1 >= −iV0|Ψ2 >

andV21 = −iV0V12 = +iV0

Degenerate perturbation theory, so since V11 = V22 = 0, we have

E(1) = ±1

2

√4V21V12 = ±V0.

The final eigenvalues are2

a2± V0

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 Schrodingerequation to show that Jr satisfies the current conservation

equation

1

r2∂

∂r(r2Jr) = 0

1

r2∂r(r

2Jr) =1

2i

(Ψ∗∂2rΨ + Ψ∗

2

r∂rΨ− (∂2rΨ

∗)Ψ− (2

r∂rΨ

∗)Ψ)

=

(Ψ∗(

l(l + 1)

r2+ U(r)− k2)Ψ− (

l(l + 1)

r2+ U(r)− k2)Ψ∗Ψ

)= 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?

∂rΨ→ (ik

r− 1

r2)Aeikr − (

ik

r+

1

r2)Be−ikr

Jr →1

2i

(1

r(A∗e−ikr +B∗eikr)(Aeikr(

ik

r− 1

r2)−Be−ikr(ik

r+

1

r2))

)

− 1

2i

((A∗e−ikr(

−ikr− 1

r2)−B∗eikr(−ik

r+

1

r2))

1

r(Aeikr +Be−ikr)

)

=k

r2(A∗A−B∗B)

∂r(r2Jr) = 0 → A∗A−B∗B is constant

Note Using the whole range of r, it can be shown that A∗A = B∗B. (not

required here)

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

+∞. The scaled Hamiltonian is given by

H ′ =2m

h̄2H = − 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.

< Ψt|H ′|Ψt >=∫ ∞−∞

Ψt(−∂2x − αe−λ|x|)Ψt

= β(β

4− α

β + λ)

NOTE: ∫ ∞−∞

Ψt − ∂2xΨtdx =∫ ∞−∞

β

2(−β

2

4+ βδ(x))e−β|x| =

β2

4,

and

−α(β

2)∫ ∞−∞

e−(λ+β)|x| = − αβ

λ+ β

item[(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.)

< H ′ > is negative for 0 < β < β1 where β1(β1 + λ) = 4α.

∂

∂β< H ′ >= 0

at β2 where β2(β2 + λ)2 = 2αλ.

(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? No

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

mean a bound state does exist? Yes

(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? No

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

h̄λx0

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

Recall the following results from class:

XI = x0(ae−iωt + a†eiωt) where x0 =

√h̄

2mω,

PI = −ip0(ae−iωt − a†eiωt) where p0 =

√h̄mω

2,

and

< OI(t) >=< ΨI(t)|OI(t)|ΨI(t) >

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

< XI(t) >= x0 < ae−iωt + a†eiωt >= x0(iλx0h̄

e−iωt − iλx0h̄

eiωt)

=iλx20h̄

(e−iωt − eiωt) =λ

2imω(eiωt − e−iωt) =

λ

mωsinωt

< PI(t) >= −ip0 < ae−iωt−a†eiωt >= −i√h̄mω

2

iλ

h̄

√h̄

2mω(e−iωt+eiωt) = λ cosωt

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

< XI(t)XI(t) >= x20 < (ae−iωt + a†eiωt)(ae−iωt + a†eiωt) >

= x20(α2e−2iωt + α∗α+ < aa† > +(α∗)2e2iωt).

Using aa† = a†a+ 1, we have

< XI(t)XI(t) >= x20(α2e−2iωt + 2α∗α + 1 + (α∗)2e2iωt)

=< XI(t) >< XI(t) > +x20

< PI(t)PI(t) >= (−ip0)2 < (ae−iωt−a†eiωt)(ae−iωt−a†eiωt) >=< PI(t)PI(t) > +p20

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

H = h̄ω(a†a+1

2)

< H >= h̄ω(α∗α +1

2) =

λ2

2m+h̄ω

2

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

by

V0(x) = − h̄2λ

2mδ(x)

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

Eb = − h̄2λ2

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.

Ek ≡(h̄k)2

2m, Eb ≡ −

(h̄λ)2

8m

Energy Conservation states h̄ω0 + Eb = Ek so the condition on ω0 is

ω0 >1

h̄(−Eb)

(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.

Solution to (b) and (c) together:

Matrix Element:

< f |U (1)|i >=−ih̄

∫ ∞−∞

dt < f |Weik0x|i > exp(i

h̄(Ek − Eb − h̄ω0))

=−ih̄< f |Weik0x|i > 2πδ(

Ek − Ebh̄

− ω0)

where

< f |Weik0x|i >=∫ ∞−∞

dxW√2πe−i(k−k0)x

√λ

2e−λ|x|/2 = W

√λ

4π

λλ2

4+ (k − k0)2

Rate: ∫dk|W |2λ

4π(

λλ2

4+ (k − k0)2

)22πδ(Ek − Eb

h̄− ω0)

Treatment of δ function:

δ(Ek − Eb

h̄− ω0) =

2m

h̄δ(k2 − k2f ) =

2m

2kf h̄(δ(k − kf ) + δ(k + kf ))

where

kf =

√2mω0

h̄− λ2

4

Rate to go to either k > 0 or k < 0:

R± =|W |2λ3m

2h̄kf

1

(λ2

4+ (k − k0)2)2

k=±kf

Ratio of rates when kf = k0.

R+

R−=

(λ

2

4+ (2k0)

2)

(λ2

4)

2

Comment: When the particle goes toward k > 0 it may pick up k0 from the perturb-

ing potential (similar to the way it picks up energy.) However, when the particle

goes toward k < 0 it must get its wave-vector (momentum) from the momenta in

the wave function of the bound state.

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: (h̄ = 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. The top

state isj = 2,m = 2 : |11 >

Lowering once we get

j = 2,m = 1 :1√2

[|10 > +|01 >]

Lowering again we get

j = 2,m = 0 :1√6

[2|00 > +|1− 1 > +| − 11 >]

(b) Construct the state with total angular momentum j = 1, and m = 0. Con-structing the state with m = 1 and orthogonal to j = 2,m = 1 we get

j = 1,m = 1 :1√2

[|10 > −|01 >]

Lowering once we obtain

j = 1,m = 0 :1√2

[|1− 1 > −| − 11 >

(c) Construct the state with total angular momentum j = 0, and m = 0. Con-structing a state with m = 0, orthogonal to j = 2,m = 0 and j = 1,m = 0 wehave

j = 0,m = 0 :1√3

[|1− 1 > +| − 11 > −|00 >]

EXPLICIT CALCULATIONS REQUIRED. NO CREDIT FOR QUOT-ING FROM TABLES