A one-dimensional quantum oscillator consisting of a mass m suspended
from a spring with spring constant k is initially in its lowest energy state.
At t = 0, the upper end of the spring is suddenly raised a distance d during a
time interval which is very short compared to an oscillator period.

(a) Give explicit expressions for the time-dependent eigenstates of the
Hamiltonian for t < 0 and those for t > 0 and discuss the relationship
between them in the context of this problem.

(b) Write down an integral for the probability that a transition has occurred to the first
excited state as a result of the disturbance at t = 0.

(c) Now consider a situation involving the above disturbance at t = 0,
followed by an interval from 0 < t < T, where T is large compared with
a ground-state period and during which the upper end of the spring remains fixed
at the disturbed or stretched position. At t = T the upper end of the
spring is suddenly returned to its lower, original position.

Derive the probability amplitude for the oscillator to be found in its
first exited state at times t > T.

The stationary states of H_{0} = p^{2}/(2m) + kx^{2}/2 are
u_{n}(x) = N_{n}H_{n}(αx)exp(-α^{2}x^{2}/2)

with N_{n} = [α/(π^{½}2^{n}n!)]^{½},
α = (mk/ħ^{2})^{¼}
and H_{n}(αx) a Hermite polynomial.

Recursion relation: H_{n+1}(αx) = 2αxH_{n}(αx)
- 2nH_{n-1}(αx).

H_{0}(αx) = 1, H_{1}(αx)
= 2αx, H_{2}(αx)
= 4(αx)^{2} - 2.

Solution:

- Concepts:

The 1-D harmonic oscillator, the sudden approximation, time-dependent perturbation theory - Reasoning:

The system is a 1-D harmonic oscillator. When the system is suddenly displaced, the sudden approximation can be used to calculate transition probabilities to the eigenstates of the new Hamiltonian. For a small perturbation that last a finite time, time-dependent perturbation theory can be use to calculate transition probabilities to different eigenstates of H_{0}. - Details of the calculation:

(a) For t < 0 we have H_{0}= p^{2}/(2m) + kx^{2}/2 = (ħ^{2}/2m)∂^{2}/∂x^{2}+ kx^{2}/2.

The eigenfunctions of H_{0}are u_{n}(x,t) = u_{n}(x)exp(-iE_{n}t/ħ), with

E_{n}= (n + ½)ħω, ω = (k/m)^{½}.

The u_{n}(x) are given, u_{n}(x) = N_{n}H_{n}(ax)exp(-a^{2}x^{2}/2).

For t > 0 the system is displaced by a distance Δx = d.

H = (ħ^{2}/2m)∂^{2}/∂x^{2}+ k(x-d)^{2}/2.

Let x' = x-d. Then (ħ^{2}/2m)∂^{2}/∂x'^{2}+ kx'^{2}/2.

The eigenfunctions are u_{n}'(x,t) = u_{n}(x')exp(-iE_{n}t/ħ) = u_{n}(x-d)exp(-iE_{n}t/ħ).

(b) Here we calculate the probability of finding the system in the first excited state of the perturbed Hamiltonian H. This probability does not change with time. We use the sudden approximation.

P_{01}= |<u_{0}(x)|u_{1}(x-d)>|^{2}= N_{0}^{2}N_{1}^{2}|∫_{-∞}^{∞}dx 2α(x-d)exp(-α^{2}x^{2}/2)exp(-α^{2}(x-d)^{2}/2)|^{2}.

= [α^{2}/(2π)]|∫_{-∞}^{∞}dx 2α(x-d)exp(-α^{2}x^{2}/2)exp(-α^{2}(x-d)^{2}/2)|^{2}.

(c) A perturbation is turned on at t = 0 and turned off at t = T. Here we calculate the probability of finding the system in the first excited state of the unperturbed Hamiltonian H_{0}at time T. This probability changes with T.

H = H_{0}+ H', H' = -kdx + ½kd^{2}.

Time-dependent perturbation theory yields

P_{01}(t > T) = (1/ħ^{2})|∫_{0}^{T}exp(iω_{10}t')H'_{10}dt'|^{2}, ω_{10}= (E_{1}- E_{0})/ħ = ω = (k/m)^{½}.

H'_{10}= -kd<u_{1}|x|u_{0}>.

<u_{1}(x)|x|u_{0}(x)> = N_{0}N_{1}∫_{-∞}^{∞}dx H_{1}(αx) x H_{0}(αx) exp(-α^{2}x^{2})

= N_{0}N_{1}(½α)∫_{-∞}^{∞}dx (H_{2}(αx) + 2H_{0}(αx)) H_{0}(αx) exp(-α^{2}x^{2})

= N_{0}N_{1}(1/α)∫_{-∞}^{∞}dx H_{0}^{2}(αx) exp(-α^{2}x^{2}) = N_{1}/(N_{0}α).

P_{01}(t > T) = (N_{1}^{2}/(N_{0}^{2}α^{2}ħ^{2})|∫_{0}^{T}exp(iωt')dt'|^{2}

= (N_{1}^{2}/(N_{0}^{2}α^{2}ħ^{2})(4/ω^{2})sin^{2}(ωT/2) = 2/(α^{2}ħ^{2}ω^{2})sin^{2}(ωT/2).

(a) An electron is located in the ground state in a 1D potential well,

U(x) = 0, 0 < x < L, U(x) = ∞ elsewhere.

Instantaneously the well becomes 1.5 times wider. What is the probability
for the electron to go directly to ground state in this new well?

(b) An electron is located in the ground state in a 1D potential well,

U(x) = 0, 0 < x < L, U(x) = ∞ elsewhere.

For a time interval ∆t, the well bottom is disturbed, and the potential
energy function becomes U(x) = 0, 0 < x < L/2, U(x) = U_{0}, L/2 < x <
L, with U_{0} << than the ground state energy. After the time interval
∆t the disturbance is removed. What is the probability for the electron to be
in the first excited state of the well at a time t = 2∆t?

Solution:

- Concepts:

The sudden approximation, time dependent perturbation theory - Reasoning:

We are asked to recognize when these approximation methods can be used. - Details of the calculation:

For a particle in an infinite well we haveψ

_{n}(x,t) = (2/L)^{½}sin(nπx/L)exp(-iE_{n}t/ħ). E_{n}= n^{2}π^{2}ħ^{2}/(2mL^{2}).

(a) Sudden approximation:

The probability that the electron to goes directly into the ground state of the new well is

P = |<ψ_{1}(1.5L)|ψ_{1}(L)>|^{2}= [(2/L)^{½}(4/3L)^{½}∫_{0}^{L}dx sin(πx/L) sin(2πx/3L)]^{2}.

∫_{0}^{L}dx sin(πx/L) sin(2πx/3L) = (3L/π)∫_{0}^{π/3}dy sin(3y) sin(2y)

= (3L/π)[sin(y)/2 - sin(5y)/10]|_{ 0}^{π/3}= (3L/π)(3√3)/10.

P = (8/3)(9/π^{2})(27/100) = 0.66.(b) Time dependent perturbation theory

During the time the perturbation is applied, the probability P(E_{2},t) of finding the system in the eigenstate |2> of H_{0}at time t, i.e . the probability of measuring the eigenvalue E_{2}, is given by

P(E_{2},t) = (1/ħ^{2})|∫_{0}^{t}exp((i/ħ)(E_{2}-E_{1})t')<Φ_{2}|W|Φ_{1}>dt'|^{2}

= |(<Φ_{2}|W|Φ_{1}>/ħ)∫_{0}^{t}exp(iω_{21}t')dt'|^{2}.

[∫_{0}^{t}exp(iω_{21}t')dt' = (-i/ω_{21})(exp(iω_{21}t) - 1)

= (-i/ω_{21})exp(iω_{21}t/2)(exp(iω_{21}t/2) - exp(-iω_{21}t/2))

= (2/ω_{21})exp(iω_{21}t/2)sin(ω_{21}t/2)]

P(E_{2},t) = (|<Φ_{2}|W|Φ_{2}>|^{2}/(ħω_{21})^{2}) 4 sin^{2}(ω_{21}t/2).

ω_{21}= 3π^{2}ħ/(2mL^{2}).

<Φ_{2}|W|Φ_{1}> = (2/L)U_{0}∫_{L/2}^{L}dx sin(2πx/L) sin(πx/L) = -4U_{0}/(3π).

P(E_{2},t) = (16/9)^{2}(m^{2}L^{4}U_{0}^{2}/(ħ^{4}π^{6}))sin^{2}(3π^{2}ħt/(4mL^{2})).

P(E_{2},∆t) = (16/9)^{2}(m^{2}L^{4}U_{0}^{2}/(ħ^{4}π^{6}))sin^{2}(3π^{2}ħ∆t/(4mL^{2})).

For t > ∆t, Φ_{2}is an eigenstate of the Hamiltonian and the probability for the electron to be in the first excited state does not change.

P(E_{2},2∆t) = (16/9)^{2}(m^{2}L^{4}U_{0}^{2}/(ħ^{4}π^{6}))sin^{2}(3π^{2}ħ∆t/(4mL^{2})).

Consider a system of two non-identical spin ½ particles. For t < 0 the particles do not interact and the Hamiltonian may be taken to be zero. For t > 0 the Hamiltonian is given by

H = (4Δ/ħ^{2}) **S**_{1}∙**S**_{2},

where Δ is a constant. For t < 0 the state of the system is |+ –>.

(a) For t > 0, find, as a function of time, the probability for finding
the system in each of the states |+ +>, |+ –>, |– +>, and |– –>, by solving the
problem exactly.

(b) For t > 0, find, as a function of time, the probability for finding
the system in each of the states |+ +>, |+ –>, |– +>, and |– –>, using
first-order time dependent perturbation theory with H a perturbation that is
switched on at t = 0.

(c) Under what condition is the perturbation calculation a bad approximation to
the exact solution, and why?

Solution:

- Concepts:

The state space of two spin ½ particles, time-dependent perturbation theory. - Reasoning:
**S**_{1}∙**S**_{2}= ½(S^{2}– S_{1}^{2}– S_{2}^{2}). The eigenstates of**S**_{1}∙**S**_{2}are the singlet state and the triplet states,

{|S, M_{s}>}, S = 0, 1.

|1,1> = |+ +>

|1,0> = ½^{½}(|+ -> + |- +>), |1,-1> = |- ->, |0,0> = ½^{½}(|+ -> - |- +>)

Therefore |+ -> = ½^{½}(|0,0> + |1,0>), |- +> = ½^{½}(|0,0> - |1,0>). - Details of the calculation:

(a) For the singlet state we have**S**_{1}∙**S**_{2}|0, 0> = -¾ħ^{2}, H|0, 0> = -3∆| 0, 0>.

For the triplet states we have**S**_{1}∙**S**_{2}|1, M_{s}> = ¼ħ^{2}, H |1, M_{s}> = ∆|1, M_{s}>.

|ψ(t)> = U(t, 0)|+ -> = 2^{-½}(exp(i3∆*t/ ħ)|0,0> + (exp(-i∆*t/ ħ)|1,0>)

P(t) of being in the state |+ +> = |1, 1> = 0.

P(t) of being in the state |- -> = |1, -1> = 0.

P(t) of being in the state |+ -> = |<+ -| ψ(t)>|^{2}.

<+ -| ψ(t)> = ½(exp(i3∆*t/ ħ) + (exp(-i∆*t/ ħ))

= ½ exp(i∆*t/ ħ) (exp(i2∆*t/ ħ) + (exp(-i2∆*t/ ħ)) = exp(i∆*t/ ħ) cos(2∆*t/ ħ).

P(t) of being in the state |+ -> = cos^{2}(2∆*t/ ħ).

For 2∆*t/ ħ << 1, P(t) ~ 1.

P(t) of being in the state |+ -> = sin^{2}(2∆*t/ ħ), since the total probability of being in one of the 4 basis states must be equal to 1.

For 2∆*t/ ω_{fi}<< 1, P(t) ~ (2∆*t/ ħ)^{2}.(b) Time-dependent perturbation theory

P_{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2}.

The initial and the final state must be different for this formula to be valid. For i = f, P(t) ~ 1.W

_{fi}= <f|H|i>. Here W_{fi}= <f|H |+ ->.

W_{fi}= 0 for |f> = |+ +> an |f> = |- ->, so P_{if}(t) = 0 for these states.For |f> = |- +> W

_{fi}= 2^{-½}<- +|H| (|00> + |10>)

= 2^{-½}[-3∆<- +| 0, 0> + ∆<- +| 1, 0>] = ½ (-3∆ - ∆) = -2∆.ω

_{fi}= (E_{f}– E_{i})/ω_{fi}= 0 for the unperturbed states.P

_{if}(t) = (1/ħ^{2})|∫_{0}^{t}(-2∆)dt'|^{2}= 4∆^{2}t^{2}/ ħ^{2}.(c) The time-dependent perturbation theory calculation result agrees with the exact result as long as x = 2∆*t/ ħ << 1, so that cos(x) = 1 and sin(x) = x are very good approximations. For large x (or t) the probability of finding the particle in the state |- +> oscillates between 0 and 1.

Consider a 3-state system with a Hamiltonian

in the {|i>} orthonormal basis, i = 1, 2, 3, with B, A > 0. For < 0, the
system is in state |1>.

At t = 0 the Hamiltonian suddenly changes to

with C << A, B, C > 0.

At t = T the Hamiltonian changes back to H_{0}.

(a) What are the eigenstates and eigenvalues if H expressed in terms of the {|i>}
basis vectors?

(b) For each of the eigenstates of H, what are the probabilities that the
system will be found in this state for 0 < t < T?

(c) Using first order time-dependent perturbation theory, what is the
probability of finding the system in state |3> at t’ > T?

(d) Solve part (c) exactly.

Solution:

- Concepts:

Transition probabilities, the sudden approximation, time-dependent perturbation theory, the exact solution - Reasoning:

We are asked to find transition probabilities, - Details of the calculation:

(a) The eigenvalues of H are E_{1}= A + C, E_{2}= B, and E_{3}= A – C.

The corresponding eigenvectors are |1’> = 2^{-½}(|1> + |3>), |2’> = |2>, |3’> = 2^{-½}(|1> - |3>}.

(b) For 0 < t < T the probabilities of finding the system in states |1’> and |3’> are ½, the probability of finding the system in state |2’> is zero.

(c) For a constant perturbation

P_{13}(t) = (|<3|W|1>|^{2}/(ħω_{31})^{2})4 sin^{2}(ω_{31}t/2).

Since ω_{31}= 0, we have P_{13}(t) = |<3|W|1>|^{2}t^{2}/ħ^{2}= C^{2}t^{2}/ħ^{2}.

The perturbation is turned off at t = T, so for t’ > T we have P_{13}(t’) = C^{2}T^{2}/ħ^{2}.

(d) |1> = 2^{-½}(|1’> + |3’>).

For 0 < t < T |Ψ(t)> = 2^{-½}(|1’> exp(-i(A + C)t/ ħ) + |3’> exp(-i(A - C)t/ ħ)).

For t’ > T |Ψ(t’)> = 2^{-½}(|1’> exp(-i(A + C)T/ ħ) + |3’> exp(-i(A - C)T/ ħ)) exp(-i(A(t’ – T)/ ħ),

since |Ψ(t’)> is an eigenstate of H_{0}.

|Ψ(t’)> = 2^{-½}(|1’> exp(-iCT/ ħ) + |3’> exp(iCT/ ħ)) exp(-i(A(t’/ ħ)

= ½[|1>(exp(-iCT/ ħ) + exp(iCT/ ħ)) + |3>(exp(-iCT/ ħ) - exp(iCT/ ħ))] exp(-i(A(t’/ ħ).

<3||Ψ(t’)> = i sin(CT/ ħ) exp(-i(A(t’/ ħ).

P_{13}(t’) = |<3||Ψ(t’)>|^{2}= sin^{2}(CT/ ħ).

If CT/ ħ << π/2 then P_{13}(t’) = C^{2}T^{2}/ħ^{2}. Then first order perturbation theory is a valid approach.

.