### Mixed method problems

#### Problem:

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 H0 = p2/(2m) + kx2/2 are un(x) = NnHn(αx)exp(-α2x2/2)
with Nn = [α/(π½2nn!)]½, α = (mk/ħ2)¼  and Hn(αx) a Hermite polynomial.
Recursion relation: Hn+1(αx) = 2αxHn(αx) - 2nHn-1(αx).
H0(αx) = 1,   H1(αx) = 2αx,   H2(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 H0.
• Details of the calculation:
(a)  For t < 0 we have H0 = p2/(2m) + kx2/2 = (ħ2/2m)∂2/∂x2 + kx2/2.
The eigenfunctions of H0 are un(x,t) = un(x)exp(-iEnt/ħ), with
En = (n + ½)ħω, ω = (k/m)½.
The un(x) are given, un(x) = NnHn(ax)exp(-a2x2/2).
For t > 0 the system is displaced by a distance Δx = d.
H = (ħ2/2m)∂2/∂x2 + k(x-d)2/2.
Let x' = x-d.  Then (ħ2/2m)∂2/∂x'2 + kx'2/2.
The eigenfunctions are un'(x,t) = un(x')exp(-iEnt/ħ) = un(x-d)exp(-iEnt/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.
P01 = |<u0(x)|u1(x-d)>|2 = N02N12|∫-∞dx 2α(x-d)exp(-α2x2/2)exp(-α2(x-d)2/2)|2.
= [α2/(2π)]|∫-∞dx 2α(x-d)exp(-α2x2/2)exp(-α2(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 H0 at time T.  This probability changes with T.
H = H0 + H',  H' = -kdx + ½kd2.
Time-dependent perturbation theory yields
P01(t > T) = (1/ħ2)|∫0Texp(iω10t')H'10dt'|2,  ω10 = (E1 - E0)/ħ = ω = (k/m)½.
H'10 = -kd<u1|x|u0>.
<u1(x)|x|u0(x)> = N0N1-∞dx H1(αx) x H0(αx) exp(-α2x2)
= N0N1(½α)∫-∞dx  (H2(αx) + 2H0(αx)) H0(αx) exp(-α2x2)
= N0N1(1/α)∫-∞dx H02(αx) exp(-α2x2) = N1/(N0α).
P01(t > T) = (N12/(N02α2ħ2)|∫0Texp(iωt')dt'|2
= (N12/(N02α2ħ2)(4/ω2)sin2(ωT/2) = 2/(α2ħ2ω2)sin2(ωT/2).

#### Problem:

(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) = U0, L/2 < x < L, with U0 << 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 have
ψn(x,t) = (2/L)½sin(nπx/L)exp(-iEnt/ħ).  En = n2π2ħ2/(2mL2).
(a)  Sudden approximation:
The probability that the electron to goes directly into the ground state of the new well is
P = |<ψ1(1.5L)|ψ1(L)>|2 = [(2/L)½(4/3L)½0L dx sin(πx/L) sin(2πx/3L)]2.
0L dx sin(πx/L) sin(2πx/3L) = (3L/π)∫0π/3 dy sin(3y) sin(2y)
= (3L/π)[sin(y)/2 - sin(5y)/10]| 0π/3 = (3L/π)(3√3)/10.
P = (8/3)(9/π2)(27/100) = 0.66.

(b)  Time dependent perturbation theory
During the time the perturbation is applied, the probability P(E2,t) of finding the system in the eigenstate |2> of H0 at time t, i.e . the probability of measuring the eigenvalue E2, is given by
P(E2,t) = (1/ħ2)|∫0texp((i/ħ)(E2-E1)t')<Φ2|W|Φ1>dt'|2
= |(<Φ2|W|Φ1>/ħ)∫0texp(iω21t')dt'|2.
[∫0texp(iω21t')dt' = (-i/ω21)(exp(iω21t) - 1)
= (-i/ω21)exp(iω21t/2)(exp(iω21t/2) - exp(-iω21t/2))
= (2/ω21)exp(iω21t/2)sin(ω21t/2)]
P(E2,t) = (|<Φ2|W|Φ2>|2/(ħω21)2) 4 sin221t/2).
ω21 = 3π2ħ/(2mL2).
2|W|Φ1> = (2/L)U0L/2L dx sin(2πx/L) sin(πx/L) = -4U0/(3π).
P(E2,t) = (16/9)2(m2L4U02/(ħ4π6))sin2(3π2ħt/(4mL2)).
P(E2,∆t) = (16/9)2(m2L4U02/(ħ4π6))sin2(3π2ħ∆t/(4mL2)).
For t > ∆t, Φ2 is an eigenstate of the Hamiltonian and the probability for the electron to be in the first excited state does not change.
P(E2,2∆t) = (16/9)2(m2L4U02/(ħ4π6))sin2(3π2ħ∆t/(4mL2)).

#### Problem:

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/ħ2) S1S2,

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:
S1S2 = ½(S2 – S12 – S22).  The eigenstates of S1S2 are the singlet state and the triplet states,
{|S, Ms>}, S = 0, 1.
|1,1> = |+ +>
|1,0> = ½½(|+ -> + |- +>),  |1,-1> = |- ->,  |0,0> = ½½(|+ -> - |- +>)
Therefore |+ -> = ½½(|0,0> + |1,0>), |- +> = ½½(|0,0> - |1,0>).
• Details of the calculation:
(a)  For the singlet state we have S1S2|0, 0> = -¾ħ2,   H|0, 0> = -3∆| 0, 0>.
For the triplet states we have S1S2|1, Ms> = ¼ħ2,   H |1, Ms> = ∆|1, Ms>.

|ψ(t)> = U(t, 0)|+ -> = 2(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 |+ -> = cos2(2∆*t/ ħ).
For 2∆*t/ ħ << 1, P(t) ~ 1.

P(t) of being in the state |+ -> = sin2(2∆*t/ ħ), since the total probability of being in one of the 4 basis states must be equal to 1.
For 2∆*t/ ωfi << 1, P(t) ~ (2∆*t/ ħ)2.

(b)  Time-dependent perturbation theory
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2.
The initial and the final state must be different for this formula to be valid.  For i = f, P(t) ~ 1.

Wfi = <f|H|i>.  Here Wfi = <f|H |+ ->.
Wfi = 0 for |f> = |+ +> an |f> = |- ->, so Pif(t) = 0 for these states.

For |f> = |- +> Wfi = 2<- +|H| (|00> + |10>)
= 2[-3∆<- +| 0, 0> + ∆<- +| 1, 0>] = ½ (-3∆ - ∆) = -2∆.

ωfi = (Ef – Ei)/ωfi = 0 for the unperturbed states.

Pif(t) = (1/ħ2)|∫0t(-2∆)dt'|2 = 4∆2t2/ ħ2.

(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.

#### Problem:

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 H0.
(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 E1 = A + C,  E2 = B,  and E3 = A – C.
The corresponding eigenvectors are |1’> = 2(|1> + |3>),   |2’> = |2>,  |3’> = 2 (|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
P13(t) = (|<3|W|1>|2/(ħω31)2)4 sin231t/2).
Since ω31 = 0,  we have P13(t) = |<3|W|1>|2t22 = C2 t22.
The perturbation is turned off at t = T,  so for t’ > T we have P13(t’) = C2 T22.
(d)  |1> = 2(|1’> + |3’>).
For 0 < t < T  |Ψ(t)> = 2(|1’> exp(-i(A + C)t/ ħ) + |3’> exp(-i(A - C)t/ ħ)).
For  t’ > T  |Ψ(t’)> = 2(|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 H0.
|Ψ(t’)> = 2(|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).
P13(t’) = |<3||Ψ(t’)>|2 = sin2(CT/ ħ).
If CT/ ħ  << π/2 then P13(t’) = C2 T22.  Then first order perturbation theory is a valid approach.

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