## Assignment 11, solutions

Problem 1:

Consider a composite system made of two spin ½ particles.  For t < 0 the Hamiltonian does not depend on time and can be taken to be zero.  For t > 0 the Hamiltonian is given by  H = (4∆/ħ2)S1S2, where ∆ is a constant.  Suppose that the system is in the state |+ -> for t ≤ 0.  Find, as a function of time, the probability for being in each of the states |+ +>, |+ ->, |- +>, and |- ->,
(a)  by solving the problem exactly, using |Ψ(t)> = U(t, t0)Ψ(t0)> and
(b)  by solving the problem assuming the validity of first-order time-dependent perturbation theory with H as a perturbation which is switched on a t = 0.
(b)  Under what conditions does the perturbation calculation disagree with the exact solution and why?

Solution:

 Concepts: The state space of two spin ½ particles, time-dependent perturbation theory. Reasoning: S1∙S2 = ½(S2 – S12 – S22).  The eigenstates of S1∙S2 are the singlet state and the triplet states, {|S, Ms>}, S = 0, 1.  These are common eigenfunctions of S2 and Sz. Details of the calculation: (a)  The common eigenfunctions of S2 and Sz are|1,1> = |++>,|1,0> = ½½(|+-> + |-+>),|1,-1> =|-->,|0, 0> = ½½(|+-> - |-+>).Therefore |+-> = ½½(|0,0> + |1,0>), |-+> = ½½(|0,0> - |1,0>).For the singlet state we have S1∙S2 |0,0> = -(3/4)ħ2,  H |0,0> = -3∆|0,0>.For the triplet states we have S1∙S2 |1, Ms> = (1/4)ħ2,  H |1, Ms> = ∆|1, Ms>.|ψ(0)> = |+-> = ½½(|0,0> + |1,0>), |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(<0,0| + <1,0|)(exp(i3∆*t/ħ)|0,0> + exp(-i∆*t/ħ)|1,0>) =  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/ħ << 1, P(t) ~ (2∆*t/ħ)2. (b)  Time-dependent perturbation theoryPif(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 = .  Here Wfi = .  Wfi = 0 for |f> = |+ +> and |f> = |- ->, so Pif(t) = 0 for these states.For |f> = |- +> we have Wfi = 2-½<- +|H| (|00> + |10>) = 2-½[-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 2:

A hydrogen atom with Hamiltonian H0(r) is placed in a time-dependent electric field E = E(t) k. The perturbed Hamiltonian is H(r,t) = H0(r) + H’(r,t).
(a)  Show that H’(r,t) = qeE(t) r cos(θ).
(b)  Assuming the electron is initially in the ground state, and recalling that the first excited state of hydrogen is quadruply degenerate, to which state of the quadruply degenerate first excited states is a dipole transition from the ground state possible?  Prove this.
(c)  If the electron is in the ground state at t = 0, find  the probability (to first order in perturbation theory) that at time t the electron will have made the transition to the state determined in (b), as a function of E(t).

Solution:

 Concepts: Time dependent perturbation theory Reasoning: We consider the interactions of the atomic electrons with the electric field perturbation to the atomic Hamiltonian. Details of the calculation: (a)  The uniform electric field changes the energy of the proton and the electron, and therefore perturbs the Hamiltonian.The energy of the proton in the field is -∫0zpqeE(t)dz = -qeE(t)zp.The energy of the electron in the field is ∫0zeqe E(t)dz = qeE(t)ze.The potential energy of both particles is qeE(t)(ze - zp) = qeE(t)z = qeE(t) r cos(θ).(b)  For a dipole transition to be allowed, we need <Φf|H’(r,t)|Φi> to be nonzero.Φnlm(r,θ,φ) = Rnl(r)Ylm(θ,φ).The ground state wave function is Φ100(r,θ,φ)  =  π-½ a0-3/2 exp(-r/a0).The wave functions of the quadruply degenerate first excited states areΦ200(r,θ,φ)  = (4π)-½ (2a0)-3/2 (2 - r/a0) exp(-r/(2a0)),Φ211(r,θ,φ)  = (8π)-½ (2a0)-3/2 (r/a0) exp(-r/(2a0)) sinθ eiφ,Φ210(r,θ,φ)  = (4π)-½ (2a0)-3/2 (r/a0) exp(-r/(2a0)) cosθ,Φ21-1(r,θ,φ) = (8π)-½ (2a0)-3/2 (r/a0) exp(-r/(2a0)) sinθ e-iφ.<Φ100| H’(r,t)|Φ2lm>  ∝  ∫0∞r3dr R00*(r)R2l(r)  ∫0πsinθ dθ∫02πdφ Y00*(θ,φ) cosθ Ylm(θ,φ)∝ ∫0πsinθ dθ∫02πdφ Y00*(θ,φ) Y10(θ,φ) Ylm(θ,φ).We can invoke the properties if the spherical harmonics.Or we can show explicitly<Φ100|H’(r,t)|Φ200> ∝ ∫0πsinθ cosθ dθ = 0,   <Φ100|H’(r,t)|Φ211> ∝ ∫02πdφ eiφ = 0,<Φ100|H’(r,t)|Φ21-1> ∝ ∫02πdφ e-iφ = 0,   <Φ100|H’(r,t)|Φ210> ∝ ∫0πsinθ dθ cos2θ ≠ 0.The only transition from the ground state to an n = 2 state that is allowed is the transition to the n = 2, l = 1, m = 0 state, Φ210(r,θ,φ). (c)  Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2,with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <Φf|H’(r,t)|Φi>, with H’(r,t)  = qeE(t) r cos(θ).Let Φi = Φ100 and Φf = Φ210.  Then ωfi = ω21 = (E2 – E1)/ħ = (10.2 eV)/ħ and Wfi(t) = <Φ210|H’(r,t)|Φ000> = [qeE(t) /(23/2a04)] ∫0∞r4dr exp(-3r/(2a0))∫0πsinθ dθ cos2θ= qeE(t) 2½ 27 a0/35.Therefore the probability that at time t the electron will have made the transition is Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2 = 0.555 qe2a02|∫0texp(iω21t')E(t')dt'|2.

Problem 3:

Conside a 2-dimensional system containing a gas of electrons completely free to move in the x direction but confined by a square-well potential of infinite depth and total width w in the y direction.  In such a system, the electrons can often be approximated as non-interacting, provided that the mass of the electron is replaced by an effective mass.  Assume for this problem that the effective mass of the electrons is about 1/10 the mass of free electrons.
(a)  This system contains states that involve quantum-mechanical motion in both the x and y directions; describe qualitatively the nature of the absorption spectrum that you expect.
(b)  Write or derive a formula for the discrete levels expected for quantized motion in the y direction in terms of the width w.
(c)  How small does the width w have to be before the transition energy between the first two discrete levels found in b is larger than the average energy available from thermal excitation at room temperature?

Solution:

 Concepts: Separation of variable, the infinite square well Reasoning: This is a two-dimensional problem, with confinement in one dimensions. Details of the calculation: (a)  H  = (-ħ2/(2m))[∂2/∂x2 + ∂2/∂y2] + U(y), H ψ(x,y) = E ψ(x,y). Separation of variables is possible. ψ(x,y) = Χ(x) Φ(y).  E = Ex + Ey.  (-ħ2/(2m))∂2Χ(x)/∂x2  = ExΧ(x). Ex is a continuous eigenvalue.  Ex = ħk2/(2m). (-ħ2/(2m))∂2Φ(y)/∂y2 + U(y) Φ(y) = EyΦ(y). Ey is quantized.  The Eyn are the eigenenergies of the infinite square well.  E = ħk2/(2m) + Eyn,  k = (2m(E – Eyn)/ħ2)½. The density of states dN/dE dk/dE = [m/(2ħ2(E – Eyn))]½ will have spikes when E = Eyn, and the absorption spectrum, for example, should reflect those spikes.  (Fermi's golden rule.) (b) Eyn = n2π2ħ2/(2mw2) (c)  3π2ħ2/(2mw2) > kT,  w < 20 nm.

Problem 4:

In a one-dimensional structure a particle of mass m is in the ground state of a potential energy function U(x) = ½ kx2.  A phase transition occurs, and the effective spring constant suddenly doubles.  What is the probability that the particle will be found in an exited state after the phase transition?  Give a numerical answer.

Solution:

 Concepts: The QM harmonic oscillator, the sudden approximation Reasoning: The sudden approximation can be used to calculate transition probabilities when the Hamiltonian changes rapidly.  The reaction time is so short that the transition amplitude <β|U(t2,t1)|α> is simply given by the overlap <β|α>.  The transition probability is |<β|α>|2.  Here |α> is the eigenstate of the Hamiltonian before the transition and |β> is the eigenstate of the Hamiltonian after the transition. Details of the calculation: The normalized ground-state wave function of the 1D harmonic oscillator for t < 0 is Φ0i(x) = (mω/(πħ))¼exp(-½mω2x2/ħ),  ω = (k/m)½. The normalized ground-state wave function for t > 0 is Φ0f(x) = (m√(2)ω/(πħ))¼exp(-½m√(2)ω2x2/ħ). The probability that the oscillator remains in the ground state is P = |<Φ0i|Φ0f>|2 =  |∫-∞∞dx Φ0i*(x) Φ0f(x)|2 = 25/4/(1 + √2) = 0.985. The probability that the particle will be found in an exited state after the phase transition is 1 - 0.985 = 0.015.

Problem 5:

In one dimension, consider a spinless particle trapped in a delta-function potential U(x) = -Cδ(x), C > 0.
At t = 0, a time dependent perturbation W(t) = Wcosωt is turned on.
Assume ω >> mC2/(2ħ3) so that the particle can be ejected from the trap.  Use perturbation theory to find the transition rate.
You can assume that the free particles will be in a box of size L,  L = very large.

Solution:

 Concepts: Fermi's golden rule, the delta-function potential Reasoning: Only one bound state exist in the delta-function potential.  This is the initial state.  The final state of the particle is a continuum state. Details of the calculation: (a)  Let H0 = p2/(2m) - Cδ(x). Normalized eigenfunction of H0:  Φ(x) = ρ½ exp(-ρ|x|),  ρ2 = -2mEi/ħ2,  Ei = -mC2/(2ħ2),  ρ = mC/ħ2. To find the transition rate we use Fermi's golden rule. If W(t) = Wcos(ωt), then the transition probability per unit time is given by w(i,E) = (π/(2ħ))ρ(E)|WEi|2δE-Ei,ħω, where WEi = <ΦE|W|Φi>. Since ħω >> -mC2/(2ħ2), we can assume that the ejected particle is a nearly free particle with energy E = ħω + Ei and k2 = 2mE/ħ2. We assume that the particle is confined to a region of width L >> a and  use periodic boundary conditions.  Then Φk(x) = L-½exp(ikx), with k = 2πn/L,  n = 0, ±1, ±2, ...  is the final state wave function. The number of states with wave vectors whose magnitudes lie between k and k + dk is dN = 2 dk/(2π/L)  = Ldk/π.  (The particle can move towards the left or to the right.) dN/dk = L/π. The density of states is dN/dE = (dN/dk)(dk/dE). With E = ħ2k2/(2m) we have ρ(E) = dN/dE = Lm/(πkħ2). WEi = (ρ/L)½W[∫0∞dx eikxe-ρx  + ∫-∞0dx eikxeρx] = (ρ/L)½W 2ρ/(ρ2 + k2). w(i,E) = (π/(2ħ))(Lm/(πkħ2))(ρ/L)W2 4ρ2/(ρ2 + k2)2 = (2mW2/ħ3) ρ3/[k(ρ2 + k2)2]. (ρ2 + k2)2 = 4m2ω2/ħ2, ρ3/[(ρ2 + k2)2] = mC3/(4ħ4ω2),  This is the transition probability per unit time.