More Problems

Problem 1:

A particle of mass m is in the ground state of an infinite square well (U = 0 for 0 < x < a and U = ∞ otherwise).
At time t = 0, the right "wall" (i.e. at x = a) shifts to x = 2a.  A measurement of the energy of the particle is made just after the wall shifts. (Assume that all this happens so quickly so the spatial wave function of the particle does not change).  What is the probability that the energy measurement yield a value EXACTLY the same as the energy of the ground state of original well?

Solution:

• Concepts:
  A particle in an infinite square well, the sudden approximation
• Reasoning:
  The reaction time is so short that the transition amplitude <β|U(t₂,t₁)|a> 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:
  For t < 0, ψ_0i(x) = (2/a)^½sin(πx/a),  E_0i = π²ħ²/(2ma²).
  For t > 0, the wave function of the first excited state is ψ_1f(x) = (1/a)^½sin(πx/a).
  E_nf = n²π²ħ²/(8ma²).   E_1f = E_0i.  If we find the particle in state ψ_1f its energy has not changed.
  The wave function just after the wall is removed is ψ_0i(x).
  The probability of finding the particle in the state ψ_1f(x) of the new well is |<ψ_1f|ψ_0i |².
  |<ψ_1f|ψ_0i>|² = (2/a²)| ∫₀^asin²(πx/a)dx|² = ½.

Problem 2:

Consider a particle of mass m in one-dimensional, infinitely deep well.  Let U(x) = 0, for 0 < x < a, and U(x) = ∞ everywhere else.  The eigenstates of H₀ = p²/(2m) + U(x) are Φ_n(x) = (2/a)^½sin(nπx/a) with eigenvalues E_n = n²π²ħ²/(2ma²).
For  t > 0 the system is subjected to a time-dependent perturnation W(x,t) = W(x) exp(-t/τ), with
W(x) = W₀, 0 < x < a/2, W(x) = 0 everywhere else, H = H₀ + W.
If at t = 0 the system is in the ground state Φ₁(x), what is the probability of finding it in the excited state Φ₂(x) at time t?   Show that as t approaches infinity (with τ finite), the limit of your expression is independent of time.

Solution:

• Concepts:
  The 1D infinite well, time dependent perturbation theory
• Reasoning:
  W is a small correction to H₀.  The problem governed by H₀ is already solved.
  W = 0  for  t < 0.  The probability P(E₂,t) of finding the system in the eigenstate Φ₂(x) of H₀ at time t, i.e. the probability of measuring the eigenvalue E₂, is
  P(E₂,t) = (1/ħ²)|∫₀^texp((i/ħ)(E₂-E₁)t')<Φ₂|W(x,t)|Φ₁>dt'|².
  This is the result of first order time dependent perturbation theory.  P(E₂,t) is the transition probability.
• Details of the calculation:
  (a) Time dependent perturbation theory:
  P₂₁(t) = (|W₂₁|/ħ)²|∫₀^texp(iω₂₁t')exp(-t/τ)dt'|²_.
  ω₂₁ = (E₂-E₁)/ħ = 3π²ħ/(2ma²).
  W₂₁ = (2W₀/π)∫₀^a/2sin(2πx/a) sin(πx/a) dx = 4W₀/(3π).
  ∫₀^texp((iω₂₁-1/τ)t')dt' = [τ/(iω₂₁τ-1)][exp((iω₂₁-1/τ)t) - 1].
  |∫₀^texp(iω₂₁t') e^-t'/τdt'|² = [τ²/(ω₂₁²τ² + 1)][1 + exp(-2t/τ) + 2exp(-t/τ)cos(ω₂₁t)].
  P₂₁(t) = [4W₀/(3πħ)]² [τ²/(ω₂₁²τ² + 1)][1 + exp(-2t/τ) + 2exp(-t/τ)cos(ω₂₁t)].
  As t -->infinity, |∫₀^texp(iω₂₁t') e^-t'/τdt'|² = [τ²/(ω₂₁²τ² + 1)].
  Since the perturbation decreases exponentially with time, we expect the probability that a transition has occurred to no longer change as t --> infinity.
  P₂₁(t --> infinity) = [4W₀/(3πħ)]² [τ²/(ω₂₁²τ² + 1)].

Problem 3:

Consider a free particle of mass M in two dimensions.  Since continuum states are not traditionally normalizable, we assume that the particle is confined to a cubical 2D box with periodic boundary conditions.  Its eigenfunctions are
ψ_nq(x,y) = (1/L)exp(ik_xx)exp(ik_yy),
with k_x = 2πn/L,  k_y = 2πq/L, n, q = 1, 2, ... .
Derive and expression for the density of states dN/dE = ρ(E).

Solution:

• Concepts:
  The density of states
• Reasoning:
  It is customary to use periodic boundary conditions for continuum states.
• Details of the calculation:
  The energy eigenvalue associated with the eigenstate
  ψ_nq(x,y) = (1/L)exp(ik_xx)exp(ik_yy) is
  E_nq = (n² + q²)4π²ħ²/(2ML²) = (k_x² + k_y²)ħ²/(2M) = k²ħ²/(2M).
  k_x = 2πn/L,  k_y = 2πq/L,  Δk_x = Δk_y = 2π/L.

  If k is large, then the number of states with wave vectors whose magnitudes lie between k and k + dk is
  dN = 2πkdk/(2π/L)²  = L²kdk/(2π).
  dN/dk = L²k/(2π).
  The density of states is dN/dE = (dN/dk)(dk/dE).
  With E = ħ²k²/(2M) we have ρ(E) = dN/dE = ML²/(2πħ²).