More Problems

Problem 1:

An operator A, representing observable A, has two normalized eigenstates Ψ₁ and Ψ₂, with eigenvalues a₁ and a₂, respectively.  An operator B, representing observable B, has two normalized eigenstates Φ₁ and Φ₂, with eigenvalues b₁ and b₂, respectively.  The eigenstates are related by
Ψ₁ = (3Φ₁ + 4Φ₂)/5,
Ψ₂ = (4Φ₁ - 3Φ₂)/5.
(a)  Observable A is measured, and the value a₁ is obtained.  What is the state of the system immediately after this measurement?
(b)  If B is now measured immediately following the A measurement, what are the possible results, and what is the probability of measuring each result?
(c)  Right after the measurement of B, A is measured again. What is the probability of getting a₁?

Solution:

• Concepts:
  The postulates of Quantum Mechanics
• Reasoning:
  When a physical quantity described by the operator A is measured on a system in a normalized state |ψ>, the probability of measuring the eigenvalue a_n is given by
  P(a_n) = Σ_i=0^gn|<u_nⁱ|ψ>|²,  where {|u_nⁱ>} (i=1,2,...,g_n) is an orthonormal basis in the eigensubspace E_n associated with the eigenvalue a_n.
  If a measurement on a system in the state |ψ> gives the result a_n, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with a_n.
• Details of the calculation:
  We assume the system has no time to evolve between measurements.
  (a)  After measuring a₁, the system is in the eigenstate ψ₁.
  (b)  If B is now measured, the probability of obtaining b₁ is 9/25 and the probability of obtaining b₂ is 16/25.
  (c)  Φ₁ = (3ψ₁ + 4ψ₂)/5,  Φ₂ = (4ψ₁ - 3ψ₂)/5.
  There are two path.  If b₁ is measured the system is in the state Φ₁ = (3ψ₁ + 4ψ₂)/5.
  The probability of measuring a₁ after measuring b₁ is 9/25.
  If b₂ is measured the system is in the state Φ₂ = (4ψ₁ - 3ψ₂)/5.
  The probability of measuring a₁ after measuring b₂ is 16/25.
  The probability of getting a₁ again after a measurement of B is (9/25)² + (16/25)² = 0.5392.

Problem 2:

Assume the wave function of a free particle at t = 0 is Ψ(x) = Nx²exp(-x²/2).
Here N is a normalization constant.
(a)  Find N so that ψ(x) is normalized.
(b)  Find the root mean square deviation Δx at t = 0.
(c)  What can you say about Ψ(x,t) for t > 0?

Solution:

• Concepts:
  The postulates of Quantum Mechanics, the mean value
• Reasoning:
  The expression for the mean value of an observable A in the normalized state |ψ> is <A> = <ψ|A|ψ>. 
• Details of the calculation:
  (a)  <Ψ|Ψ> = 1.  ∫_-∞^+∞Ψ*(x)Ψ(x) dx = 1.
  N²∫_-∞^+∞x⁴ exp(-x²) dx = N²3π^½/4 = 1.
  N = 2/(9π)^¼.

  [∫₀^+∞x²ⁿ exp(-x²) dx = [(1*3*5*...*(2n - 1))/2ⁿ⁺¹]π^½.]

  (b)  Δx = (<x²> - <x>²)^½.
  <x> = 0 from symmetry.
  <x²> =  ∫_-∞^+∞Ψ*(x)x²Ψ(x) dx = N²∫_-∞^+∞x⁶ exp(-x²)dx = N²15π^½/8 = 5/2.
  Δx = (5/2)^½.

  (c)  We do not know p, but we know that Δp is not zero, Δp_min ≈ ħ/Δx.  The wave function is a superposition of plane waves traveling with different velocities.
  Therefore the wave function has to change shape.  Δx may increase or decrease initially, but eventually it will increases, the wave function will spread out, we will lose position information.

Problem 3:

Consider the one-particle Schroedinger equation in one dimension.
The probability per unit length of finding a particle described by the normalized wave function ψ(x,t) is given by dP(x,t)/dx = |ψ(x,t)|².  The total probability of finding the particle at time t anywhere in space is ∫_{all space} |ψ(x,t)|²dx = <ψ|ψ> = 1.
Local conservation of a classical quantity is usually expressed through the equation ∇∙j = -(∂/∂t)ρ.  Here ρ(r,t) is the volume density and j(r,t) is the current density.  In one dimension this becomes ∂j(x,t)/∂x = -(∂/∂t)ρ(x,t).
By assuming that probability is locally conserved, derive an expression for the probability current density j(x,t).

Solution:

• Concepts:
  The Schroedinger equation, ρ(x,t) = |ψ(x,t)|².
• Reasoning:
  Using the Schroedinger equation we can find an expression for ∂ρ(r,t)/∂t in terms of space derivatives.
• Details of the calculation:
  Assuming local conservation of probability we write
  ∂ρ/∂t = ∂ψ^*ψ/∂t = ψ^*∂ψ/∂t + ψ∂ψ^*/∂t
  = (1/(iħ))[ψ^*((-ħ²/(2m))∂²ψ/∂x² + Vψ) - ((-ħ²/(2m))∂²ψ/∂x² + Vψ)^*ψ]
  = (-ħ/(2im))[ψ^*∂²ψ/∂x² -(∂²ψ/∂x²)^*ψ], since V is real.
  [We use iħ∂ψ/∂t = Hψ,  ∂ψ/∂t = (1/(iħ))Hψ,  ∂ψ^*/∂t = (-1/(iħ))(Hψ)^*.]
  ∂ρ/∂t = (-ħ/(2im))(∂/∂x)[ψ^*∂ψ/∂x - ψ∂ψ^*/∂x] = -∂j(x,t)/∂x.
  j(x,t) = (ħ/(2im))[ψ^*∂ψ/∂x - ψ∂ψ^*/∂x] = (-ħ/m) Im(ψ^*∂ψ/∂x) = (1/m)Re(ψ^*(ħ/i)∂ψ/∂x).
  We interpret ρ(x,t) as the probability density and j(x,t) as the probability current density.