Assignment 8

Problem 1: 

A particle in some unspecified potential is in the initial normalized state

ψ(x) = a x e^-x/b, x ≥ 0,
ψ(x) = 0, x < 0.

(a)  Where are you most likely to find the particle if you were to measure its position?
(b)  What is the probability of finding the particle in the region x > 0?

Solution:

• Concepts:
  The fundamental assumptions of quantum mechanics
• Reasoning:
  The wave function ψ(x) is interpreted as a probability amplitude of the particles presence.
  Here ψ(x) is given.  If it can be normalized, then |ψ(x)|² is the probability density.
  The position x for which |ψ(x)|² has a maximum is the most likely position.
• Details of the calculation:
  (a)  |ψ(x)|² = a² x² e^-2x/b.
  d|ψ(x)|²/dx = 2a² x e^-2x/b - 2(a²/b) x² e^-2x/b = 0.
  1  - x/b = 0, x = b.
  If you were to measure the position of the particle you are most likely to find it in an interval dx about x = b.
  (b)   The probability of finding the particle in the region x > 0 is 1.

Problem 2:

Consider the one-dimensional time-independent Schroedinger equation for some arbitrary potential U(x).  Prove that if a solution ψ(x) has the property that ψ(x) → 0 as x → ±∞, then the solution must be non-degenerate and therefore real, apart from a possible overall phase factor.
Hint:  Show that the contrary assumption leads to a contradiction.

Solution:

• Concepts:
  The time-independent Schroedinger equation
• Reasoning:
  If ψ(x) is a solution of the time-independent Schroedinger equation, i.e. of the eigenfunction of the Hamiltonian H(x),
  then ∂²ψ(x)/∂x² - (2m/ħ²)(E - U(x))ψ(x) = 0.
• Details of the calculation:
  Assume two solutions ψ₁(x) and ψ₂(x) exist.
  The Schroedinger equation tells us that
  ∂²ψ₁(x)/∂x² - (2m/ħ²)(E - U(x))ψ₁(x) = 0,
  ∂²ψ₂(x)/∂x² - (2m/ħ²)(E - U(x))ψ₂(x) = 0.
  Multiplying the first equation by ψ₂(x) and the second equation by ψ₁(x) and subtracting we obtain
  ψ₂(x)∂²ψ₁(x)/∂x² - ψ₁(x)∂²ψ₂(x)/∂x² = 0.
  ∂/∂x(ψ₂(x) ∂ψ₁(x)/∂x - ψ₁(x) ∂ψ₂(x)/∂x) = ψ₂(x) ∂²ψ₁(x)/∂x² - ψ₁(x)∂²ψ₂(x)/∂x² = 0.
  Therefore ψ₂(x) ∂ψ₁(x)/∂x - ψ₁(x) ∂ψ₂(x)/∂x = C - constant.  (Integration)
  For bound states the wave function must vanish as x --> ± ∞.  Therefore C = 0.
  ψ₂(x) ∂ψ₁(x)/∂x - ψ₁(x) ∂ψ₂(x)/∂x = 0.
  (1/ψ₁(x)) ∂ψ₁(x)/∂x = (1/ψ₂(x)) ∂ψ₂(x)/∂x.
  ∂/∂x(lnψ₁(x)) = ∂/∂x(lnψ₂(x)).
  This integrates to lnψ₁(x) = lnψ₂(x) + c,  ψ₂(x)  = γ ψ₁(x), where γ = e^c.
  ψ₂(x) can differ from ψ₁(x) only by a multiplicative constant.
  If ψ₁(x) and ψ₂(x) are normalized, then |γ|² = 1, γ = e^iδ, where δ is a real number.  If we choose δ = 0, then c = 1 and ψ is a real function.

Problem 3:

For a certain system, the operator corresponding to the physical quantity A does not commute with the Hamiltonian. 
It has eigenvalues a₁ and a₂, corresponding to the eigenfunctions
Φ₁ = 2^-1/2(u₁ + u₂) and Φ₂ = 2^-1/2(u₁ - u₂),
where u₁ and u₂ are normalized eigenfunctions of the Hamiltonian with eigenvalues E₁ and E₂, respectively.
If the system is in the state ψ = Φ₁ at time t = 0, calculate the time evolution of the expectation value of A.

Solution:

• Concepts:
  Fundamental assumptions of QM, the evolution operator
• Reasoning:
  The evolution operator is a unitary operator defined through |ψ(t)> = U(t,t₀)|ψ(t₀)>.
  If H does not explicitly depend on time, then U(t,t₀) = exp(-(i/ħ)H(t - t₀)).
  If |Ψ(t₀)> is an eigenfunction of H with eigenvalue E,
  then |Ψ(t)> = exp(-(i/ħ)E(t - t₀))|Ψ(t₀)>.
• Details of the calculation:
  ψ(t) = 2^-½[u₁exp(-(i/ħ)E₁t) + u₂exp(-(i/ħ)E₂t)]
  = ½[Φ₁(exp(-iE₁t/ħ) + exp(-(i/ħ)E₂t)) + Φ₂(exp(-iE₁t/ħ) - exp(-(i/ħ)E₂t))].
  Aψ(t) = ½[a₁Φ₁(exp(-iE₁t/ħ) + exp(-(i/ħ)E₂t)) + a₂Φ₂(exp(-iE₁t/ħ) - exp(-(i/ħ)E₂t))].
  <A(t)> = <ψ(t)|A|ψ(t)>
  = (1/4)[2a₁<Φ₁|Φ₁> + 2a₂<Φ₂|Φ₂> + 2a₁<Φ₁|Φ₁>cos((E₁ - E₂)t/ħ) - 2a₂<Φ₂|Φ₂>cos((E₁ - E₂)t/ħ)]
  =  ½[a₁ + a₂ + (a₁ - a₂)cos((E₁ - E₂)t/ħ)].

Problem 4:

In one dimension, at t = 0 the normalized wave function of a free particle of mass m in k-space is
Φ(k,0) = Nexp(-k²/(2b²))n with N = 1/(b√π) ^½.
Its FWHM in of |Φ(k,0)|² in k-space is 2√(ln2)b, and <p> = ħ<k> = 0.
The corresponding wave packet Ψ(x,0) in coordinate space is
Ψ(x,0) = [1/(2π)^½][1/(b√π)^½] ∫_-∞^+∞exp(-k²/(2b²))exp(ikx') dk = (b²/π)^¼exp(-x'²b²/2),
with x' = x - x₀, where x₀ depends on the choice of the origin of the coordinate system.
Choose x₀ = 0.
The FWHM in of |Ψ(x,0)|² in coordinate space is 2√(ln2) (1/b) and <x> = 0.

(a)  Find the FWHM in of |Ψ(x,t)|² an some later time t.  Does it change with time?
(b)  Find the FWHM in of |Φ(k,t)|.  Does it change with time?

Hint:  ∫_-∞^+∞exp(-a²(x + c)²)dx = √π/a

Solution:

• Concepts:
  Postulates of QM, the Fourier transform
• Reasoning:
  We investigate the properties of a Gaussian wave packet representing a free particle.
• Details of the calculation:
  (a)  The wave function of a free particle is a linear superposition of plane waves.
  Ψ(x,t) = [1/(2π)^½][1/(b√π)^½] ∫_-∞^+∞exp(-k²/(2b²))exp(i(kx - ωt)) dk.
  For a plane wave E = ħω = ħ²k²/(2m),  ω = ħk²/(2m).
  We can group the k-dependent terms into a perfect square
  -k²/(2b²) + i(kx - ħk²t/(2m)) = -k²[1/(2b²) + iħt/(2m)] + ikx
  = -A(k² - ikx/A + (ix/2A)² - (ix/2A)²]
  = -A[k - ix/2A]² - x²/(4A),
  where A = 1/(2b²) + iħt/(2m), |A|² = 1/(4b⁴) + ħ²t²/(4m²).                                                                                  =
  ∫_-∞^+∞exp(-A[k - ix/2A]² - x²/(4A)) dk = exp(-x²/(4A)∫_-∞^+∞exp(-A[k - ix/2A]²) dk
  = exp(-x²/(4A)√(π/A), using ∫_-∞^+∞exp(-a(x+b)²)dx = √(π/a).
  Ψ(x,t) = [1/(2π)^½][1/(b√π)^½] √(π/A) exp(-x²/(4A)).
  Ψ(x,t) = (1/(b²π))^¼[1/(1/b² + iħt/m)]^½ exp(-x²/(4A)).
  
  |Ψ(x,t)|² = (1/(b²π))^½[(1/(1/b⁴ + (ħt/m)²) ^½ exp(-(b²x²/(1 + (b²ħt/m)²)). 
  = b/(π(1 + (b²ħt/m)²)^-½ exp(-(b²x²/(1 + (b²ħt/m)²)). 
  To find the FWHM we use
  exp(-(b²x²/(1 + (b²ħt/m)²)) = ½.   b²x² = ln(2)(1 + (b²ħt/m)²),
  FWHM = 2√(ln2)(1/b² + (bħt/m)²)^½ = 2√(ln2) (1/b)(1 + (b²ħt/m)²).
  The FWHM of |Ψ(x,t)|² increases with time.
  
  (b)  iħ∂Φ(k,t)/∂t = [p²/(2m)]Φ(k,t) = ħω Φ(k,t).
  Φ(k,t) = exp(-iωt)) Φ(k,0).  FWHM of |Φ(k,t)| = 2√(ln2)b.  It does not change with time.

Problem 5:

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)  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.