Wave packets and the Fourier transform

Fourier transform

Problem:

The wave function of a particle at t = 0 is 
Ψ(x) = 1/L^½,  |x| < L/2,  Ψ(x) = 0 otherwise. 
At t = 0, what possible values of the momentum of the particle can be found, and with what probability?

Solution:

• Concepts:
  The postulates of quantum mechanics
• Reasoning:
  P is a Hermitian operator.  The possible results of a measurement are the eigenvalues of the P, -∞ < p < ∞.
  When a physical quantity described by the operator P is measured on a system in a normalized state |Ψ>, the probability of measuring the eigenvalue p is given by dP(p) = |<p|Ψ>|²dp = |Ψ(p)|²dp, where |p> is the eigenvector corresponding to the eigenvalue p; we assume p is a non-degenerate continuous eigenvalue of P. 
• Details of the calculation:
  Ψ(p)  = (2πħ)^-½∫Ψ(x,0) exp(-ipx/ħ)dx,
  Ψ(p)  = (2πħL)^-½∫_-L/2^+L/2 exp(-ipx/ħ)dx
  = (2πħL)^-½(iħ/p) [exp(-ipL/(2ħ)) - exp(-pL/(2ħ))]
  = (2πħL)^-½(2ħ/p) sin(pL/(2ħ)).
  The probability of finding the particle with momentum between p and p + dp  is |Ψ(p)|²dp.
  |Ψ(p)|²  = (2ħ/(πLp²) )sin²(pL/(2ħ)).

Problem:

Consider a free particle which is described at t = 0 by the normalized Gaussian wave function ψ(x,0) = Nexp(ik₀x)exp(-ax²).
(a)  Normalize the wave function.
(b)  Find the probability density |ψ(x,0)|² of the particle. 
(c)  Find its Fourier transform Φ(k,0) of the wave function and the probability density
|Φ(k,0)|² in k-space.

Solution:

• Concepts:
  The Fourier transform
• Reasoning:
  We are asked to find the Fourier transform of a wave packet.
• Details of the calculation:
  (a)  N²∫_-∞^+∞ exp(-2ax²) dx = N² √(π/2a)   = 1
  ψ(x,0) = (2a/π)^¼exp(ik₀x)exp(-ax²).
  (b)  |ψ(x,0)|² = (2a/π)^½ exp(-2ax²).
  (c)  ψ(x,0) = (2π)^-½∫_-∞^+∞ Φ(k,0) exp(ikx) dk.
  Φ(k,0) = (2π)^-½∫_-∞^+∞ ψ(x,0) exp(-ikx) dx
  = (a/(2π³))^¼∫_-∞^+∞ exp(-ax²) exp(-i(k-k₀)x) dx.
  ax² + i(k-k₀)x = ax² + i(k-k₀)x - (k-k₀)²/(4a) + (k-k₀)²/(4a)
  = (a^½x + i(k-k₀)/(2a^½)² + (k-k₀)²/(4a).  (completing the square)
  Φ(k,0) = (a/(2π³))^¼exp(-(k-k₀)²/(4a))∫_-∞^+∞ exp(-(ax² + i(k-k₀)x  - (k-k₀)²/(4a))dx
  = (a/(2π³))^¼exp(-(k-k₀)²/(4a))∫_-∞^+∞ exp(-(a^½x + i(k-k₀)/(2a^½)²) dx
  = (1/(a2π³))^¼exp(-(k-k₀)²/(4a))∫_-∞^+∞ exp(-x'²) dx'
  = (a2π)^-¼ exp(-(k-k₀)²/(4a)).
  |Φ(k,0)|² = (a2π)^-½ exp(-(k-k₀)²/(2a)) is the probability density in k-space.

Evolution of a wave packet

Problem:

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)).
(a)  Find the normalization constant N.  Find the expectation value <p> = ħ<k>.
(b)  Find the FWHM in of |Φ(k,0)|² in k-space.
(c)  Find the corresponding wave packet Ψ(x,0) in coordinate space.  Find <x>.
(d)  Find the FWHM in of |Ψ(x,0)|² in coordinate space.
(e)  Find the FWHM in of |Ψ(x,t)|² an some later time t.  Does it change with time?
(f)  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)  N²∫_-∞^+∞exp(-k²/b²)dk = 1.  ∫_-∞^+∞exp(-k²/b²)dk = b√π
  N = 1/(b√π) ^½.  <p> = <k> = 0.
  
  (b)  exp(-k²/b²) = ½,   k = b√(ln2),  FWHM = 2√(ln2)b.
  
  (c)  Ψ(x',0) = [1/(2π)^½][1/(b√π)^½] ∫_-∞^+∞exp(-k²/(2b²))exp(ikx') dk.
  ∫_-∞^+∞exp(-k²/(2b²))exp(ikx') dk = ∫_-∞^+∞exp(-k²/(2b²))cos(kx') dk
  = 2 ∫₀^+∞exp(-k²/(2b²))cos(kx') dk = 2b(π/2)^1/2 exp(-x'²b²/2).
  We can also evaluate the integral by grouping the k-dependent terms into a perfect square.
  -k²/(2b²) + ikx' = [-1/(2b²)][k - ix'b²]² - x'²b²/2.
  Now we can evaluate the integral using  ∫_-∞^+∞exp(-a(x+b)²)dx = √(π/a).
  Ψ(x',0) = (b²/π)^¼exp(-x'²b²/2).  <x'> = 0.
  x' = x - x₀ depends on the choice of the origin of the coordinate system.
  
  (d)  exp(-(x - x₀)²b²) = ½.   FWHM = 2√(ln2) (1/b).
  
  (e)  Choose x₀ = 0.
  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:

In one dimension, at t = 0 the normalized wave function of a free particle in momentum space is
Φ(p,0) = Nexp(-(p - p₀)²/(2b²ħ²))exp(-ipx₀/ħ).
(a)  Find the normalization constant N.  Find the expectation value <p>.
(b)  Find the FWHM in of |Φ(p,0)|² in momentum space.
(c)  Find the corresponding wave packet Ψ(x,0) in coordinate space.  Find the expectation value <x>.
(d)  Find the FWHM in of |Ψ(x,0)|² in coordinate space.

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)  N²∫_-∞^+∞exp(-(p - p₀)²/(b²ħ²))dp = 1.  ∫_-∞^+∞exp(-(p - p₀)²/(b²ħ²))dp = bħ√π.
  N = 1/(bħ√π) ^½.  <p> = p₀.
  (b)  exp(-(p - p₀)²/(b²ħ²)) = ½,   (p - p₀) = aħ√(ln2),  FWHM = 2√(ln2)bħ.
  (c)  Ψ(x,0) = [1/(2πħ)^½][1/(bħ√π)^½] ∫_-∞^+∞exp(-(p - p₀)²/(2b²ħ²))exp(ip(x-x₀)/ħ) dp.
  Let us group the p-dependent terms into a perfect square.
  -(p - p₀)²/(2b²ħ²) + ip(x-x₀)/ħ
  = [-1/(2b²ħ²)][p - p₀ - i(x-x₀)b²ħ]² - (x-x₀)²b²/2 + ip₀(x-x₀)/ħ.
  Then, using ∫_-∞^+∞exp(-a²(x+b)²)dx = √π/a, we have
  Ψ(x,0) = (b²/π)^¼exp(-(x-x₀)²b²/2) exp(ip₀(x-x₀)/ħ.  <x> = x₀.
  (d)  exp(-(x-x₀)²b²) = ½.   FWHM = 2√(ln2) (1/b).

Problem:

The wave packet for a quantum mechanical particle of mass m in one dimension is described by

Ψ(x,t) = [1/(2π)^½] lim(R-->∞) ∫_-R^+R dk Φ(k) exp(i(kx - ω(k)t)),

where  Φ(k) = N exp(-(k - k₀)²/(4(Δk)²)) is a "strongly peaked" distribution around k = k₀ with ΔkΔx = ½ at t = 0.
(a)  Show that  ψ(x, 0) =  [Δx^-½(2π)^-¼] exp(ik₀x)exp(-x²/(4Δx²)) (evaluate the integral),
and find the probability density |ψ(x, 0)|² of the particle.  Use ΔkΔx = ½  to eliminate Δk.  Show your work!
(b)  For a free particle E = ħω = ħ²k²/(2m).  Show that for a free particle
|ψ(x, t)|² = exp[-½ (x - ħk₀t/m)²/Δx(t)²]/[ (2π)^½ Δx(t)],
with Δx(t) = (Δx² + ħ²t²/(4m²Δx²)^½.
(c)  Determine the group velocity of the wave packet.
(d)  Evaluate the time it takes for the wave-packet to double in spatial extent, specifically if the particle is an electron and Δx ~10 nm, at t = 0.

Useful integrals: 
∫_-∞^+∞exp(-αx²)dx = √(π/α) 
∫_-∞^+∞exp(-αx² + iβx))dx = (π/α)^½exp(-β²/(4α))
Note:  You can complete some parts of the problem by using the given results without evaluating the integrals.

Solution:

• Concepts:
  Time evolution of a free wave packet
• Reasoning:
  We are asked to evaluate integrals to better visualize the time evolution of a free wavepacket.
• Details of the calculation:
  (a)  Normalize Φ(k):
  N²∫_-∞^+∞exp(-(k - k₀)²/(2(Δk)²))dk = 1.    ∫_-∞^+∞exp(-(k - k₀)²/(2(Δk)²))dk = Δk√(2π).
  N = 1/(Δk√(2π))^½.
  ψ(x, 0) = [exp(ik₀x)N/√(2π) ]∫_-∞^+∞exp(-(k - k₀)²/(4(Δk)²) + ix(k - k₀))dk.
  (k - k₀)²/(4(Δk)²) - ix(k - k₀)) - x²(Δk)² + x²(Δk)²
  = ((k - k₀)/(2 Δk) - ixΔk)² + x²(Δk)².  (completing the square)
  ψ(x, 0) = [N/√(2π)]exp(ik₀x)exp(-x²(Δk)²)∫_-∞^+∞exp(-k'²)k'.
  ψ(x, 0) = [(Δk)^½2^¼/π^¼]exp(ik₀x)exp(-x²(Δk)²) = [Δx^-½(2π)^-¼] exp(ik₀x)exp(-x²/(4Δx²)).
  |ψ(x, 0)|² = [1/(Δx(2π)^½] exp(-x²/(2Δx²)).
  This is a Gaussian with σ = Δx centered at x = 0.
  
  (b)  ψ(x, t) = (2π)^-½∫_-∞^+∞ Φ(k, t) exp(ikx) dk
  Φ(k, t) = N exp(-(k - k₀)²/(4(Δk)²) - iħk²t/(2m ))
  N = 1/(Δk√(2π))^½.
  -(k - k₀)²/(4(Δk)²) + ixk  - iħk²t/(2m)
  = -(k - k₀)²/(4(Δk)²) + ix(k - k₀) - iħ(k - k₀)²t/(2m) + iħk₀²t/(2m) - iħkk₀t/m + ik₀x
  = -(k - k₀)²[1/(4(Δk)²) + iħt/(2m)] + i(k - k₀)[x - ħk₀t/m] + iħk₀²t/(2m) - iħkk₀²t/m + ik₀x
  = -(k - k₀)²[Δx² + iħt/(2m)] + i(k - k₀)[x - ħk₀t/m] - iħk₀²t/(2m)  + ik₀x
  ψ(x, t) =  [N/√(2π)]exp(ik₀x - iħk₀²t/(2m)) ∫_-∞^+∞exp(-αk'² + iβk'))dk'
  = [N/√2] exp(ik₀x - iħk₀²t/(2m))[(Δx² + iħt/(2m))]^-½ exp[-¼(x - ħk₀t/m)²/(Δx² + iħt/(2m))].
  ψ(x, t) = Δx^½ [(Δx² + iħt/(2m))]^-½(2π)^-¼ ]*
  exp(ik₀x - iħk₀²t/(2m)) exp[-¼(x - ħk₀t/m)²/(Δx² + iħt/(2m))].
  |ψ(x, t)|² = exp[-½(x - ħk₀t/m)²/Δx(t)²]/[ (2π)^½Δx(t)], with Δx(t) = (Δx² + ħ²t²/(4m²Δx²)^½.
  (c) |ψ(x, t)|² is a Gaussian with  σ = Δx(t) centered at x = ħk₀t/m.  The center of the wave packet moves with speed ħk₀t/m.  This is the group velocity v_g = dω/dk of the wave packet.
  
  (d)  Δx(t) = (Δx² + ħ²t²/(4m²Δx²)^½ = 2Δx
  Δx² + ħ²t²/(4m²Δx²) = 4Δx².  t² = 12 Δx⁴ m²/ ħ².
  t² = 12*10^-32*(9.1*10^-31)²/(1.05*10^-34)² s² =  ~10^-23 s²,  t ~ 3*10^-12 s.
  We get the same order of magnitude result by just using the uncertainty principle.
  (Δv ~ ħΔk/m = ħ/(2mΔx),  t ~ 2Δx/Δv = 4mΔx²/ħ).