Spreading of a wave packet
We can terminate the Taylor series expansion
β(k) = β(k0) + (k - k0)dβ(k)/dk|k0 + ½(k - k0)2d2β(k)/dk2|k0  + ...
after the first term if all higher order terms are much smaller than 2π.  However, as t increases, the higher order terms increase steadily with t.  Therefore the second order term must be included for times greater than some t1.
We have
d2β(k)/dk2|k0 = d2α(k)/dk2|k0 + d2ω(k)/d2k|k0t,   d2ω(k)/d2k =  ħ/m.
Therefore d2β(k)/dk2|k0 depends on t, and for large enough times t its magnitude increases with time.
The wave packet is the of the form
Ψ(x,t) ∝ ∫|g(k)| exp(i((k - k0)(x - x0 - vgt) - (k - k0)2x1) dk,   x1 = -½d2β(k)/dk2|k0.
x1 depends on t, and for large enough times t its magnitude increases with time.
Ψ(x,t) ∝ ∫|g(k)| exp(i2π(k - k0)/τ) dk,  with  τ = 2π/((x - x0 - vgt) - (k - k0)x1).
Now |x - x0 - vt |>> Δk-1 is no longer sufficient for |τ| << Δk.
We need |(x - (k - k0)x1) - x0 - vt | >> Δk-1 for the integrand to oscillate rapidly and the contribution to the integral to be zero.
The width of the wave packet increases with time.

#### Example: Let |g(k)| = 1 for k0 < k < (k0 + π).  Then Ψ(x,t) ∝ ∫k0k0+π exp(i(k - k0)(x - x0 - vgt) - (k - k0)2x1) dk = ∫k0k0+π cos((k - k0)(x - x0 - vgt) - (k - k0)2x1) dk + i∫k0k0+π sin((k - k0)(x - x0 - vgt) - (k - k0)2x1) dk. P(x,t) = |ψ(x, 0)|2 = [∫k0k0+π cos((k - k0)(x - x0 - vgt) - (k - k0)2x1) dk]2 + [∫k0k0+π sin((k - k0)(x - x0 - vgt) - (k - k0)2x1) dk]2. Let x - x0 - vgt = a, x1 = b.Compare P(x,t) at t = 0 when b = 0 and at some later time when b ≠ 0.

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) = N exp(-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)1/2][1/(b√π)1/2] ∫-∞+∞exp(-k2/(2b2))exp(ikx) dk.
Let us group the k-dependent terms into a perfect square.
-k2/(2b2) + ikx = [-1/(2b2)][k – ixb2]2 - x2b2/2,
Integrating we find
Ψ(x,0) = (b2/π)1/4exp(-x2b2/2).  <x> = 0.
(d)  exp(-(x-x0)2b2) = ½.   FWHM = 2√(ln2) (1/b).
(e)  Ψ(x,t) = [1/(2π)1/2][1/(b√π)1/2] ∫-∞+∞exp(-k2/(2b2))exp(i(kx – ωt)) dk.
E = ħω = ħ2k2/(2m),  ω = ħk2/(2m).  dω/dk = ħk/(m) = vg.
-k2/(2b2) + i(kx – ħk2t/(2m))  = -A[k – ix/2A]2 – x2/(4A),
where A = 1/(2b2) + iħt/(2m).
Integrating we find
Ψ(x,t) = [1/(2π)1/2][1/(b√π)1/2] √(π/A) exp(-x2/(4A))
= (4/(b2π))1/4 e/(4/b4 + 4(ħt/m)2)1/4 exp(-x2/(2/b2 + 2iħt/m)),
where φ is real and independent of x.
Here
[1/(2π)1/2][1/(b√π)1/2] √(π/A) = (4/(b2π))1/4 √(1/(4A)) = (1/(b2π))1/4 |√(1/(4A))|e.
|Ψ(x,t)|2 = (4/(b2π))1/1 1/(4/b4 + 4(ħt/m)2)1/2 exp(-(4x2/b2)/(4/b4 + 4(ħt/m)2)).
exp(-(4x2/b2)/(4/b4 + 4(ħt/m)2)) = ½.   FWHM = 2√(ln2)(1/b)(1 + b4(ħt/m)2)1/2.
The FWHM of |Ψ(x,t)|2 increases with time
(f)  Φ(k,t) = exp(-iωt)) Φ(k,0).
FWHM of |Φ(k,t)| = 2√(ln2)b.  It does not change with time.