**Spreading of a
wave packet
**We can terminate the Taylor series expansion

β(k) = β(k

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 t

We have

d

Therefore d

The wave packet is the of the form

Ψ(x,t) ∝ ∫|g(k)| exp(i((k - k

x

Ψ(x,t) ∝ ∫|g(k)| exp(i2π(k - k

Now |x - x

We need |(x - (k - k

The width of the wave packet increases with time.

Ψ(x,t) ∝ ∫

= ∫

P(x,t) = |ψ(x, 0)|

+ [∫

Let x - x

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(-k

(a) Find the normalization constant N. Find the expectation value <p> = ħ<k>.

(b) Find the FWHM in of |Φ(k,0)|

(c) Find the corresponding wave packet Ψ(x,0) in coordinate space. Find <x>.

(d) Find the FWHM in of |Ψ(x,0)|

(e) Find the FWHM in of |Ψ(x,t)|

(f) Find the FWHM in of |Φ(k,t)|. Does it change with time?

Hint: ∫_{-∞}^{+∞}exp(-a^{2}(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) N^{2}∫_{-∞}^{+∞}exp(-k^{2}/b^{2})dk = 1. ∫_{-∞}^{+∞}exp(-k^{2}/b^{2})dk = b√π,

N = 1/(b√π)^{ ½}. <p> = <k> = 0.

(b) exp(-k^{2}/b^{2}) = ½, k = b√(ln2), FWHM = 2√(ln2)b.

(c) Ψ(x,0) = [1/(2π)^{1/2}][1/(b√π)^{1/2}] ∫_{-∞}^{+∞}exp(-k^{2}/(2b^{2}))exp(ikx) dk.

Let us group the k-dependent terms into a perfect square.

-k^{2}/(2b^{2}) + ikx = [-1/(2b^{2})][k – ixb^{2}]^{2}- x^{2}b^{2}/2,

Integrating we find

Ψ(x,0) = (b^{2}/π)^{1/4}exp(-x^{2}b^{2}/2). <x> = 0.

(d) exp(-(x-x_{0})^{2}b^{2}) = ½. FWHM = 2√(ln2) (1/b).

(e) Ψ(x,t) = [1/(2π)^{1/2}][1/(b√π)^{1/2}] ∫_{-∞}^{+∞}exp(-k^{2}/(2b^{2}))exp(i(kx – ωt)) dk.

E = ħω = ħ^{2}k^{2}/(2m), ω = ħk^{2}/(2m). dω/dk = ħk/(m) = v_{g}.

-k^{2}/(2b^{2}) + i(kx – ħk^{2}t/(2m)) = -A[k – ix/2A]^{2}– x^{2}/(4A),

where A = 1/(2b^{2}) + iħt/(2m).

Integrating we find

Ψ(x,t) = [1/(2π)^{1/2}][1/(b√π)^{1/2}] √(π/A) exp(-x^{2}/(4A))

= (4/(b^{2}π))^{1/4}e^{iφ}/(4/b^{4}+ 4(ħt/m)^{2})^{1/4}exp(-x^{2}/(2/b^{2}+ 2iħt/m)),

where φ is real and independent of x.

Here

[1/(2π)^{1/2}][1/(b√π)^{1/2}] √(π/A) = (4/(b^{2}π))^{1/4}√(1/(4A)) = (1/(b^{2}π))^{1/4}|√(1/(4A))|e^{iφ}.

|Ψ(x,t)|^{2}= (4/(b^{2}π))^{1/1}1/(4/b^{4}+ 4(ħt/m)^{2})^{1/2}exp(-(4x^{2}/b^{2})/(4/b^{4}+ 4(ħt/m)^{2})).

exp(-(4x^{2}/b^{2})/(4/b^{4}+ 4(ħt/m)^{2})) = ½. FWHM = 2√(ln2)(1/b)(1 + b^{4}(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.