 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/j/1/b/solution

Put

s = 1 + i t

. Logarithmic differentiation gives

\frac{\partial _{t} Ψ}{Ψ} = - \frac{i}{2 s} + \frac{im x ^{2}}{2ℏ s ^{2}}, \frac{\partial _{x}^{2} Ψ}{Ψ} = - \frac{m}{ℏ s} + \frac{m ^{2} x ^{2}}{ℏ ^{2} s ^{2}} .

Therefore

i ℏ \partial_{t} Ψ = (\frac{ℏ}{2 s} - \frac{m x ^{2}}{2 s ^{2}}) Ψ = - \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} Ψ,

so the appropriate potential energy is

V (x) = 0

. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus is

∣Ψ (x, t) ∣^{2} = \frac{1}{1 + t ^{2}} exp (- \frac{m x ^{2}}{ℏ ( 1 + t ^{2} )}) .

It changes width and height with time, so this is not a stationary state. Its integral is the constant

π ℏ/ m

; multiplying the wavefunction by

(m / (π ℏ))^{1/4}

normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance

ℏ (1 + t^{2}) / (2 m)

. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

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