Solution

Complete the square, keeping the equilibrium displacement signed:

x_{0} = - \frac{γ}{m ω ^{2}}, U_{m i n} = - \frac{γ ^{2}}{2 m ω ^{2}}, H = \frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} (x - x_{0})^{2} + U_{m i n} .

Write

X = x - x_{0}

for position relative to the new minimum. By the sudden approximation, the state immediately after the quantum quench is still the old vacuum. Its initial means are

⟨ X (0^{+})⟩ = - x_{0} = \frac{γ}{m ω ^{2}}, ⟨ p (0^{+})⟩ = 0 .

The old ground state has

⟨ H_{0} ⟩ = ℏ ω /2

and

⟨ x ⟩ = 0

. Thus

⟨ H ⟩ = ℏ ω /2

, but the energy measured above the new potential minimum is

⟨ H - U_{m i n} ⟩ = \frac{ℏ ω}{2} + \frac{γ ^{2}}{2 m ω ^{2}} = \frac{ℏ ω}{2} + \frac{1}{2} m ω^{2} x_{0}^{2} .

The extra term is the excitation energy above the new ground state; the zero-point term remains part of the energy above the potential minimum.

The variables

X, p

obey the oscillator equations in part (a). Using the above initial means gives

⟨ X (t)⟩ = - x_{0} cos ω t, ⟨ x (t)⟩ = x_{0} (1 - cos ω t), ⟨ p (t)⟩ = mω x_{0} sin ω t .

The wavepacket's centre therefore oscillates about the new equilibrium, beginning at the old origin with zero mean momentum.

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