= Solution
Complete the square, keeping the equilibrium displacement signed:
$$
x_0=-\frac\gamma{m\omega^2},\qquad
U_{\min}=-\frac{\gamma^2}{2m\omega^2},\qquad
H=\frac{p^2}{2m}+\frac12m\omega^2(x-x_0)^2+U_{\min}.
$$
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
$$
\boxed{\langle X(0^+)\rangle=-x_0=\frac\gamma{m\omega^2},\qquad
\langle p(0^+)\rangle=0}.
$$
The old <ground state> has $\langle H_0\rangle=\hbar\omega/2$ and $\langle x\rangle=0$. Thus $\langle H\rangle=\hbar\omega/2$, but the energy measured above the \i[new potential minimum] is
$$
\boxed{\langle H-U_{\min}\rangle=\frac{\hbar\omega}{2}+\frac{\gamma^2}{2m\omega^2}
=\frac{\hbar\omega}{2}+\frac12m\omega^2x_0^2}.
$$
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
$$
\boxed{\langle X(t)\rangle=-x_0\cos\omega t,\qquad
\langle x(t)\rangle=x_0(1-\cos\omega t),\qquad
\langle p(t)\rangle=m\omega x_0\sin\omega t}.
$$
The wavepacket's centre therefore oscillates about the new equilibrium, beginning at the old origin with zero mean momentum.
Back to article page
