Displaced quantum harmonic oscillator ($E_n=\hslash \omega (n+1/2)-F^2/(2m{\omega }^2)$)

Adding $-Fx$ to the quantum harmonic oscillator potential gives $\frac{1}{2}m{\omega }^2(x-d{)}^2-F^2/(2m{\omega }^2)$, where $d=F/(m{\omega }^2)$. Translation preserves the oscillator spectrum, apart from the displayed constant shift. Eigenfunctions are translated by $d$ and the energy spacing is unchanged.

Displacing a one-dimensional quantum harmonic oscillator produces a linear force term and a scalar shift relative to the unshifted Hamiltonian. The scalar adds only a common phase. The linear term couples number states separated by one excitation. Arbitrary linear driving of the ground state creates a coherent state; a first-order expansion of that state contains only the vacuum and first excited state.^[ref]

A sinusoidal trap displacement with amplitude $\epsilon$ and duration $T$ produces a resonant first-order excitation amplitude proportional to $\epsilon \omega T/a_0$, where $a_0$ is the oscillator length. Keeping its square much smaller than one is the long-time condition for weak excitation. The small-displacement condition $\epsilon /a_0\ll 1$ alone does not ensure uniform validity as $T$ grows. Exact linear driving remains solvable as a coherent state after this first-order approximation fails.

The ladder identity $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ proves the matrix element in the title. In time-dependent perturbation theory, a linear force excites only $n=1$ directly from $n=0$. Parity alone rules out even excited states; the ladder identity also rules out odd $n>1$. Higher-order evolution can reach other levels, so the driven oscillator is not an exact two-level system.

For $H(t)=\hslash \omega (a^{†}a+1/2)-Fx{e}^{t/\tau }$ on $t\le 0$, start in the unforced ground state at $t=-\infty$. Define $x_{01}=\sqrt{\hslash /(2m\omega )}$ and $\eta =F{x}_{01}/(\hslash \omega )$. Time-dependent perturbation theory givesThus the first excited-state probability in the original oscillator basis tends to zero for a rapid turn-on and to ${\eta }^2$ for slow turn-on, within the weak-force approximation $|\eta |\ll 1$.

This finite adiabatic limit is not a failure of the adiabatic theorem: the instantaneous ground state is displaced and has a nonzero projection onto the original excited states. In fact a linear force preserves a coherent state, whose exact original-basis displacement at $t=0$ is $\alpha =i\eta \omega \tau /(1+i\omega \tau )$. It follows either from $\dot{\alpha }=-i\omega \alpha +iF{x}_{01}{e}^{t/\tau }/\hslash$ or from the forced oscillator equations. Consequently the exact original-basis probability is $P_1^{(0)}=|\alpha {|}^2{e}^{-|\alpha {|}^2}$. The instantaneous oscillator basis has displacement $\eta$, so its mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/[1+(\omega \tau {)}^2]$, which vanishes adiabatically.

In the sudden approximation, an initial oscillator ground state overlaps the displaced ground state by $\exp [-m\omega d^2/(4\hslash )]$. Completing the square in the product of the two normalized Gaussian wavefunctions and using the Gaussian integral proves this amplitude. Squaring it gives the displayed probability. Only an irrelevant overall phase evolves before the sudden switch.