= Coherent-state oscillator expectation values
{title2=$\langle x\rangle_t=\sqrt{2\hbar/(m\omega)}\operatorname{Re}(\alpha e^{-i\omega t})$}
A normalized oscillator <coherent state> obeys $a|\alpha\rangle=\alpha|\alpha\rangle$, so its mean <occupation number> is $N=|\alpha|^2$. Under the <quantum harmonic oscillator> <Hamiltonian operator> $\hbar\omega(a^\dagger a+1/2)$, its evolution is $|\alpha(t)\rangle=e^{-i\omega t/2}|\alpha e^{-i\omega t}\rangle$. Inserting $x=\sqrt{\hbar/(2m\omega)}(a+a^\dagger)$ and $p=-i\sqrt{m\hbar\omega/2}(a-a^\dagger)$ gives
$$
\langle x\rangle_t=\sqrt{\frac{2\hbar}{m\omega}}\operatorname{Re}(\alpha e^{-i\omega t}),\qquad
\langle p\rangle_t=\sqrt{2m\hbar\omega}\operatorname{Im}(\alpha e^{-i\omega t}).
$$
These mean values obey the classical <harmonic oscillator> equations. The displacement amplitude is $\sqrt{2\hbar N/(m\omega)}$, whose classical orbit has <energy> $\hbar\omega N$. The full quantum mean <energy> has the additional <zero-point energy> $\hbar\omega/2$. Indeed the variances are $\hbar/(2m\omega)$ and $m\hbar\omega/2$, and their kinetic and potential contributions sum to $\hbar\omega/2$.
Back to article page
