Time evolution of a coherent state

Question

$\newcommand\norm[1]{\lVert#1\rVert}$ $\newcommand\ket[1]{|#1\rangle}$

I consider an Hamiltonian of the Harmonic Oscillator $\hat{H} = \frac{P^2}{2m}+\frac{1}{2}m\omega^2 X^2$.

I proved already if the initial state of the Harmonic Oscillator is given by the coherent state $\ket{\psi}(t=0) = \ket{\alpha}$, then at any time $t\geq t_0=0$, the coherent state must be equal to $\ket{\psi}(t) = e^{-i\omega t/2}\ket{\alpha e^{-i\omega t}}$.

This proof uses several well-known relations. But to do it, I had to "guess" a relation: it is quite similar to a relation I have had the opportunity to prove earlier, but I have seen it nowhere. Can someone fact-check my proof?

Here it is.

\begin{align} \ket{\psi}(t) &= \sum_{n\geq 0}e^{-i\frac{\omega t}{2}}e^{-i\omega tn}\ket{\alpha}\notag\\ &= e^{-i\frac{\omega t}{2}}\sum_{n\geq 0}e^{-i\omega tn}e^{-\frac{1}{2}\norm{\alpha}^2+\alpha a^\dagger}\ket{0}\notag\\ &= e^{-i\frac{\omega t}{2}}\sum_{n\geq 0}e^{-i\omega tn}e^{-\frac{1}{2}\norm{\alpha}^2+\alpha a^\dagger}\frac{a^n}{\sqrt{n!}}\ket{n}\notag\\ &= e^{-i\frac{\omega t}{2}}\sum_{n\geq 0}e^{-\frac{1}{2}\norm{\alpha}^2+\alpha a^\dagger}\frac{\left(ae^{-i\omega t}\right)^n}{\sqrt{n!}}\ket{n}\notag\\ &= e^{-i\frac{\omega t}{2}}\sum_{n\geq 0}e^{-\frac{1}{2}\norm{\alpha}^2+\alpha a^\dagger}\ket{0e^{-i\omega t}}\notag\\ \ket{\psi}(t) &= e^{-i\frac{\omega t}{2}}\ket{\alpha e^{-i\omega t}} \end{align}

Where I introduced the relation $\ket{0} = \frac{a^n}{\sqrt{n!}}\ket{n}$, which makes intuitively sense. Is this correct?

Answer 1

Your answer is actually correct, from the definition of the coherent state in the basis of the harmonic oscillator, $$|\alpha \rangle =e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle$$

When applying the phase to each eigenvector, we get what you deduced, $\alpha(t)=\alpha e^{-i\omega t}$, with the normalized ket,

$$|\alpha (t)\rangle =e^{\frac{i \omega t}{2}}|e^{-i\omega t} \alpha \rangle$$

From this expression you can see some interesting things about the dynamics of the coherent states,

As the energy $\hat{H}$ and so $\hat{N}$ conserve, we have that $\alpha(t)$ only is going to rotate without changing its module,

$$ \langle \hat{N} \rangle _\alpha (t)=\langle \alpha |\hat{a}^{\dagger}\hat{a}|\alpha\rangle=|\alpha|^2=ct $$

And if we look at the expected values of momentum and position,

$$\langle \hat{x}\rangle_\alpha (t)=x_0 \cos{\omega t}+\frac{p_0}{m\omega}\sin{\omega t} \\ \langle \hat{p}\rangle_\alpha (t)=p_0 \cos{\omega t}-m\omega x_0\sin{\omega t} $$

With the constant defined as,

$$x_0=Re(\alpha(0))\sqrt{\frac{2 \hbar}{m \omega}}\\ p_0=Im(\alpha(0))\sqrt{2\hbar m \omega} $$

As we expected for coherent states the uncertainties do not change in time,

$$\Delta x(t)=\Delta p(t)=ct$$