# Time evolution of a coherent state

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

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?

• Hi and welcome to Physics SE! The physics package cannot be used on this site, but you can define new commands at the beginning of your post (as you did). See Why don't we have the physics MathJax extension enabled? Should we? for more information. New commands must be in . The amsmath package should work, as far as I know. Commented Oct 31, 2021 at 20:23
• Hi @Jonas. I see. Thank you a lot! Commented Oct 31, 2021 at 20:38
• Are you asking about the last line relation? Of course: dot by $\langle 0|$. The rest is self-evident: $\alpha(t)=e^{-i\omega t} \alpha(0)$, and $a|\alpha(t)\rangle= \alpha(t)|\alpha(t)\rangle$. What is you question? Commented Oct 31, 2021 at 21:08
• your proof is incorrect. You have a sum over $n$ but $\vert\alpha\rangle$ as you have written it does not depend on $n$ so you also need to expand $\vert \alpha\rangle$ into its $vert n\rangle$ components to get this $\sum_n$. see also physics.stackexchange.com/q/158849/36194 Commented Oct 31, 2021 at 21:20

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,

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

• that $ct$ on your last line can’t be right. See physics.stackexchange.com/a/353893/36194, or else you should define what it means. Commented Oct 31, 2021 at 21:37
• The $ct$ means constant. But looking what you send is possibly that this only happens in a concrete case, i have not the deduction in my notes so i assumed that. Thanks for that! Commented Nov 1, 2021 at 6:48