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So I have an issue understanding how to solve problems with time-dependent operators (Heisenberg picture).

In this case I have a resonator prepared in a coherent state: $|\psi (t=0)\rangle = |\alpha \rangle$. Then I need to find the time evolution of the system that has the following Hamiltonian, $H/\hbar = \omega a ^\dagger a$: $$|\psi (t) \rangle = e^{-i|\alpha|^2\omega t}|\alpha\rangle$$

But then I need to find the "time-dependent expectation value of the operator $Q = \frac{a + a^\dagger}{\sqrt{2}}$". Now I don't understand how you are supposed the time dependence here:

$$\langle \psi (t)|Q|\psi (t)\rangle = \langle\alpha|e^{i|\alpha|^2\omega t}Qe^{-i|\alpha|^2\omega t}|\alpha\rangle = \langle\alpha|Q|\alpha\rangle = \frac{\alpha + \alpha^*}{\sqrt{2}}$$

Am I missing something here, this feels like it is very simple but can't wrap my head around it.

Many thanks!

bsaoptima
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    You have the wrong time dependence. Since $\vert \alpha\rangle$ is not a h.o. eigenstate there is no way the time dependence is as you state. Near dupe of this post – ZeroTheHero Oct 07 '23 at 16:46

2 Answers2

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I will develop Mike Stone's answer. Actually, coherent states are eigenstates of the annihilation operator only, i.e. $\hat{a}|\alpha\rangle = \alpha|\alpha\rangle$, hence $\langle\alpha|\hat{a}^\dagger = \alpha^*\langle\alpha|$ by taking the hermitian conjugate, but $\hat{a}^\dagger|\alpha\rangle \neq \alpha^*|\alpha\rangle$. Indeed, if the creation and annihilation operators were simultaneously diagonalizable $-$ in other words, if they possessed the same eigenstates $-$, then they would commute, but we know that $[\hat{a},\hat{a}^\dagger] = 1$. In consequence, the relation $\hat{a}^\dagger\hat{a}|\alpha\rangle = |\alpha|^2|\alpha\rangle$ is not true, nor $e^{-i\omega t\hat{a}^\dagger\hat{a}}|\alpha\rangle = e^{-i\omega t|\alpha|^2}|\alpha\rangle$.

Now, since the eigenbasis of the number operator $\hat{n} = \hat{a}^\dagger\hat{a}$ corresponds to the Fock states $\{|n\rangle\}_{n\in\mathbb{N}}$ and given that the coherent state $|\alpha\rangle$ is represented by $e^{-\frac{1}{2}|\alpha|^2} \displaystyle\sum_{n\ge0} \frac{\alpha^n}{\sqrt{n!}} |n\rangle$ in that basis, one has : $$ |\psi(t)\rangle = e^{-i\hat{H}t/\hbar}|\alpha\rangle = e^{-\frac{1}{2}|\alpha|^2} \sum_{n\ge0} \frac{\alpha^n}{\sqrt{n!}} e^{-i\omega t\, \hat{a}^\dagger\hat{a}}|n\rangle = e^{-\frac{1}{2}|\alpha|^2} \sum_{n\ge0} \frac{(\alpha e^{-i\omega t})^n}{\sqrt{n!}} |n\rangle = |\alpha e^{-i\omega t}\rangle, $$ hence $$ \langle\hat{q}\rangle = \langle\psi(t)|\hat{q}|\psi(t)\rangle = \langle\alpha e^{-i\omega t}| \frac{\hat{a}^\dagger+\hat{a}}{\sqrt{2}} |\alpha e^{-i\omega t}\rangle = \frac{\alpha^*e^{i\omega t} + \alpha e^{-i\omega t}}{\sqrt{2}}, $$ where $\hat{a}^\dagger$ has been applied on the left and $\hat{a}$ on the right.

Abezhiko
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Your mistake is that $a^\dagger a |\alpha\rangle \ne |\alpha^2||\alpha \rangle$, so $$ e^{-i \omega a^\dagger a t}|\alpha\rangle\ne e^{-i \omega |\alpha|^2 t}|\alpha\rangle, $$ but rather (I may have sign wrong) $$ e^{-i \omega a^\dagger a t}|\alpha\rangle= |\alpha e^{-i\omega \alpha}\rangle, $$ which is a rather different expression

mike stone
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  • I have never seen such an expression, from my knowledge, coherent states will obey: $a|\alpha\rangle = \alpha |\alpha \rangle$ and then I used the fact that: $A|a\rangle=a|a\rangle \Leftrightarrow f(A)|a\rangle = f(a)|a\rangle$ – bsaoptima Oct 05 '23 at 19:25
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    But you also said that $a^\dagger |\alpha\rangle = a^* |\alpha\rangle $, which is not true. – mike stone Oct 06 '23 at 13:40