# Why is $\hat{a}^{\dagger 2}\hat{a}^2|\alpha\rangle=|\alpha|^4|\alpha\rangle$ for $|\alpha\rangle$ a coherent state of a QM harmonic oscillator?

I'm currently trying to understand the following:

Consider a quantum harmonic oscillator with a coherent state $$|\alpha\rangle$$. Show that $$\langle E^2\rangle=\hbar\omega (|\alpha|^4+2|\alpha|^2+\frac{1}{4})$$.

The solution in my notes goes as follows:

$$\langle E^2 \rangle = \langle\alpha | (\hat{a}^{\dagger}\hat{a}+\frac{1}{2})^2\hbar^2\omega^2|\alpha\rangle = \hbar\omega\langle\alpha| (\hat{a}^{\dagger}(\hat{a}^{\dagger}\hat{a}+1)\hat{a}+\hat{a}^{\dagger}\hat{a}+\frac{1}{4})|\alpha\rangle=\hbar\omega\langle\alpha|(\hat{a}^{\dagger 2}\hat{a}^2+2\hat{a}^{\dagger}\hat{a}+\frac{1}{4})|\alpha\rangle=\hbar\omega(|\alpha|^4 + 2|\alpha|^2+\frac{1}{4})$$

Question: Why does $$\hat{a}^{\dagger 2}\hat{a}^2|\alpha\rangle=|\alpha|^4|\alpha\rangle$$ hold? As far as I know, $$\hat{a}$$ is not normal so I'm guessing this is some property of coherent states? (Please excuse if this is a trivial question for you but I really don't know why this holds.)

## 2 Answers

The relation $$\hat{a}^{\dagger 2} \hat{a}^2 |\alpha\rangle \stackrel{!}{=} |\alpha|^4 |\alpha\rangle$$ is not true in general. What is true, however is that for a coherent state $$\langle \alpha | \hat{a}^{\dagger 2} \hat{a}^2 |\alpha\rangle = |\alpha|^4 \langle \alpha |\alpha\rangle.$$

The coherent states by definition are the eigenstates of the lowering operator $$\hat{a}$$. In other words, by definition:

$$\hat{a} |\alpha\rangle = \alpha |\alpha\rangle.$$

Therefore, it should be quite simple to see that: $$\hat{a}^2 |\alpha\rangle = \alpha^2 |\alpha\rangle,$$ and consequently (by taking the Hermitian conjugate) that

$$\langle \alpha| \hat{a}^{\dagger2} = \langle \alpha |\alpha^{*2} .$$

Combining this together, it should be trivial for you to show that

$$\langle \alpha | \hat{a}^{\dagger 2} \hat{a}^2 |\alpha\rangle = |\alpha|^4 \langle \alpha|\alpha\rangle.$$

$$(a^{\dagger}a)^2=a^{\dagger}aa^{\dagger}a$$, not $$a^{\dagger 2}a^2$$. Then use $$a^{\dagger}a|\alpha$$>=$$\alpha^2|\alpha$$> twice.

• Thanks but $(a^{\dagger}a)^2=a^{\dagger 2} a^2$ is never claimed, I'm using $(a^{\dagger}a)^2=a^{\dagger}(a^{\dagger}a+1)a$ by $[a,a^{\dagger}]=1$. Jun 19, 2021 at 15:59
• And I'm also not sure why $a^{\dagger}|\alpha\rangle=\alpha |\alpha\rangle$ should hold since $a$ is in general not normal, no? Jun 19, 2021 at 16:00
• You wrote $\alpha$ is a coherent state, so $a|\alpha$>=$\alpha|\alpha$>. You don't even need to expand $(a^{\dagger}a)^2$, just use it as a single operator. And yes, I forgot the absolute value of course. Jun 19, 2021 at 16:16