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I'm new to quantum mechanics and I just have a doubt.

If $\hat a$ is the annihilation operator of quantum harmonic oscillator and $ \hat a^{\dagger}$ is the creation operator, what is the value of $\langle 0|\hat{a} \, \hat{a}^{\dagger}|0 \rangle$?

I think it should be $0$, since the annihilation operator acts on the left bra. But I just realized if I work on the right first bra-ket, I get: $\langle 1|1 \rangle=1$.

I'm just confused at the moment. Thanks in advance for your help.

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    $\begingroup$ The annihilation operator acts as a creation operator when it acts to the left on the bra. $\endgroup$
    – march
    Apr 5 at 15:07
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    $\begingroup$ I understand finally. Thanks a lot. $\endgroup$
    – Shanks Red
    Apr 5 at 15:09
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    $\begingroup$ Another way to look at it: since $\langle 0|\hat{a}^\dagger\hat{a}|0\rangle=0$, $\langle0|\hat{a}\hat{a}^\dagger|0\rangle=\langle 0|[\hat{a},\,\hat{a}^\dagger]|0\rangle=\langle 0|1|0\rangle=1$. $\endgroup$
    – J.G.
    Apr 5 at 15:17

2 Answers 2

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The key point to remember is that $$\left(A|\psi\rangle\right)^\dagger=\left(\langle\psi|A^\dagger\right),$$ where I am using the $^\dagger$ for Hermitian conjugation. Equivalently, $$\left(A^\dagger|\psi\rangle\right)^\dagger=\left(\langle\psi|A\right).$$ We can use this in your expression to simplify $$\langle 0|a=(a^+|0\rangle)^\dagger$$ (your notation's $^+$ is the same as my $^\dagger) and the contradiction is resolved.

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  • $\begingroup$ thanks a lot. its' clear now. $\endgroup$
    – Shanks Red
    Apr 6 at 7:36
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These operators has the following properties: $a|n\rangle=\sqrt{n}|n-1\rangle$ and $a^\dagger|n\rangle=\sqrt{n+1} |n+1\rangle$, so: $$\langle0|a a^\dagger |0\rangle=\langle0|a\sqrt{0+1}|1\rangle=\langle 0|\sqrt{1}|0\rangle=1$$ The annihilation operator acts on left bra as following: $$\langle0|a=(a^\dagger|0\rangle)^\dagger=(|1\rangle)^\dagger=\langle1| $$

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