# Annihilation and creation operators in quantum harmonic oscillator

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.

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

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.
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: $$\langle0|a=(a^\dagger|0\rangle)^\dagger=(|1\rangle)^\dagger=\langle1|$$