quantum mechanics - Why particle number operator $hat{N}$ is $hat{a}^daggerhat{a}$ rather than $hat{a}hat{a}^dagger$?

Both $\hat{a}^\dagger\hat{a}$ and $\hat{a}\hat{a}^\dagger$ are Hermitian, how do we know which one represents the particle number?

Answer

We require that the number operator have the following property:

$$\hat n |0\rangle = 0.$$

We know that

$$\hat a |0\rangle = 0$$

and we know that

$$\hat a |1\rangle = |0\rangle$$

and we know that

$$\hat a^{\dagger} |0\rangle = |1\rangle. $$

Thus, it follows that

$$\hat a \hat a^{\dagger} \ne \hat n$$

since

$$\hat a \hat a^{\dagger} |0\rangle = \hat a|1\rangle \ne 0.$$

Now, it remains to be shown that

$$\hat a^{\dagger}\hat a = \hat n. $$

Can you take it from here?