quantum mechanics - Annihilation and Creation operators not hermitian

The annihilation and creation operators are given below $$ \hat a|n\rangle=\sqrt{n}|n-1\rangle\qquad\text{and}\qquad\hat a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle $$

Now I know the operator $a^\dagger$ is not Hermitian but as to why I'm a little confused. Is it because they 'create' and 'annihilate' photons hence they're not hermitian? I don't think that explanation would be good for an exam if this question was asked though...

Answer

Creation and annihilation operators are ladder operator in the sense that they raise and lower respectively the quantum numbers of a state (such as e.g. the number of particles in an harmonic oscilattor, the angular momentum for spins, etc...). If they were hermitian, that is $a=a^\dagger$, the same operator $a$ should lower and raise the quantum number at the same time spoiling its very definition. Perhaps, you can think of the ``number'' operator $a^\dagger a$ as a sort of hermitian cousin of $a$ or $a^\dagger$, and indeed it does not raise or lower the quantum numbers but it simply counts.