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...
 A: 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.
A: We have the annihilation and creation operators $a$ and $a^\dagger$, respectively (we know that they are the hermitian conjugates of each other, but we won't assume that fact). So let's rename $a^\dagger=b$.
I'll try to show that $a=a^\dagger$ and $b=b^\dagger$, gives a contradiction. We know: 
$$ba\vert n\rangle=n\vert n\rangle \quad, \quad ab\vert n\rangle=(n+1)\vert n\rangle$$We are supposing that $a^\dagger$ is not the hermitian conjugate of $a$.
With this you could prove: $[a,b]=\mathbb{I}$
Then:
$$n^2\langle n \vert n\rangle=\langle n\vert a^\dagger b^\dagger ba\vert n \rangle=\langle n\vert a b ba\vert n \rangle=$$
$$=\langle n\vert[( b^\dagger a^\dagger+\mathbb{I}) ba]\vert n \rangle=n(n+1) \langle n\vert n \rangle+n \langle n\vert n \rangle$$
Where I have used the commutation relation and the hermiticy property. This yields: $n=0$ or $\vert n\rangle =0$.
So unless $\vert n\rangle=\vert 0\rangle$ or $0$, $a$ and $a^\dagger$ won't be hermitian (I don't think it holds  if the only state is $\vert 0 \rangle $ or $0$ but I'm not sure).

Is it because they 'create' and 'annihilate' photons

This reasoning would be very sloppy. Ladder operators arise in the context of the harmonic oscillator, angular momentum... But in Quantum Mechanics, particle number is conserved (you could say something like it modifies the energy, hence it emmits a photon or something like that). These operators create and destroy photons in Quantum Field Theory.
A: My first reaction is to give an argument along the lines of the other answers, if $a=a^\dagger$ then saying that $a$ raises and $a^\dagger$ lowers doesn't make sense. 
However, if you don't want to assume that the hermitian conjugate of $a$ is a raising operator, you could be more direct than that though and show $a$ is not hermitian given just that $a$ is a lowering operator, without knowing $a^\dagger$ is a raising operator. All you have to do is construct the matrix elements of $a$ in some convenient basis, and show that the resulting matrix is not hermitian. The $|n\rangle$ basis is a pretty convenient one!
In order for $a$ to be hermitian, it must be that $a_{mn}^*= a_{nm}$, or equivalently
\begin{equation}
\langle m | a | n \rangle^* {=}^? \langle n | a| m \rangle 
\end{equation}
But this is clearly false given the definition of $a$, because
\begin{equation}
LHS = \langle m | a | n \rangle^* = \sqrt{n} \langle m | n-1 \rangle^* = \sqrt{n} \delta_{m,n-1} = \sqrt{n} \delta_{m+1,n}
\end{equation}
while
\begin{equation}
RHS = \langle n | a | m \rangle = \sqrt{m} \langle n-1 | m \rangle = \sqrt{n-1} \delta_{m-1,n}
\end{equation}
Clearly $LHS \neq RHS$, so $a$ is not hermitian. [To see they don't equal just try a few examples, any time $LHS\neq 0$ then $RHS=0$. If you wrote them out in matrix notation, LHS would have entries one slot above the diagonal, RHS would have entries one slot below.]
In fact, this is essentially a proof that $a^\dagger$ is a raising operator (since $\langle n | a^\dagger | m \rangle = \langle m | a | n \rangle^*$), so this argument is not fundamentally that different from the other answers.
A: If $a$ is a lowering operator then by def*: $$\left[N,a\right]=-a$$
Now if $a$ is Hermitian then the left hand side of the above is anti-Hermitian, contradicting its equality to the left.
Taking the Hermitian conjugate of this equation gives:$$\left[N,a^{\dagger}\right]=a^{\dagger}$$
Making $a^{\dagger}$ a raising operator.
*Here's why: If $N$ is some operator with a basis of Eigenstates and $a$ is a ladder operator which shifts $N$'s value by $\lambda$ then on any state in the Eigenbasis with Eigenvalue $n$: $$ Na \left|n\right>=(n+\lambda)a\left|n\right> = \lambda a\left|n\right>+ a n \left|n\right>=\lambda a\left|n\right>+ a N \left|n\right> $$
$$\Rightarrow (Na-aN) \left|n\right>= \lambda a \left|n\right> $$
Since $\left|n\right>$ is a general basis element we find: $$\left[N,a\right]= \lambda a$$
A: Lets give an answer based on the physical meaning of the operator $\hat{a}$
The operator $\hat{a}$, is the quantum-mechanical operator which represents a measure of the classical complex representation of the electric field. $\hat{a}$ in classical physics is well-known as the analytic signal, which is in short a generalization of the complex representation of any time-varying signal.
For example:
$$ 
2*E_0cos(\omega t) = E_0e^{i\omega t} + E_0e^{-i \omega t}
$$
The complex representation of the real harmonic signal $E_0 cos(\omega t)$ can be taken as $E_0/2e^{i\omega t}$. But what to pick as the complex representation for any time-varying electric field ? Well, one can see from the simple above example that the complex representation can be defined by keeping only the positive frequency part of the Fourier transform of the signal.
This complex number (the analytic signal a) for any real signal x(t), can be written as $a = x - ip$ where $p$ deduces from the Hilbert transform of $x(t)$
Quantum mechanics is simply the mathematical reinterpretation of the cinematics of the classical physics quantities and by construction the eigenvalues of the operator $\hat{a}$ are complex.
