I am doing a calculation of an amplitude in QFT, not an expert in the subject so this may be a trivial question but cannot find the answer.
What is the normalization of the vacuum state of the electromagnetic field? Is it just $$ \langle 0 | 0 \rangle = 1 $$ or is there some Dirac delta function, like $$\langle 0 | 0 \rangle = \delta(0) $$ that can be interpreted as the quantization volume?
The vacuum state is denoted $|0\rangle$ in the analogy with the $0$-quanta ground state $|0\rangle$ of a quantum SHO, and in both cases $\langle 0|0\rangle=1$. Unsurprisingly, this implies $3$-momentum eigenstates satisfy$$\left\langle\vec{k}|\vec{q}\right\rangle=\left\langle0|a_{\vec{k}}a^\dagger_{\vec{q}}|0\right\rangle=\left\langle0|(2\pi)^32\omega_\vec{k}\delta^3(\vec{k}-\vec{q})\mathbb{I}|0\right\rangle=(2\pi)^32\omega_\vec{k}\delta^3(\vec{k}-\vec{q})$$in the Lorentz-invariant normalization. Of course, the case $\vec{k}=\vec{q}=\vec{0}$ then satisfies$$\left\langle\vec{0}|\vec{0}\right\rangle=(2\pi)^32\omega_\vec{k}\delta^3(\vec{0}),$$but now the zeros mean something different.
The normalization is just 1, or Kronecker delta if you are thinking about single-mode quantum states in general (e.g. 0,1,2... bosons in the same excitation state). The vacuum state is defined inside a Fock space and has length one.
As I understand it, the $\delta (k-k')$ arises naturally when you are computing the amplitude of a k, k' transition, $\langle k \vert k' \rangle$, from the commutation relations of the k, k' "ladder" operators (you can see, for example this document, which I believe is a good starting reference, in equations 2.97 and 2.98). That is what can be interpreted as a quantization volume as you suggested, at least according to my notes on this from the degree (which may not be the best reference).
