How is the state $|a_0 a_i\rangle$ physical? For a state $|\psi\rangle$ to be physical we require that:
$$\langle\psi|a_0^\dagger a_0|\psi\rangle = \langle\psi|a_i^\dagger a_i|\psi\rangle$$
It is always said that physical state must contain equal numbers of longitudial and temporal photons so let us try $|\psi\rangle = a^\dagger_0 a^\dagger_i |0\rangle$ as a state that is to be physical. (is it ? Why wouldn't it be ?)
The left hand side becomes:
$$\langle 0| a_0 a_i a^\dagger_0 a_0 a^\dagger_0 a^\dagger_i | 0\rangle = \langle 0|a_i a^\dagger_ia_0  a^\dagger_0 a_0 a^\dagger_0  | 0\rangle$$
$$\langle 0|(1+a^\dagger_ia_i)(-1+a^\dagger_0a_0)(-1+a^\dagger_0a_0)|\rangle$$
$$\langle 0|(1)(-1+a^\dagger_0a_0)(-1)|\rangle = \langle 0|(1)(-1)(-1)|\rangle = 1$$
The right hand side becomes:
$$\langle 0| a_0 a_i a^\dagger_i a_i a^\dagger_0 a^\dagger_i | 0\rangle = \langle 0|a_0 a^\dagger_0 a_i  a^\dagger_i a_i a^\dagger_i  | 0\rangle$$
$$\langle 0|(-1+a^\dagger_0 a_0)(1+a^\dagger_ia_i)(1+a^\dagger_ia_i)|\rangle$$
$$\langle 0|(-1)(1+a^\dagger_ia_i)(1)|\rangle = \langle 0|(-1)(1)(1)|\rangle = -1$$
Such that $rhs \neq lhs$ and the state appears to by unphysical.
The question
One of the following statements must be true but I can't figure out which one and why. I'm sorry in advance if this is trivial (it should be) but I'm really confused at the moment:


*

*the state that I tested is indeed unphysical, but why would that be? It contains an equal number of longitudinal and temporal photons so it it should be physical.

*I have made a sign or conceptual error in my calculation that I fail to spot.
 A: I am confused by the first statement in your question.
Normally, in the Gupta-Bleuler method we require physical states to satisfy
$$ (a_0 - a_3) \left| \Psi \right> = 0, $$
which is a quantized version of the Lorentz gauge constraint in the momentum space
$$ p_{\mu} A^{\mu} = 0, $$
provided that we choose the $x^3$ axis along the photon's momentum.
The space of solutions of this constraint decomposes into physical states generated by $a_{1,2}^{\dagger}$ and spurious states generated by $a_0^{\dagger} - a_3^{\dagger}$ which can be shown to have zero norm:
$$ \left< 0 \right| (a_0 - a_3) (a_0^{\dagger} - a_3^{\dagger}) \left| 0 \right> = 0. $$
These can be artificially excluded from the inner product space, after which we obtain the physical Hilbert space.
The states generated by $a_0^{\dagger} + a_3^{\dagger}$ don't satisfy the constraint and are thus unphysical.
Note that your state which is $a_0^{\dagger}a_3^{\dagger} \left| 0 \right>$ doesn't satisfy the constraint:
$$ (a_0 - a_3) a_0^{\dagger} a_3^{\dagger} \left| 0 \right> \neq 0, $$
and thus it is unphysical. This is because we can't represent the operator $a_0^{\dagger} a_3^{\dagger}$ solely as a function of $a_1^{\dagger}$, $a_2^{\dagger}$ and $a_0^{\dagger} - a_3^{\dagger}$.
