Lorenz gauge in the Gupta-Bleuler Method

Question

Greiner in his book Field Quantization page 180 & 181 wrote:

As shown in (7.24) the Lorenz condition cannot be enforced as an operator identity. Instead we will use it as a condition for the state vectors in Hilbert Space $|\Psi\rangle$. Only those state vectors are admitted for which the expectation value of the gauge condition is satisfied:
$$\langle\Psi|\partial^\mu\hat{A}_\mu(x)|\Psi\rangle=0\tag{7.48} $$

And finally he constructed a condition to pick out the physical states $|\Psi\rangle$

$$\partial^\mu\hat{A}^{(+)}_\mu(x)|\Psi\rangle=0$$

from which he achieved (7.48).

Below is my question:
Suppose all that we will discuss is in the physical state space $V=\{|\Psi\rangle\}$, now (7.48) would mean $\partial^\mu\hat{A}_\mu(x)=0$ in this space $V$, but this is apparently not true, since even the most trivial condition failed $$\partial^\mu\hat{A}_\mu(x)|0\rangle=\partial^\mu\hat{A}^{(-)}_\mu(x)|0\rangle\neq0$$ since $\hat{A}^{(-)}_\mu(x)$ consists of creation operators.
So what happened?

Answer 1

To test whether $|0\rangle$ is physical, you would apply $\partial^\mu A_\mu^+$ to it. So $|0\rangle \in V$.

Note that 7.48 does not mean that $\partial_\mu A^\mu = 0$ as an operator identity in this space. It means that all it's matrix elements in this space are zero. You have noticed that the state you created, $\partial_\mu A^\mu |0\rangle$, lives not in $V$ but in the space orthogonal to it. So it has zero overlap with any states in $V$, so all the matrix elements of $\partial_\mu A^\mu$ are still zero in $V$.

Long story short, we can't impose $\partial_\mu A^\mu = 0$ as an operator identity without losing the canonical commutation relations.