by letting $\omega' \to -\omega'$ in Eq. (\[acommu\]) . By using the Klein-Gordon inner products computed above, we find the following commutators: $$\begin{aligned}
 \left[ a^{(3n)}_{\omega lm}, a^{(3n')\dagger}_{\omega' l'm'}\right]
= -\left[ a^{(0n)}_{\omega lm}, a^{(3n')\dagger}_{\omega' l'm'}\right]
&=& \delta_{nn'}\delta_{ll'}\delta_{mm'}\delta(\omega-\omega')\,, \\
\left[ a^{(1n)}_{\omega lm}, a^{(1n')\dagger}_{\omega' l'm'}\right]
= \left[ a^{(2n)}_{\omega lm}, a^{(2n')\dagger}_{\omega' l'm'}\right]
& = & \delta_{nn'}\delta_{ll'}\delta_{mm'}\delta(\omega-\omega')\end{aligned}$$ with all other commutators vanishing. The Gupta-Bleuler condition [@IZ] requires that any physical state $|{\rm phys}\rangle$ satisfy $$\hat{G}^{(+)}|{\rm phys}\rangle = 0\,,$$ where $\hat{G}^{(+)}$ is the positive-frequency part of $\hat{G} = \nabla^\mu \hat{A}_\mu + K^\mu \hat{A}_\mu$. Since this quantity is nonvanishing only for $A^{(0n;\,\omega lm)}_\mu$, this condition is equivalent to $$a^{(0n)}_{\omega l m}|{\rm phys}\rangle = 0\ \ \ {\rm for\ all\ }
(n,\omega,l, m)\,\ ({\rm with}\ \omega > 0)\,.$$ (We let $\omega > 0$ below.) The Boulware vacuum $|0\rangle$ [@Boul] is defined by requiring that it be annihilated by all $a^{(\lambda n)}_{\omega lm}$ operators $(\lambda=0,1,2,3)$. Note that the states obtained by applying any number of creation operators excluding $a^{(3n)\dagger}_{\omega lm}$ are all physical states. Any state of the form $a^{(3n)\dagger}_{\omega lm}|{\rm phys}\rangle$ is unphysical because $$a^{(0n')}_{\omega'l'm'}a^{(3n)\dagger}_{\omega lm}|{\rm phys}\rangle
= - \delta_{nn'}\delta_{ll'}\delta_{mm'}\delta(\omega-\omega')
|{\rm phys}\rangle \neq 0\,.$$ Note also that the physical states of the form $a^{(0n)\dagger}_{\omega lm}|{\rm phys}\rangle$ have zero 