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I'd like to know what the identity operator in the canonical quantization of EM field. Here's my guess: \begin{equation} 1=\int\frac{d^3 \vec{p}}{(2 \pi)^3}\frac{1}{2|\vec{p}|}\sum_{\lambda=0}^{3}|\vec{p}, \lambda \rangle\langle \vec{p},\lambda| \end{equation} But there's something wrong when I tested my guess.

\begin{equation} \begin{split} |\vec{k},\lambda> &= \int \frac{d^3 \vec{p}}{(2\pi)^3} \frac{1} {2|\vec{p}|}\sum_{\lambda'=0}^{3}|\vec{p}, \lambda' \rangle\langle \vec{p},\lambda'|\vec{k},\lambda \rangle \\ & = \int \frac{d^3 \vec{p}}{(2\pi)^3} \frac{1} {2|\vec{p}|}\sum_{\lambda'=0}^{3}|\vec{p}, \lambda' \rangle\langle 0|\sqrt{2|\vec{p}|}\sqrt{2|\vec{k}|}\, a_{\vec{p}}^{\lambda'} a^{\lambda \dagger}_{\vec{k}}|0 \rangle \\ & = \int \frac{d^3 \vec{p}}{(2\pi)^3} \sqrt{\frac{|\vec{k}|} {|\vec{p}|}}\sum_{\lambda'=0}^{3}|\vec{p}, \lambda' \rangle\langle 0|[a_{\vec{p}}^{\lambda'}, a^{\lambda \dagger}_{\vec{k}}]|0 \rangle \\ & = \int \frac{d^3 \vec{p}}{(2\pi)^3} \sqrt{\frac{|\vec{k}|} {|\vec{p}|}}\sum_{\lambda'=0}^{3}|\vec{p}, \lambda'\rangle \left[-g^{\lambda\lambda'}(2\pi)^3\delta^3(\vec{p}-\vec{k}) \right] \\ & = -\sum_{\lambda'=0}^{3}g^{\lambda\lambda'}|\vec{k}, \lambda' \rangle \end{split} \end{equation}

This is correct when $\lambda=1,2,3$, but there's an extra minus sign when $\lambda=0$. Does any one know how to fix my guess?

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  • $\begingroup$ λ =0? ....... ? $\endgroup$ – Cosmas Zachos Oct 26 '17 at 16:42
  • $\begingroup$ If $\lambda = 0$, $|\vec{k},0> \rightarrow -|\vec{k},0> $ after the identity acts on it. $\endgroup$ – LY3000 Oct 26 '17 at 16:44
  • $\begingroup$ How did you define your polarization vectors and how many are there? $\endgroup$ – Cosmas Zachos Oct 26 '17 at 16:53
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You should write \begin{equation} 1=-\int\frac{d^3 \vec{p}}{(2 \pi)^3}\frac{1}{2|\vec{p}|}\sum_{\lambda', \lambda^{\prime\prime}} g_{\lambda' \lambda^{\prime\prime}}\, |\vec{p}, \lambda' \rangle\langle \vec{p},\lambda^{\prime\prime}| \;, \end{equation} then you end up with $$ \sum_{\lambda', \lambda^{\prime\prime}} g_{\lambda' \lambda^{\prime\prime}} g^{\lambda^{\prime\prime} \lambda}\, | \vec k, \lambda' \rangle = |\vec k, \lambda \rangle \;. $$ Note that this is only the identity operator on the 1-particle subspace of the Fock space (the Fock space before imposing the Gupta-Bleuler condition!).

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