# Commutation relations in Gupta-Bleuler formalism

When quantising the EM field thanks to the Gupta-Bleuler formalism, Itzykson and Zuber assume that the canonical commutation rules are

$$[\hat{A}_\rho (t,\vec{x}), \hat{\pi}^\nu(t,\vec{y})]= i \, g_\rho^{\, \,\nu}\, \delta^3(\vec{x}-\vec{y})$$

I don't quite understand where the metric tensor comes from. When quantising the scalar field, we assume commutation rules between the field and its conjugate momentum to be

$$[\hat{\phi} (t,\vec{x}), \hat{\pi}(t,\vec{y})]= i \, \delta^3(\vec{x}-\vec{y})$$

because of the Poisson bracket to commutator classical to quantum correspondance

$$\{\phi (t,\vec{x}), \pi(t,\vec{y})\}_{Poisson} \rightarrow -i \, [\hat{\phi} (t,\vec{x}), \hat{\pi}(t,\vec{y})]$$

For what reason does $$g$$ appear in this case?

• FWIW, $g_\rho^{\, \,\nu}$ is the Kronecker delta function. – Qmechanic Apr 9 '19 at 21:37

The fact is that $$\hat{\pi}^{\nu}$$ is the conjugate momentum to $$\hat{A}_{\nu}\;$$, so the presence of the tensor $$g_{\rho}^{\nu}=g_{\rho\tau}g^{\tau\nu}$$ ensures that the commutator is taken between the field (component) and its respective conjugate momentum.
• Of course, $\hat{\pi}^{\rho}= \dfrac{\partial \mathcal{L}}{\partial(\partial_0 A_\rho)}$... I feel silly for not seeing this! Thanks a lot – takunitoche Apr 10 '19 at 7:38