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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?

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

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  • $\begingroup$ Of course, $\hat{\pi}^{\rho}= \dfrac{\partial \mathcal{L}}{\partial(\partial_0 A_\rho)}$... I feel silly for not seeing this! Thanks a lot $\endgroup$ – Dima Fontaine Apr 10 at 7:38
  • $\begingroup$ Well it was good anyways to see gupta-bleuler to be brought up in a question! :-) $\endgroup$ – AoZora Apr 10 at 21:19

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