Quantization of the free electro-magnetic field has essential differences in comparison to quantization of say scalar or massive vector fields. In fact there are different approches to it.

One of them is a covariant approach due to Gupta and Bleuler, see Section 3-2-1 in "Quantum Field Theory" by Itzykson and Zuber. First one modifies the usual Lagrangian $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ to $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} +\frac{\lambda}{2}(\partial\cdot A)^2.$$ Then the conjugate momenta to the four components of $A$ are $$\pi^\rho=\frac{\partial \mathcal{L}}{\partial(\partial_0A_\rho)}=F^{\rho 0}-\lambda g^{\rho 0}(\partial\cdot A).$$

Then one assumes canonical commutation relations (see (3-102) in the book): $$[A_\rho(\vec x,t),\pi_\nu(\vec y,t)]=i g_{\mu\nu}\delta^3(\vec x-\vec y).$$

For $\mu=\nu=0$ this prescription is different from the standard one known from QM as the commutator has the wrong sign. In other words the standard prescription tells to take $\delta_{\mu\nu}$ in the right hand side rather than $g_{\mu\nu}$. This new prescription seems somewhat arbitrary and unmotivated to me. I would be happy to get any explanation.

REMARK. In quantization of the Dirac field one also changes the prescription known from QM by replacing commutators with anti-commutators. However this significant change is usually motivated in text books, including the one mentioned above.


1 Answer 1


The standard prescription has the contravariant form of the conjugate momentum: $$[A_\mu(\vec x,t),\pi^\nu(\vec y,t)]=g^{\nu\alpha}[A_\mu(\vec x,t),\pi_\alpha(\vec y,t)]=i g^{\nu\alpha}g_{\mu\alpha}\delta^3(\vec x-\vec y)=i\delta^\nu_\mu\delta^3(\vec x-\vec y)$$

  • $\begingroup$ Does that mean that I wrote the standard prescription known from QM after lowering one index, i.e. there was no change of the prescription at all? In that case my understanding was completely wrong. $\endgroup$
    – MKO
    Commented Jun 20, 2019 at 18:49
  • $\begingroup$ @MKO You seem to have neglected the tensor identity. On the other hand I don't see how it can make a big difference, but I have not studied this much. $\endgroup$
    – user195162
    Commented Jun 20, 2019 at 18:59

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