When canonical quantizing the electromagnetic field in the Lorenz gauge, the equal time commutator is written as: $$[A^\mu(\vec{x},t), \pi^{\nu}(\vec{y},t)]=ig^{\mu\nu} \delta^3(\vec{x}-\vec{y}).\tag{1}$$ This is a little bit confusing to me.

The Lagrangian of the free EM field is $$\mathcal{L}_{EM}=-\frac{1}{4}F_{\mu\nu}F^{\mu \nu}.\tag{2}$$ Thus the canonical momentum is: $$\pi^{\nu} = \frac{\partial \mathcal{L}}{\partial \dot{A}_{\nu}}=-F^{0\nu} = -\partial^0A^{\nu} + \partial ^{\nu}A^0.\tag{3}$$

So, if we write the $A_{\mu}$ field in Fourier mode expansions, it is :

$$A_{\mu}(x)=\int\frac{d^3\vec{p}}{(2\pi)^3} \frac{1}{\sqrt{2|\vec{p}}|}\sum_{\lambda=0}^3\epsilon_{\mu}^{\lambda}\{a_{\vec{p}}^{\lambda}e^{-ipx} + a_{\vec{p}}^{\lambda \dagger}e^{ipx} \}.\tag{4}$$

By the definition of the canonical momentum, its mode expansion should be $$ \pi_{\mu}(x) = i\int\frac{d^3\vec{p}}{(2\pi)^3} \sqrt{\frac{|\vec{p}|}{2}}\sum_{\lambda=0}^3\epsilon_{\mu}^{\lambda}\{a_{\vec{p}}^{\lambda}e^{-ipx} - a_{\vec{p}}^{\lambda \dagger}e^{ipx} \} - i \int\frac{d^3\vec{p}}{(2\pi)^3} \frac{p_{\mu}}{\sqrt{2|\vec{p}}|}\sum_{\lambda=0}^3\epsilon_{0}^{\lambda}\{a_{\vec{p}}^{\lambda}e^{-ipx} - a_{\vec{p}}^{\lambda \dagger}e^{ipx} \},\tag{5}$$ where only the first term in $\pi_{\mu}(x)$ fits the commutation relation. Is the commutation relation wrong or is my canonical momentum wrong?


1 Answer 1


Comments to the post (v1):

  1. Eq. (1) is the defining property of the Hamiltonian momenta $\pi^{\nu}$.

  2. Eq. (3) is the definition of the Lagrangian momenta $\pi^{\nu}$. Note that $\pi^{0}= 0$.

  3. In order to harmonize these two definitions, one should perform a Dirac-Bergmann analysis to introduce adequate constraints.


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