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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.