# Canonical equal time commutation relations in QED

I understand that to quantize the classical electromagnetic field one needs to impose commutation relations and express the field in terms of creation and annihilation operators. I notice that the canonical equal-time commutation rule comes in to use here. How do you derive this? What is it meant to represent? Also, how do you prove the following commutation relation? In classical Hamiltonian mechanics one has the generalized coordinates $q^{i}(t)$ and momenta $p_{i}(t)$. The Poisson bracket is defined as, $$[F,G]_{PB}=\frac{\partial F}{\partial q^{k}}\frac{\partial G}{\partial p_{k}}-\frac{\partial F}{\partial p_{k}}\frac{\partial G}{\partial q^{k}}\ .$$ Using the q's and p's in place of $F$ and $G$ one has the fundamental PBs, $$[q^{i}(t),p_{j}(t)]_{PB}=\delta^{i}_{k}\delta^{k}_{j}=\delta^{i}_{j}$$ $$[q^{i}(t),q^{j}(t)]_{PB}=0$$ $$[p_{i}(t),p_{j}(t)]_{PB}=0 \ .$$ In order to set up a quantum theory, Dirac says that the phase space functions $F$ and $G$ are changed to operators $\hat{F}$ and $\hat{G}$ and the PB becomes the commutator, $$[\hat{F},\hat{G}]_{-}=\hat{F}\hat{G}-\hat{G}\hat{F}=i\widehat{[F,G]_{PB}} \ .$$ So, going over to the quantum theory in this way, the fundamental PBs become commutators, $$[\hat{q}^{i}(t),\hat{p}_{j}(t)]_{-}=i\delta^{i}_{j}$$ $$[\hat{q}^{i}(t),\hat{q}^{j}(t)]_{-}=0$$ $$[\hat{p}_{i}(t),\hat{p}_{j}(t)]_{-}=0 \ .$$ These three commutators are the ones in the question. This is because the generalized coordinates for the classical electromagnetic field are, $$q^{i}(t)\rightarrow q^{(\mu,x)}(t) \rightarrow A_{\mu}(t,x) \ .$$ The generalized momenta $p_{i}(t)$ will be the values of a vector field $\pi^{\mu}(t,x)$ and so the first fundamental commutator becomes, $$[\hat{A}_{\mu}(t,x),\hat{\pi}^{\nu}(t,y)]_{-}=i\delta^{\nu}_{\mu}\delta(x-y)$$ and this is the first commutator in the question. The other two commutators in the question follow from the remaining two commutators for the q's and p's with a bit of raising/lowering indices of four-vectors.