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? enter image description here

Also, how do you prove the following commutation relation?

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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.

  • $\begingroup$ Thanks. Im only 15 so it's nice to see how all this comes together in QFT. $\endgroup$ – user34039 Dec 21 '13 at 23:21
  • 2
    $\begingroup$ @user34039 Wow, that's impressively young to be studying QFT. $\endgroup$ – Stephen Blake Dec 22 '13 at 0:01
  • $\begingroup$ @user34039 You might find good company in user Dimensio1n0 - look him up and also his blogs where he gives videos of discussions of the Kerr metric, foundations of mathematics and quite a bit more! $\endgroup$ – WetSavannaAnimal Dec 22 '13 at 12:22

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