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I have the QED lagrangian: $$ L = \bar {\Psi}(i \gamma^{\mu }\partial_{\mu} + q\gamma^{\mu}A_{\mu} - m)\Psi + \frac{1}{16 \pi}F_{\alpha \beta}F^{\alpha \beta} . $$ I tried to get hamiltonian by getting zero component of energy-momentum tensor: $$ T^{\mu}_{\quad \nu} = i\bar {\Psi}\gamma^{\mu}\partial_{\nu}\Psi + \frac{1}{4 \pi}F^{\mu \gamma}\partial_{\nu}A_{\gamma} - \frac{1}{4 \pi}J^{\mu}A_{\nu}\Rightarrow $$ $$ T^{0}_{\quad 0} = i\Psi^{\dagger}\partial_{0}\Psi + \frac{1}{4 \pi}F^{0\gamma}\partial_{0}A_{\gamma} - \frac{1}{4 \pi}J^{0}A_{0} = i\Psi^{\dagger}\partial_{0}\Psi + \frac{1}{4 \pi}F^{0\gamma}\partial_{0}A_{\gamma} - \frac{1}{4 \pi}\Psi^{\dagger}A_{0}\Psi = H_{density}. $$ But it seems that it's incorrect, because I never get by this expression term $\bar {\Psi} \gamma^{\mu}\Psi A_{\mu}$, which refer to interaction part.

So how to find the true hamiltonian?

Thank you.

Added. Hmm, I find the mistake in expression of energy-momentum tensor. Fixed.

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1 Answer 1

You can only find the Hamiltonian if you do a so-called 'gauge fixing' procedure, since the Dirac field couples (minimally, but uniquely) to a gauge field. To get the Hamiltonian (density) you need to perform the full Dirac constraint analysis and at the end 'gauge fix'. See the books by Sundermayer or Henneaux+Teitelboim for details regarding the Hamiltonian formalism of constraint systems.

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