# How to get Hamiltonian of QED from lagrangian?

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.

What follows is what you could describe as naiively applying the Legendre transform for constructing the Hamiltonian from the Lagrangian. Weinberg's "Quantum Theory of Fields" Vol. I chapter 8 goes over the canonical quantization of the electromagnetic field that correctly handles constraint equations.

Rather than finding the $T^{00}$ component of the stress-energy tensor, which often needs to be modified using the equations of motion to bring them into a recognizable form, you can construct the Hamiltonian using the canonical formalism. First, take the Lagrangian (using that instead of the density so that I can be free to move back and forth from mode space - also using Heaviside-Lorentz units with $\hbar=c=1$, and particle physicists' metric signature $(+,-,-,-)$), \begin{align} L & = \int \operatorname{d}^3x \left(\bar{\psi}\left[i\gamma^\mu D_\mu - m\right]\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} \right) \\ & = \int \operatorname{d}^3x \left( \bar{\psi}\left[i\gamma^\mu D_\mu - m\right]\psi + \frac{1}{2} E_iE_i - B_iB_i \right) \\ & = \int \operatorname{d}^3x \left( \vphantom{\frac{1}{2}} \bar{\psi}i\gamma^0\partial_0\psi + \bar\psi i \gamma^i D_i \psi - m\bar{\psi}\psi - e \phi \bar{\psi}\gamma^0\psi \right. \\ & \hphantom{=\int \operatorname{d}^3x}\ \ \left. + \frac{1}{2}\left[-\partial_i\phi-\partial_0A_i\right] \left[-\partial_i\phi-\partial_0A_i\right] - \frac{\epsilon_{ijk}\epsilon_{inm}}{2} \partial_jA_k \partial_nA_m \right). \end{align} The reason for expanding the the covariant inner products is to make explicit the next step: defining the canonically conjugate momenta. \begin{align} \pi_\psi &\equiv \frac{\delta L}{\delta \partial_0 \psi} = \bar{\psi}i\gamma^0 = i\psi^\dagger \\ \Pi_i &\equiv \frac{\delta L}{\delta \partial_0 A_i} = \partial_i\phi + \partial_0 A_i \\ \Pi_\phi &\equiv \frac{\delta L}{\delta \partial_0 A_0} = 0 \end{align} Notice how the momentum canonically conjugate to $\phi=A_0$ doesn't appear in the Lagrangian. This means that, without gauge fixing, $\phi$ is a non-dynamical field that doesn't have any momentum. In classical physics, it plays the roll of a Lagrange multiplier that enforces $\rho - \nabla\cdot\mathbf{E}= e\bar{\psi}\gamma^0\psi + \partial_i \Pi_i =0$, the first of Maxwell's equation.

Now, $H$ is defined as \begin{align} H &= \int \operatorname{d}^3x \left[\pi_\psi \partial_0\psi + \Pi_i \partial_0 A_i\right] - L \\ &= \int \operatorname{d}^3x \left[ \vphantom{\frac{\epsilon_i}{2}} \Pi_i (\Pi_i - \partial_i\phi) - \pi_\psi \gamma^0 \gamma^i D_i\psi -i m\pi_\psi \gamma^0\psi - ie\phi\pi_\psi \psi \right. \\ &\hphantom{= \int \operatorname{d}^3x}\ \ \left.- \frac{1}{2} \Pi_i \Pi_i + \frac{\epsilon_{ijk}\epsilon_{inm}}{2}\partial_jA_k \partial_nA_m \right]. \end{align} Notice that the term containing the time derivative of $\psi$ vanishes entirely because the Dirac equation is first order in time, so all of the needed time derivatives are supplied by the canonical equations of motion.

Now we move the gauge invariant parts of the electromagnetic field into mode space (Fourier transform over space, but not time) to highlight some interesting structure in the Hamiltonian. \begin{align} H & = \int \operatorname{d}^3k \left[ \frac{1}{2}\Pi_i\Pi_i + \frac{k^2}{2} \left(\delta_{ij} - \frac{k_i k_j}{k^2}\right)A_iA_j \right] \\ &\hphantom{=}+\int \operatorname{d}^3x \left[- \pi_\psi \gamma^0 \gamma^i D_i\psi -i m\pi_\psi \gamma^0\psi - ie\phi\pi_\psi \psi - \Pi_i\partial_i\phi \right] \\ &= \int \operatorname{d}^3k \left[ \frac{k_ik_j}{2 k^2}\Pi_i\Pi_j + \frac{1}{2} \left(\delta_{ij} - \frac{k_i k_j}{k^2}\right)\left(\Pi_i\Pi_j + k^2 A_iA_j\right) \right] \\ &\hphantom{=}+\int \operatorname{d}^3x \left[- \pi_\psi \gamma^0 \gamma^i D_i\psi -i m\pi_\psi \gamma^0\psi - ie\phi\pi_\psi \psi + \phi \partial_i \Pi_i\right] \end{align} Note that the electromagnetic field momentum, $\Pi_i$, is gauge invariant (it's just $-E_i$). The quantity $\left(\delta_{ij} - \frac{k_i k_j}{k^2}\right)A_j$ is also gauge invariant since it is the Fourier transform of the solenoidal part of $A_i$; in other words, the purpose of that last line was to collect the solenoidal components of the electromagnetic field (the photons) and move the derivative from $\phi$ to $\Pi_i$ in that term. The longitudinal kinetic term, $\frac{k_i k_j}{2k^2}\Pi_i\Pi_j$, is likewise nicely isolated.

It is also interesting to note that the only terms that aren't manifestly gauge invariant are $$-ie\phi\pi_\psi \psi + \phi \partial_i \Pi_i = e\phi\bar{\psi}\gamma^0\psi - \phi\partial_i E_i,$$ which vanishes by the equations of motion. It strikes me as likely that it is dealing with these terms in a canonical fashion that leads to the requirement for gauge fixing and constraints.

• Not sure if this is a typo or a notational misunderstanding on my part, but when you expand the covariant derivative in the Lagrangian, why is there no relative minus between the 0 and the $i$ terms? I.e., with the signature $(+---)$, I would write $\gamma^\mu \partial^\mu = \gamma^0 \partial_0 - \gamma^i \partial_i$. Nov 24, 2021 at 8:53
• @Durd3nT it's because the derivative with respect to index up coordinates is naturally index down and the gamma matrices are naturally index up. You only get the metric signs when you need to use the metric to raise or lower one set of indices. Note that you can only contract index up with index down - if you contract both up or both down you've made a mistake. In other words it's $\gamma^\mu \partial_\mu = \gamma^0\partial_0 + \gamma^i\partial_i = \gamma_0\partial_0 - \gamma_i\partial_i$. Nov 24, 2021 at 17:53
• @Durd3nT Also, consider $\mathrm{d}x^\nu\, \partial_\nu$, for example. Nov 24, 2021 at 17:55

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 Sundermeyer or Henneaux+Teitelboim for details regarding the Hamiltonian formalism of constraint systems.