Given a Lagrangian, to switch to a Hamiltonian, we do a Legendre transform.

Suppose the Lagrangian has fermions, say a term like $\frac{i}{2}(\bar{\psi} \dot{\psi} - \dot{\bar{\psi}} \psi)$, then I believe the two conjugate momenta would be

\begin{align*} \Pi_{\psi} &= \frac{\partial L}{\partial \dot{\psi}} = \frac{i}{2} \bar{\psi}\\ \Pi_{\bar{\psi}} &= \frac{\partial L}{\partial \dot{\bar{\psi}}} = -\frac{i}{2} \psi. \end{align*}

Perhaps there is already an ambiguity above with minus signs, and therefore the conjugate momenta are correct only up to a sign. Regardless, when we want the Hamiltonian, it makes a difference whether we have

\begin{align*} H &= ..\ \Pi_{\psi}\dot{\psi} + \Pi_{\bar{\psi}}\dot{\bar{\psi}} - L \end{align*}


\begin{align*} H &= ..\ \dot{\psi}\Pi_{\psi}+ \dot{\bar{\psi}}\Pi_{\bar{\psi}} - L \end{align*}

Could someone explain to me what the understood conventions are please?


1 Answer 1


The rule with fermions is as follows.

You get the momentum as a left-derivative of $L$; in other words, write $\dot{\psi}$ on the left of each factor so$$L=-\frac{i}{2}(\dot{\psi}\overline{\psi}+\dot{\overline{\psi}}\psi)\implies\Pi_\psi=-\frac{i}{2}\overline{\psi},\,\Pi_\overline{\psi}=-\frac{i}{2}\psi.$$

You get $H+L$ as a sum of $\dot{q}p$ terms, not $p\dot{q}$. So$$H=\dot{\psi}\Pi_\psi+\dot{\overline{\psi}}\Pi_\overline{\psi}-L=0.$$

Unfortunately, this example isn't very helpful pedagogically; $H=0$ doesn't happen in general. The problem here is (i) you only used time derivatives in $L$, not spacetime derivatives as expected in a field theory, & (ii) each term is proportional to a time derivative, so $L$ is just the value of $H+L$ from a Legendre transform. A more realistic $L$, assuming Cartesian spacetime coordinates in Minkowski space with the $+---$ convention, is$$L=-i\partial_\mu\psi\partial^\mu\overline{\psi}=i\partial_\mu\overline{\psi}\partial^\mu\psi\implies\Pi_\psi=-i\partial^0\overline{\psi},\,\Pi_\overline{\psi}=i\partial^0\psi\implies H=i\Pi_\overline{\Psi}\Pi_\Psi+i\partial_j\psi\partial^j\overline{\psi}.$$Note that, unlike in discrete mechanics, a field theory's Hamiltonian depends on not just canonical fields and conjugate momentum densities, but also space (but not time) derivatives of canonical fields. Meanwhile, because the Legendre transform introduces two $\dot{q}p$ terms but $-L$ only cancels one of them, I've kept one, which has had its $\dot{q}$ factor rewritten in terms of $p$s (you always have to do that when obtaining $H$, although sometimes $\partial_jq$ is also needed to do it).

You get Hamilton's equations with right-derivatives of $H$. For example, $\dot{\psi}=i\Pi_\overline{\Psi}$, but to obtain $\dot{\overline{\Psi}}$ by differentiating $H$ with respect to $\Pi_\overline{\Psi}$ the two-momenta term must be rewritten with that momentum on the write, which as expected gives an extra $-$ sign. Note these results match what $L$ gave us.

  • $\begingroup$ Thanks. Is there any source you can recommend that explains this sort of thing carefully? Especially for the quantum mechanical case, rather than field theory $\endgroup$
    – Gleeson
    Commented Dec 18, 2023 at 12:04
  • $\begingroup$ And what about when calculating $\delta L$? In Tong's notes on page 22, damtp.cam.ac.uk/user/tong/susy/susyqm.pdf, he doesn't seem to worry about these sorts of things $\endgroup$
    – Gleeson
    Commented Dec 18, 2023 at 12:16
  • $\begingroup$ Also, if the fermions are 2-component spinors (which I had meant, but accidentally didn't say), then I don't see how your answer can make sense, because wouldn't the barred fermion need to go on the left? $\endgroup$
    – Gleeson
    Commented Dec 18, 2023 at 12:22
  • $\begingroup$ I wish I could suggest a better source off the top of my head than this. Are you sure that's on page 22? If you're worried about spinors, write everything in terms of components (e.g. $\psi_a^\ast(\gamma^0)_{ab}\psi_b$) so you can rearrange products as needed. $\endgroup$
    – J.G.
    Commented Dec 18, 2023 at 12:35
  • $\begingroup$ Yes, page 22 where he is calculating $\delta L$. It seems like the "delta" variation when applied to L just moves past terms with no minus sign issues? And then he seems to just be careful when plugging in the various formula from 1.19 for $\delta \psi, \delta \dot{\psi}$ etc to place the formulae in the precise place where the $\delta \psi, \delta \dot{\psi}$ was in $\delta L$. Is this consistent with your suggestion if one instead wrote $\delta L = \frac{\partial L}{\partial \psi}\delta \psi + \ ..$ etc? $\endgroup$
    – Gleeson
    Commented Dec 18, 2023 at 12:48

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