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Suppose one has a real scalar field $\phi$ and its conjugate momentum field $\pi := \partial_t \phi$. If the scalar is free and using metric $(-+++)$, the action is $$ \int d^4x\ \mathscr{L}_{\mathrm{free}} = - \frac{1}{2} \int d^4 x\ ( \partial_\mu \phi \partial^{\mu} \phi + m^2 \phi^2 ) $$ which means the free Hamiltonian is $$ H_{\mathrm{free}} = \frac{1}{2} \int d^3 \mathbf{x}\ \big( \pi^2 + ( - \nabla^2 + m^2 ) \phi \ \big) \ . $$

My question is: For what reason is it not allowed that $H_{\mathrm{free}}$ includes terms of the form $\pi \phi$? I can see how this is related to $\partial_t(\phi^2)$, and feel like this may be why, but can't quite see the reason.

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    $\begingroup$ Who says it is not allowed? $\endgroup$
    – Qmechanic ♦
    Commented Nov 17, 2021 at 22:53
  • $\begingroup$ @Qmechanic Maybe it's incorrect to say its not allowed, but it is surely true that terms of this form are usually not written down in an action for a real scalar - why not? $\endgroup$
    – QuantumEyedea
    Commented Nov 18, 2021 at 1:48
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    $\begingroup$ A reason you wouldn't literally see the coupling $\pi \phi$ is that then the Hamiltonian would be unbounded from below (since the term $\pi \phi$ could become arbitrarily large in absolute value and negative if $\pi$ and $\phi$ had opposite signs). However there's no reason you couldn't have terms like $(\pi \phi)^2$. Sometimes you can remove these kinds of "mixed" terms with a canonical transformation, and if so people may also prefer to choose variables where these mixed terms don't arise. $\endgroup$
    – Andrew
    Commented Nov 18, 2021 at 2:07
  • $\begingroup$ A term like $\phi \partial_t \phi$ would not be Lorentz invariant $\endgroup$
    – Sal
    Commented Nov 18, 2021 at 17:45

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$\partial_t(\phi^2)$ is why $H_{free}$ includes terms of the form $\pi \phi$.

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