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I have read some questions (and the Wikipedia article) about the hamiltonian formulation of a QFT, but the only example that seems to be brought up is the scalar case, saying that $$\mathcal{H}_S=\Pi\partial_0\phi-\mathcal{L}_S.$$ Can I write the Hamiltonian for a general theory in the same way? For example, for Yang-Mills theory is the following true? $$\mathcal{H}_{YM}=\pi_\mu^a\partial_0W^{a\mu}-\mathcal{L}_{YM}.$$ What about for an interacting theory like Yang-Mills coupled with a scalar, can I write as follows? $$\mathcal{H}=\pi_\mu^a\partial_0W^{a\mu}+\Pi\partial_0\phi-\mathcal{L}.$$

I don't see why not, after all the two functions should exist for all these theories, and I can't think of another way to find the Hamiltonian knowing the Lagrangian.

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    $\begingroup$ One subtlety is if you include fermions you need $\dot{q}p$ terms rather than $p\dot{q}$ terms (bosons don't care either way). You then get equations of motion from the Lagrangian with left-derivatives, or the Hamiltonian with right-derivatives. $\endgroup$
    – J.G.
    Commented Nov 15, 2021 at 15:22
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    $\begingroup$ @J.G. I don't understand your comment. If I consider the Dirac Lagrangian $\mathcal{L} = \bar{\psi} (i \gamma^{\mu} \partial_{\mu} - m) \psi$, the conjugate momentum to $\psi$ is $\pi = i \psi^{\dagger}$, and computing $\pi \partial_0 \psi - \mathcal{L}$ gives the correct Dirac Hamiltonian. Is your point somehow related to treating the elements of your Lagrangian as Grassmann variables? $\endgroup$
    – Zack
    Commented Nov 15, 2021 at 15:49
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    $\begingroup$ @Zack Yes, I had Grassmann variables in mind. $\endgroup$
    – J.G.
    Commented Nov 15, 2021 at 16:34
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    $\begingroup$ In general, a QFT may not even have a lagrangian. I assume you meant QFTs with a lagrangian description, but be aware that the answer to the question as asked is a trivial "no." $\endgroup$
    – d_b
    Commented Nov 15, 2021 at 16:47
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    $\begingroup$ Every QFT mentioned here for one: arxiv.org/abs/1802.09626 $\endgroup$
    – Connor Behan
    Commented Nov 16, 2021 at 14:51

1 Answer 1

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In general the Legendre transformation$^1$ from the Lagrangian to the Hamiltonian formulation may be singular, which leads to primary constraints. This is e.g. the case for gauge theories like Yang-Mills (YM) theory with or without matter, which OP mentions.

However, in case of a singular Legendre transformation, by performing a so-called Dirac-Bergmann analysis (which may lead to secondary constraints), it is still possible in principle to define a corresponding Hamiltonian formulation. Typically, the canonical Hamiltonian $H_0=p\dot{q}-L$ gets amended with terms of the form 'constraint times Lagrange multiplier'. For details, see e.g. Refs. 1 & 2.

References:

  1. P.A.M. Dirac, Lectures on QM, 1964.

  2. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994.

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$^1$ Concerning fermions, see e.g. this Phys.SE post.

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  • $\begingroup$ So I cannot use the second and third equation I wrote, right? $\endgroup$
    – Mauro Giliberti
    Commented Nov 16, 2021 at 15:12
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    $\begingroup$ It might be worth mentioning that this has nothing to do with the theory being quantum, but that this issue already exists classically. $\endgroup$
    – ACuriousMind ♦
    Commented Nov 16, 2021 at 16:37
  • $\begingroup$ @ACuriousMind: Good point. $\endgroup$
    – Qmechanic ♦
    Commented Nov 16, 2021 at 16:38

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