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I know that given the Hamiltonian of a theory, there can be many different associated Lagrangians, or even none at all, but why is that so? In classical mechanics the Hamiltonian and Lagrangian formulations are related by a Legendre transformation, what happens in QFT with this transformation to make it so ill-behaved?

I understand that Hamiltonians are not covariant, and if I understand correctly, a given Hamiltonian can be better understood as the time evolution operator of a given observer after fixing the Poincare symmetry. This, as we have to choose a explicit time coordinate, that determines the spatial coordinates and conjugate momenta that appear in the Legendre transformation (up to a rotation and translation, of course). Wouldn't that Legendre transformation then be unique? (As rotations and translations are still symmetries of our chosen frame of reference and thus return the same Lagrangian)

What's the reason we can have several different Lagrangians or maybe even none at all?

The only examples of no-Lagrangian QFTs that I have found in my studies are some CFTs , but I am not sure if those theories actually have a Hamiltonian at all, so I do not know if they can be a good working example.

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  • $\begingroup$ You know from which example/reference? $\endgroup$ – Qmechanic Apr 12 '20 at 16:15
  • $\begingroup$ I clarified that my experience so far only involves some CFTs, but I do not know I they constitute good examples. $\endgroup$ – edmateosg Apr 12 '20 at 17:54
  • $\begingroup$ About the theories with several Lagrangians I have seen, for example, in the talk by Judži Tačikawa in indico.ipmu.jp/indico/event/134/contribution/17/material/slides/… Should I provide another source? I haven't found any paper talking about it yet, but I can keep looking $\endgroup$ – edmateosg Apr 12 '20 at 18:51
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  1. The usual physics lore is that there is a bijective correspondence between Lagrangian and Hamiltonian formulations via a (possibly singular) Legendre transformation, cf. e.g. this Phys.SE post.

  2. It seems that OP's counterexamples come from the family of ${\cal N}=2$ SUSY field theories in 4D, cf. Refs. 1 & 2.

    Such theories are typically defined via (i) a twisted compactification of a 6D (2,0) SCFT (so-called class ${\cal S}$), or (ii) a field theory limit of string theory on a Calabi-Yau singularity (aka. geometric engineering).

    A Hamiltonian operator can usually be defined via a representation of the SUSY algebra.

    Not all such theories have a Lagrangian action formulation. Some have several Lagrangian action formulations linked via duality/plurality.

References:

  1. Y. Tachikawa, ${\cal N}=2$ supersymmetric dynamics for pedestrians, arXiv:1312.2684.

  2. D. Gaiotto, Families of ${\cal N}=2$ field theories, arXiv:1412.7118.

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  • $\begingroup$ So when is the Legendre transformation degenerate and why? Given a Hamiltonian, how is it exactly that you can obtain several inequivalent Lagrangians? Thanks $\endgroup$ – edmateosg Apr 12 '20 at 22:55
  • $\begingroup$ Via duality/plurality. $\endgroup$ – Qmechanic Apr 12 '20 at 23:03
  • $\begingroup$ I do not really understand it and I didn't find an explicit answer in the linked references. 1. Do you have that duality because you can choose different equally-valid coordinates and conjugate momenta when performing the Legendre transformation? 2. Or maybe the formula gets some extra terms when we are working with operators in a Quantum theory? 3. Can a classical theory also show duality/plurality? I appreciate any insight into where the difference with the regular case is $\endgroup$ – edmateosg Apr 12 '20 at 23:17
  • $\begingroup$ 1. No. 2. Yes. 3. Yes. $\endgroup$ – Qmechanic Apr 12 '20 at 23:54

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