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Consider a $d=0+1$ theory of fermions, i.e., fermionic QM: $$ L=i\psi\partial_t\psi-V(\psi) $$ The Hamiltonian is just $H=V$.

What is the definition of a symmetry here? I can construct transformations that commute with $V$ but not with the kinetic term, so they leave $H$ invariant but not $L$. Are they symmetries?

From the QFT pov they should not be symmetries, but according to the standard QM definition they are. In fact, I believe they do have some effect on the spectrum of the theory, so they are not entirely meaningless, but I can't quite formalize my intuition.

Nihar Karve
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AccidentalFourierTransform
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    Note your canonical momentum cannot be inverted to give the 'velocity' in terms of the momentum, so the Hamiltonian you calculated is the 'canonical Hamiltonian', not the 'total Hamiltonian' which should also incorporate constraints. This even happens in the usual Dirac equation (See Das: "Quantum Field Theory", Sec. 10.5) but there you can end up back with the same canonical Hamiltonian - not sure if that happens here but it might account for the issue regarding symmetries? – bolbteppa Sep 17 '21 at 15:13
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    The Hamiltonian is not just $V$. The canonical momentum is $\pi \sim \psi$ so this is a constrained system. For such systems you need to construct the so-called "primary Hamiltonian" which is given by $H = H_c+\sum_i a_i C_i$. $H_c$ is the canonical Hamiltonian given here by $H_c = V$ and $C_i$ is the set of all (first and second class) constraints of the theory. The coefficients $a_i$ are determined by requiring that the constraints are preserved under time evolution. – Prahar Sep 17 '21 at 15:15
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    Hmm I don't know. What about e.g. the SYK model? See 2.1 vs 2.4 in https://arxiv.org/abs/1601.06768 – AccidentalFourierTransform Sep 17 '21 at 15:19
  • (For the record, those two equations cited above are not specific to that paper in particular, they are quite universal in the study of 0+1 dimensional models of fermions.) – AccidentalFourierTransform Sep 17 '21 at 15:33
  • https://physics.stackexchange.com/a/352595/25851 – bolbteppa Sep 17 '21 at 15:36
  • Assuming it's not the constraints, i.e. that the 'total Hamiltonian' reduces to the one in the paper even after dealing with constraints, is it not simply the fact that the Hamiltonian, being a function of more variables, can simply allow for more symmetries, and that thinking of Lagrangian Noether symmetries with a Hamiltonian analog is too constraining, e.g. the answer above, or are those symmetries 'really' in the Lagrangian formalism too somehow? – bolbteppa Sep 17 '21 at 15:51
  • @bolbteppa That seems reasonable. I would be quite happy if that turns out to be the right way to think about it. – AccidentalFourierTransform Sep 17 '21 at 15:53
  • This says even the Hamiltonian version of Noether is more powerful (covers more symmetries). I think it's safe to assume (from the usual QM proof) the quantities commuting with $\hat{H}$ are symmetries in QM regardless of the Lagrangian relation. – bolbteppa Sep 17 '21 at 15:54

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