The bosonic sigma model in ordinary QM (i.e. a 'free' particle trapped on a curved manifold $\mathcal{M}$), has a Hamiltonian which is just the negative Laplacian on $\mathcal{M}$.

For any $\mathcal{M}$ there is a supersymmetric extension where the Hilbert space is extended to the space of differential forms on $\mathcal{M}$ (as opposed to just scalar functions, which are 0-forms), and the Hamiltonian is extended in the natural way to the Laplacian on differential forms (sometimes called the Laplace-de Rham operator).

Crucially, the Hamiltonian acts in the exact same way in the bosonic sector. If a 0-form is an eigenfunction in the bosonic sigma model, it will be an eigenfunction with the exact same eigenvalue in the SUSY sigma model. This should mean that the correlation functions of the bosonic fields which generate these 0-forms are exactly the same in either model.

If this is true, how is this seen in the path integral approach?

In the path integral, the Lagrangian is $$\frac{1}{2}g_{ab}\dot{\phi}^a\dot{\phi}^b+ig_{ab}\left(\bar\psi^a\dot\psi^b+\Gamma^b_{cd}\bar\psi^a\psi^c\dot{\phi^d}\right)-\frac{1}{2}R_{abcd}\psi^a\bar{\psi}^b\psi^c\bar{\psi}^d.$$ $\phi$ are the bosonic coordinate fields on $\mathcal{M},$ and the geometric quantities $g,\Gamma,R$ all have $\phi$ dependence. If we integrate out the fermionic fields $\psi$ it seems we will get a complicated determinant in terms of $\phi$ and this would modify the correlation functions, which from the canonical quantization point of view we expect to be the same.

Is there any way to see from the path integral alone that these correlation functions are the same?

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    $\begingroup$ Are you positive the bosonic correlation functions are the same? this is not obvious to me, not in the path integral formalism, not in the operator formalism. Is this a well-known property of the system? Do you by any chance have any reference? $\endgroup$ – AccidentalFourierTransform Feb 26 at 1:42
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    $\begingroup$ @AccidentalFourierTransform, I am positive the energy of the bosonic states is the same in the two systems. This is because the Hamiltonian of the SUSY model is just the Laplace-de Rham operator. The original reference on the QM sigma model which shows this is section 10 of Witten, Nucl Phys B 202 (1982) 253. If the Hamiltonian acting on bosonic states is the same whether there is SUSY or not, then given the correlation functions are just expectation values of the time translations of bosonic fields, they must be the same (unless there is something subtle I am missing). $\endgroup$ – octonion Feb 26 at 16:08
  • $\begingroup$ Sure the energies are the same, but that does not imply the correlation functions are also the same, does it? E.g., inserting a complete set of eigenstates, you get different answers in the regular theory, and the SUSY one, because the latter contains more states, and I don't expect the extra states to magically add up to zero -- although I definitely could be very wrong here. Witten doesn't say anything about correlators though, as far as I can see. Anyway, best of luck finding the answer! $\endgroup$ – AccidentalFourierTransform Feb 27 at 0:25
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    $\begingroup$ @AccidentalFourierTransform, the extra states magically add up to zero because the matrix element of a bosonic field between a fermionic state and the bosonic vacuum is zero $\endgroup$ – octonion Feb 27 at 18:19
  • $\begingroup$ I find the question confusing... "This should mean that the correlation functions of the bosonic fields which generate these 0-forms are exactly the same in either model" presumably either model refers to SUSY vs nonSUSY QM? If that is the case and the quoted claim is indeed true then it had better follow from a SUSY localisation argument. But prima facie it looks like the 0-form observables aren't BRST closed so it seems you're out of luck $\endgroup$ – alexarvanitakis Feb 29 at 16:07

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