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On-shell symmetry from a path integral point of view

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Normally supersymmetric quantum field theories have Lagrangians which are supersymmetric only on-shell, i.e. with the field equations imposed. In many cases this can be solved by introducing auxilary fields (field which don't carry dynamical degrees of freedom, i.e. which on-shell become a function of the other fields). However, there are cases where no such formulation is known, e.g. N=4 super-Yang-Mills in 4D.

Since the path integral is an integral over all field configurations, most of them off-shell, naively there is no reason for it to preserve the on-shell symmetry. Nevertheless the symmetry is preserved in the quantum theory.

Of course it is possible to avoid the problem by resorting to a "Hamiltonian" approach. That is, the space of on-shell field configurations is the phase space of the theory and it is (at least formally) possible to quantize it. However, one would like to have an understanding of the symmetries survival in a path integral approach. So:

How can we understand the presence of on-shell symmetry after quantization from a path integral point of view?

This post has been migrated from (A51.SE)
asked Nov 5, 2011 in Theoretical Physics by Squark (1,700 points) [ no revision ]
The equivalence between N=1 and N=0 is by integrating over the auxiliary fields, as far as I see. Hence it is not quite manifest in the N=0 language. For N=4 you can sure use the N=1 superspace, moreover the GIKOS approach apparently allows making an N=3 sub supergroup manifest. However this doesn't prove the whole symmetry is preserved.

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OK, I think I see the answer. Once you can prove the equivalence between N=0 and N=1 you can get N=4 by choosing different N=1 subsupergroups

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Well, I need to chew a bit more on the N=4 SYM case, but, in the meanwhile, consider N=1 SYM in 10D. Already we have no off shell formulation and the N is minimal.

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Dear @Squark, surely you may write $N=4$ in the $N=1$ superspace, making the $N=1$ subalgebra manifest even off-shell and even in the path integral, can't you? The path integral for the $N=1$ language is trivialy equivalent to the $N=0$ "in components" formulation – the only slightly nontrivial statement behind this assertion is that the measure flip including the aux. fields doesn't spoil SUSY. So in this sense, I think that SUSY is manifest even in the non-SUSY $N=0$ "in components" formalism of the path integral, off-shell. If you see some problems with this conclusion, tell me details.

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1 Answer

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How can we understand the presence of on-shell symmetry after quantization from a path integral point of view?

One can derive a Schwinger-Dyson equation associated with the current conservation, also known as a Ward identity; see e.g. Peskin and Schroeder, An Introduction to Quantum Field Theory, Section 9.6; or Srednicki, Quantum Field Theory, Chapter 22.

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answered Nov 5, 2011 by Qmechanic (2,580 points) [ no revision ]
OK, and where is the answer to my question in Srednicki's book? In fact it seems to me he doesn't go beyond N=1 4D supersymmetry

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I interpreted the question(v1) as asking about on-shell symmetry on general grounds, cf. the title(v1). The references do not have any specific mentioning of $N=4$ 4D supersymmetry.

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OK. So Peskin and Schroeder have a treatment of on-shell symmetry on general grounds? Which example they consider if not supersymmetry? Also, I still have no access to the book.

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@Squark: Srednicki's book (up to some small changes) is available on [his webpage](http://web.physics.ucsb.edu/~mark/qft.html).

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I don't have access to the book. How do you derive it using the path integral if the symmetry only exists on shell?

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