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How do you derive Noether's theorem when the action combines chiral, antichiral, and full superspace?

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2 Answers 2

The full superspace terms are the "most general ones" but you may convert the chiral and antichiral terms to the full superspace form, too. In particular, $$\int d^2\bar\theta = \int d^2\theta\, d^2 \bar\theta\, \theta_1\theta_2$$ and similarly for its complex conjugate. Sorry if a sign is wrong. Note that $\theta_1\theta_2$ may be written as $\epsilon^{ab}\theta_a \theta_b/2$ up to a sign if you happen to be annoyed by the explicit components.

The fact that chiral superfields only depend on $\theta$ but not $\bar\theta$ doesn't matter: it just means that it is a more constrained field. But you may still imagine that it's a function of the full superspace that just happens to have a special form.

Once you convert the action to an integral over the full superspace, you may proceed just like you would proceed for the full superspace, D-term-like terms. I am actually a bit unfamiliar with this thing - so I would convert the action to the components (no superspace at all) and proceeded just like in non-supersymmetric theories. Of course, the resulting conserved quantities wouldn't be nicely organized in supermultiplets - which they can be in a supersymmetric theory.

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You can think of a generalized innerproduct, where terms get integrated over the "correct" superspace. In my view, the best way of deriving the propagators for chiral fields. It's much cleaner than lifting everything up to full superspace.

An example of this occurs in section 4.8 of Ideas and methods of supersymmetry and supergravity and probably other papers by the authors. In this section only chiral and antichiral spaces are used, but it can be generalized to include a full superspace sector.

Once you've set up your conventions for this type of innerproduct, then everything basically works as you'd expect.

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