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What decides the signs and coefficients of terms in superfield?

+ 6 like - 0 dislike
159 views

I'm working on a problem in 3d field theory and I'm confused about how to write the superfields. Specifically, I'm not sure if the signs and coefficients of terms are purely a matter of convention or if they need to be chosen more carefully.

For example in 0806.1519v3, page 18 we have the 3d chiral, anti-chiral and vector superfields respectively:

(1) $\Phi = \phi(x_{L}) + \sqrt{2} \theta \psi(x_{L}) + \theta^{2} F(x_{L})$

(2) $\Phi^{\dagger} = \bar{\Phi} = \bar{\phi}(x_{R}) - \sqrt{2} \bar{\theta} \bar{\psi}(x_{R}) - \bar{\theta}^{2} \bar{F}(x_{R})$

(3) $V = 2i\theta \bar{\theta} \sigma (x) + 2 \theta \gamma^{\mu} \bar{\theta} A_{\mu}(x) + \sqrt{2} i \theta^{2} \bar{\theta} \bar{\chi}(x) - \sqrt{2} i \bar{\theta}^{2} \theta \chi(x) + \theta^{2} \bar{\theta}^{2} D(x) $

However one could take the alternative approach of beginning with the 4d superfields of Wess and Bagger in their book "Supersymmetry and Supergravity":

(4) $\Phi = \phi(x_{L}) + \sqrt{2} \theta \psi(x_{L}) + \theta^{2} F(x_{L})$

(5) $\Phi^{\dagger} = \bar{\Phi} = \bar{\phi}(x_{R}) + \sqrt{2} \bar{\theta} \bar{\psi}(x_{R}) + \bar{\theta}^{2} \bar{F}(x_{R})$

(6) $V = - \theta \gamma^{\mu} \bar{\theta} A_{\mu}(x) + i \theta^{2} \bar{\theta} \bar{\chi}(x) - i \bar{\theta}^{2} \theta \chi(x) + \dfrac{1}{2} \theta^{2} \bar{\theta}^{2} D(x) $

Its clear that simply performing a dimensional reduction of (4), (5) and (6) will not give us (1), (2) and (3).

In fact one could look at "Introduction to Supersymmetry" by Mller-Kirsten and Wiedemann and see that, even sticking to the 4d case, different signs and coefficients are used compared to (4), (5) and (6).

Now normally I'd simply accept that there's some difference in notation and not consider it a big deal. However, when we write the superfields in actions we have superfield multiplications (e.g: $\Phi^{\dagger} e^{qV} \Phi$). These give important terms when we multiply out the consituents of the superfields. Some such terms are also present in the non-susy case. Having performed a few calculations it has become clear that different conventions for the superfields result in different signs and coefficients for these resulting terms (unsurprisingly). I would have thought such differences to be physically important since they arise in the action. When calculating probability amplitudes, for example, would such signs and coefficients not play a crucial role? In non-supersymmetric field theory the terms tend to have very specific signs and coefficients, I would have thought the superfields would have had to make sure such exact terms are reproduced when superfields are multiplied together.

This post imported from StackExchange Physics at 2014-07-09 07:30 (UCT), posted by SE-user Siraj R Khan
asked Jul 8, 2014 in Theoretical Physics by Siraj R Khan (105 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

I'm not sure it will help you, but, perhaps http://arxiv.org/ftp/hep-th/papers/0108/0108200.pdf may be of use. There is a long chapter on the Toy model of 3D-SUSY starting at page 15. I think their expansion has yet another sign choice. See page 18 equation 2.2.8.

answered Jul 9, 2014 by James S. Cook (95 points) [ no revision ]

Thanks for your help James, that source has proven very useful.

+ 2 like - 0 dislike

Signs and normalisations are matters of convention and physics doesn't depend on it. It is possible to dimensionally reduce $\mathcal{N}=1$ theories in $d=4$ to get $\mathcal{N}=2$ theories in $d=3$ and $(2,2)$ theories in $d=2$.  For instance, to go from (6) to (3), you need to remember that the sigma field is the fourth-component of the four-dimensional gauge field. Further, you need to get the gamma matrix conventions and conventions of conjugation correct.

answered Jul 10, 2014 by suresh (1,535 points) [ no revision ]

Thank you for the help Suresh!

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