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Since the Susy variation of a scalar field is itself a (Lorentz?) scalar, shouldn't the variation of the variation of the scalar field again be the same, that is - the "dot product" of the Susy parameter and the spinor field?

+ 2 like - 0 dislike
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I am trying to study Susy from Patrick Labelle's book. While I see that we want the Susy variation to take \(\phi\) into \(\chi\) and vice versa, I don't see why I cannot argue that since \(\delta\phi=\zeta.\chi\) is itself a scalar, another variation by a Susy parameter (say, \(\beta\)) won't just produce \(\delta_\beta \delta_\zeta \phi=(\beta.\zeta)(\zeta.\chi)\), or something like that, since \[\delta_\beta \delta_\zeta \phi=\delta_\beta(\zeta.\chi)=\delta_\beta(\phi')=\beta.\chi\]again? Here I have defined \[\phi'=\zeta.\chi\] to be the new scalar field. Is it right to think of \(\zeta.\chi\) as a scalar field?

Put differently, what stops me from thinking of \(\zeta.\chi\) as another scalar, the Susy variation of which should again give something similar to the Susy variation of \(\phi\)?

I have a "feeling" that I am perhaps not clear about the fact that \(\phi\) is a scalar under Lorentz transformation. I am also not clear about whether it is the scalar or the bosonic nature of \(\phi\) that I should be thinking about.

Any help would be much appreciated!

asked Dec 29, 2015 in Theoretical Physics by Chinmay Mishra (30 points) [ revision history ]
edited Dec 29, 2015 by Chinmay Mishra

Lorentz invariance itself is not restrictive enough for you to determine the form of variation. In the simple example you gave, since you are doing two Susy transformations, you final result should contain two fermionic parameters, so it cannot be $\beta \cdot \chi$.

Thanks, but could you elaborate on what you mean by "Lorentz invariance itself is not restrictive enough"?

Your example suffices to demonstrate why "Lorentz invariance itself is not restrictive enough", $\delta_\beta \delta_\zeta \phi=\beta\cdot\chi$ is a Lorentz scalar, so Lorentz invariance doesn't rule it out, but it's still wrong due to the previously mentioned reason.

Hi @JiaYiyang. I have modified my question to further bring out the doubt that I have.

1 Answer

+ 2 like - 0 dislike

I guess I didn't get my previous point across in the comment and I think your confusion is simply a psychological one. The point is, there is simply no reason to require non-Lorentz transformations on a field to be determined by its Lorentz transformation properties. For example, consider a scalar matter field in a gauge theory, let's say Higgs doublet $\phi$ in a $SU(2)$ gauge theory. Now the scalar is in fundamental rep and the gauge field $W_\mu$ is in the adjoint rep, however, you simply cannot say $\partial_\mu \phi$ should also live in adjoint just because it has the same Lorentz index structure as $W_\mu$, and instead it clearly lives in fundamental because $\phi$ does. 

In your case, the situation is exactly the same: $\zeta \cdot \chi$ and $\phi$ are simply two different fields even if they have the same Lorentz index structure. In fact, maybe as an even more pertinent example , in Wess-Zumino model with auxiliary scalar fields, those auxiliary scalar fields transform differently from the propagating scalar fields, which is easy to see by counting their mass dimensions.

answered Dec 29, 2015 by Jia Yiyang (2,465 points) [ revision history ]

Thanks Jia!

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