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  Is a $SU(2)$ supergauge theory really a $SU(2)$ gauge theory?

+ 3 like - 0 dislike

Consider $SU(2)$ supergauge theory with $A$, a doublet of two chiral superfields in the fundamental representation.

where $\Phi_1$ and $\Phi_2$ are chiral superfields. Since we have said it is in the fundamental representation it transforms


where $\sigma_j$ are Pauli matrices. If $e^{i\sigma_j\phi_j}$ were to be a $SU(2)$ matrix then the $\phi_j$ ought to be real. Nonetheless, this would make $A'$ no longer chiral unless $\phi_j$ are themselves chiral superfields, that is complex functions making $e^{i\sigma_j\phi_j}$ not a $SU(2)$ matrix.

Therefore this is no longer a proper $SU(2)$ theory. What is going on?

asked Nov 27, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]
retagged Nov 27, 2015 by dimension10

Why would this transformation make $A'$ not chiral? 

take the super-covariant derivative $\bar{D}_{\dot{\alpha}}$. It needs not vanish in general for such a transformation.

Why can't $\phi$ be both chiral and real?

@JiaYiyang I am a newbie with these things and i would love to hear a better reasoning to the question you pose Jia, but I do know that in 4 space-time dimensions and $\mathcal{N}=1$ SUSY  a (scalar) chiral superfield $\Phi$ accomodates a complex scalar $\phi$, a left chiral Weyl spinor $\psi$ an a auxiliary field $F$ in the following way $\Phi=\phi+\sqrt{2}\theta\psi+\theta^2F+\ldots$  (the dots denote that there are more terms but they are irrelevant for my argument). You see that the fact that $\phi$ being complex makes $\Phi$ complex. Now, I have been wondering if it might be possible to have supermultiplets containnig real scalars, actually I posted a question not so long ago in overflow http://www.physicsoverflow.org/32944/have-supersymmetry-using-real-scalar-instead-complex-scalars and the answer seems to be no.

Yeah I guess reality+chirality=chirality+ antichirality which would make the $\phi$ a constant field, which is too trivial and no longer a local transformation. Let me think about it, and I'm also quite new to Susy.

The peculiarity goes away if we look at the theory in a gauge where only the physical x-space fields remain in the gauge superfield $V$, i.e. Wess-Zumino gauge. This gauge choice completely determines the $\phi$ in your $e^{i\sigma_j\phi_j}$ up to a real part, and this remaining real part is the gauge transformation you can still perform on the matter fields.

1 Answer

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The kinetic term in the Lagrangian density for a chiral superfield $A$ coupled to a vector superfield $V$ is $A^\dagger e^V A$. The correct gauge transformation of $A$ is given by $A \rightarrow e^{i \phi} A$ where $\phi$ is indeed a chiral superfield to preserve the chirality of $A$, and so with values in the complexified Lie algebra of the gauge group. Under this transformation, $A^\dagger \rightarrow A^\dagger e^{-i \phi^\dagger}$ and so

$A^\dagger e^V A \rightarrow A^\dagger e^{-i \phi^\dagger } e^{V} e^{i \phi} A$.

As the kinetic term has to be gauge invariant, this shows that the correct transformation of the vector superfield $V$ under a gauge transformation is $V \rightarrow V'$ such that $e^{V'} = e^{-i \phi^\dagger } e^{V} e^{i \phi}$ i.e.  $V' = V+i(\phi - \phi^\dagger)+...$ (the ... are here if the gauge group is non-abelian because then one has to be careful about exponentials) and this is indeed the correct gauge transformation which preserves the reality condition satisfied by the gauge superfield, $V=V^\dagger$, which guarantees that the gauge group of the theory is indeed the compact gauge group and not its complexification.

Conclusion: in $\mathcal{N}=1$ $4d$ superfield formalism, gauge transformations are parametrized by chiral superfields, thus preserving chirality of the various charged matter fields, but this chiral superfield enters in the gauge transformation of the gauge field with its conjugate in a way preserving the reality condition of the gauge field.

One standard reference for this kind of questions is the book by  Wess and Bagger, "Supersymmetry and supergravity".

answered Dec 6, 2015 by 40227 (5,140 points) [ no revision ]

But how about the gauge transformation on the matter field? I imagine it should at least reduce to a real-parameter gauge transformation in the x-space, but it's not immediately obvious to me.

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