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When I first learned gauge theories in my introductory quantum field theory course, I was taught that the Faraday (field-strength) tensor can be constructed by computing the commutator of the gauge-covariant derivative:


Now, I am studying supersymmetry following Martin's SUSY primer, and in chapter 4.8, the author immediately writes down the super-symmetric field strength chiral superfield out of the vector superfield $V$:

$$\mathcal{W}_\alpha=-\frac{1}{4}D^\dagger D^\dagger D_\alpha V.$$

I would have liked a more gentle introduction to this in terms of something I am already familiar with: is there a way for me to have constructed this using the commutator of some 'gauge super-covariant derivative'?

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up vote 2 down vote accepted

As far as I know, it is defined in that form in order to satisfy chirality \begin{equation}D^{\dagger}_{\dot{\alpha}}W_{\alpha}=0\end{equation} and gauge invariance \begin{equation}\delta W_\alpha=0.\end{equation}

I have not seen a definition by a commutator anywhere.

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Ok! is there a reason why the supersymmetric field strength tensor must be a chiral superfield? Is there a geometric way of constructing the supersymmetric field strength tensor? Should I post a new question? – QuantumDot Jan 14 '13 at 17:18
@QuantumDot Read Wess-Bagger. The answer is there. – Yuji Jan 17 '13 at 11:03
QuantumDot, it's more appropriate to say that the field strength may be required to be a chiral superfield, so if the constraint is possible, one should clearly impose it to work with the minimal possible - maximally constrained - representations. Quite generally, your attempt to write the field strength as a commutator of covariant derivatives is misguided. That's how the gauge-covariant objects are constructed in non-SUSY theory but there's no reason that this is the right universal template for all theories, e.g. for SUSY. – Luboš Motl Jan 20 '13 at 10:22
The universal conditions we expect from field strength is not that it is a commutator - this is just a solution in the case of non-SUSY gauge theories or theories in non-SUSY formalism. If we want field strength in general, it is a field that transforms covariantly - for U(1), it is gauge-invariant - and that is minimum possible so that one may conveniently construct free Lagrangians from it etc. For non-SUSY, it's solved via your commutator route, in N=1 SUSY, Frederic wrote you the next step in the solution. – Luboš Motl Jan 20 '13 at 10:24
Ordinary covariant derivatives are bad in SUSY because the $\partial_\mu$ derivative is no longer the fundamental "minimal" derivative. Instead, one may find a square root of it - and the superderivatives $D_\alpha$ etc. are square roots of the ordinary derivatives, and they're therefore more fundamental. So you haven't really started to think in the SUSY way if you still want to place ordinary derivatives everywhere. – Luboš Motl Jan 20 '13 at 10:25

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