I've got a question concerning non-Abelian supersymmetric gauge theories.
Consider supersymmetric non-Abelian theory realized on chiral superfields Φi in a representation R with matrix generators Taji. Let us define supergauge transformation as Φi→(e2ıgaΩaTa)ijΦj.
This is the moment that I don't get. What does it mean? Strictly speaking, the latter expression is complicated non-linear equation on components of Va superfield.
I guess they mean, that since the second term on r.h.s. doesn't depend on Va, it's possible to solve it within the framework of perturbation theory in the coupling constant(s) ga. Is it correct? If so, how to prove it strictly in all orders?
Answer
I) The gauge transformation of the real gauge field V reads
e˜V = eXeVeY,X := iΩ†,Y := −iΩ.
We next use the following BCH formulas
eXeV BCH= eV+B(adV)X+O(X2),eVeY BCH= eV+B(−adV)Y+O(Y2).
Keeping only linear orders in Ω, we get
˜V (1)+(2)= B(adV)X+V+B(−adV)Y (4)= V+12[V,Y−X]+B+(adV)(X+Y),
where
B(x) := xex−1 = ∞∑m=0Bmm!xm = B+(x)−x2 = 1−x2+x212−x4720+x630240+O(x8)
and
B+(x) := B(x)+B(−x)2 = x/2tanhx2 = 1+x212−x4720+x630240+O(x8)
are generating functions of Bernoulli numbers.
II) We would like ˜V to be in WZ gauge
˜V = O(θ2).
For given V, ˜V, and X−Y, the eqs. (3+6) is an affine1 equation in X+Y=iΩ†−iΩ. This has formally a solution if the operator
B+(adV) = 1+…
is invertible, which is true, at least perturbatively. To finish the proof, one should write out the equation in its superfield components to check that the above affine shift mechanism really is realized at the component level. Recall e.g. that the gauge field ˜V can not be gauged away completely (= put to zero), since Ω is a chiral superfield with not enough θ's to reach all components of ˜V, so to speak.
References:
- S.P. Martin, A Supersymmetry Primer, arXiv:hep-ph/9709356; p.43.
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1 An affine equation is a linear equation with an inhomogeneous term/source term.
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