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  Showing closure of the SUSY algebra of a free abelian gauge multiplet

+ 3 like - 0 dislike

Given the complete supersymmetric lagrangian of a free abelian gauge multiplet $$ \mathcal{L} = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} + i \bar{\lambda} \bar{\sigma}^\mu \partial_\mu \lambda + \frac{1}{2} D^2 + \xi D $$ With the following variations $$ \delta_\zeta A_\mu = \bar{\zeta} \bar{\sigma}^\mu \lambda + \bar{\lambda} \bar{\sigma}^\mu \zeta $$ $$ \delta_\zeta \lambda_\alpha = \frac{i}{2} F_{\mu\nu} (\sigma^\mu \bar{\sigma}^\nu \zeta)_\alpha + D \zeta_\alpha $$ $$ \delta_\zeta D = -i \bar{\zeta} \bar{\sigma}^\mu \partial_\mu \lambda + i \partial_\mu \bar{\lambda} \bar{\sigma}^\mu \zeta $$

Where $\zeta_\alpha$ is an infinitesimal Grassmann number.

Now to show that the algebra closes I took the commutator of the variations, acting on each of the fields.

Showing that the algebra closes for $\lambda$ and $D$ is relatively start forward algebra, resulting in $$ [\delta_\psi, \delta_\zeta] \lambda_\alpha = -2i(\bar{\zeta} \bar{\sigma}^\mu \psi - \bar{\psi} \bar{\sigma}^\mu \zeta)\partial_\mu \lambda_\alpha $$

$$ [\delta_\psi, \delta_\zeta] D = -2i(\bar{\zeta} \bar{\sigma}^\mu \psi - \bar{\psi} \bar{\sigma}^\mu \zeta)\partial_\mu D $$

However when I do it for the gauge fields I find, $$ [\delta_\psi, \delta_\zeta] A_\mu = -2i(\bar{\zeta} \bar{\sigma}^\alpha \psi - \bar{\psi} \bar{\sigma}^\alpha \zeta)(\partial_\alpha A_\mu - \partial_\mu A_\alpha) $$

Which almosts looks like the translation that one would expect. However it has that extra piece there. Is there something that makes that $\partial_\mu A_\alpha$ term zero that I am missing?

In http://arxiv.org/abs/hep-th/0109172 Jose suggests (pg 12) that the auxiliary fields are added to close the algebra off-shell, which in the case of the gauginos it did. He also mentions on (pg 21) that in the case of having gauge fields the algebra will only close up to a gauge transformation. Now he did that before adding auxiliary fields, so maybe it isn't true...

However if this is true, it seems odd to me that the algebra would only close up to a gauge transformation. To me that would imply that gauge fixing is required and that doesn't really seem kosher. For we would need to choose a gauge for the theory to be valid, otherwise we don't have a true gauge symmetry. How do we get around that problem?


@Qmechanic I believe a good reference is http://arxiv.org/abs/hep-ph/9709356, just a note in advanced he uses the mostly plus convention where I use the mostly minus, essentially the translations and sigma matrices just differ by an overall minus sign, but the spirit is there. I would look at sections 2 and 3 in particular section 3.3 (for the abelian case set a=1 and the covariant derivatives just become standard derivatives). Also note that I absorbed the factor $\frac{1}{2}$ into the infinitesimal parameter and he's choice of sign is different on the variation of the gauge field.

Notes of Convention: My conventions are nearly identical to the SUSY Primer reference except for the metric. $$ \eta^{\mu\nu} = diag(+ - - -) $$

Clarification of the Question Provided that my arithmetic is correct, how can one see that the $\partial_\mu A_\alpha$ term dies?

After that, even without checking my work (thank you if you do!), why is it for the non-abelian case equation (3.3.9), do we see a covariant derivative? If the algebra were to truly be closed I would expect that to be a partial derivative which are the spacetime translations that we expect to result (see eq 3.1.29). If we should see a covariant derivative, then why?

Additional Comment

It turns out that the turn doesn't die and is indeed correct. It gives the form $$ [\delta_\psi, \delta_\zeta] A_\mu^a = -2i(\bar{\zeta} \bar{\sigma}^\alpha \psi - \bar{\psi} \bar{\sigma}^\alpha \zeta)F_{\alpha\mu}^a $$

Which holds in the case of the Non-Abelian case (the other terms go from being of the form in (3.3.9) of SUSY Primer)

In Freedman's and De Wit's paper journals.aps.org/prd/abstract/10.1103/PhysRevD.12.2286 they discuss the implications of what having a gauge symmetry does to the commutator of two covariant SUSY transformations. It now generates a new transformation which can be described as a translation plus a gauge transformation, or what they call a gauge-covariant translation.

How is this considered closing the graded lie algebra, which should consist of Translations, Boosts/Rotations, and a SUSY transformation?

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user Peter Anderson
asked May 10, 2015 in Theoretical Physics by Peter Anderson (205 points) [ no revision ]
retagged May 17, 2016
Comment to the question (v1): Consider adding references to make the question more accessible to the reader and focus the answers.

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user Qmechanic
I added a reference and believe I clarified the question, let me know if you have any other suggestions. I could add my algebra if that would be helpful?

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user Peter Anderson
Your SUSY primer reference says after (3.3.9) that the closure equation shall hold for $F$, not for $A$, and since you obtained $F$ as the RHS of your equation for $A$, it seems plausible your equation for $F$ would actually yield be in the right form. Are you sure you are supposed to obtain the equation for $A$ you are trying to obtain?

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user ACuriousMind
I think you may be correct. I must have missed that when cross referencing... However it seems strange that the covariant derivative comes in. I have some leads as to why that is, journals.aps.org/prd/abstract/10.1103/PhysRevD.12.2286. It seems that it is a result of choosing the Wess-Zumino gauge in the superspace formalism. If I find a more concrete answer, I'll post it =)

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user Peter Anderson

1 Answer

+ 2 like - 0 dislike

The bracket you have written is of the form $$[\delta,\delta] A_\mu = v^\nu \partial_\nu A_\mu + \partial_\mu (v^\nu A_\nu).$$ As you pointed out, the first term corresponds to a translation by $v^\mu$. The second term corresponds to a gauge transformation $\delta A_\mu = \partial_\mu \lambda$ with $\lambda = v^\nu A_\nu$. So the algebra closes up to a translation and a gauge transformation, taking the form $[\delta,\delta]A_\mu = v^\mu F_{\mu\nu}$, which is typical of these kinds of calculations.

The non-abelian case is similar. The commutator of two supersymmetry transformations on, say, a scalar $\phi$ in the vector multiplet gives a translation $v^\nu\partial_\nu \phi$ plus a gauge transformation $[v^\nu A_\nu,\phi]$ (since fields in the vector multiplet transform in the adjoint of the gauge group). Thus, $$[\delta,\delta]\phi = v^\nu \partial_\nu \phi + [v^\nu A_\nu, \phi] = v^\nu D_\nu \phi,$$ where $D_\nu = \partial_\nu + [A,\cdot]$.

To answer your follow-up question: Consider the vector field $A_\mu$. It belongs to the space $\mathcal{A}$ of gauge fields on $\mathbb{R}^{1,3}$. One would like to write down a representation of the supersymmetry algebra on the space of functionals of the fields $\{A_\mu,\ldots\}$, where the $\ldots$ denote other fields in the multiplet like scalars and fermions. But this space does not give a representation of the algebra. That's not surprising, since field configurations which differ by a (small) gauge transformation are physically equivalent. Thus, to be more precise, we should say that the physical gauge field corresponds to an equivalence class $[A_\mu]$ in $\mathcal{A}/\mathcal{G}$, where $\mathcal{G}$ denotes the space of small gauge transformations. It is the functionals of $\{[A_\mu],\ldots\}$ which gives a representation of the algebra, which requires that $$\big[[\delta,\delta]A_\mu\big] = \big[v^\sigma \partial_\sigma A_\mu\big].$$ In terms of fields in $\mathcal{A}$, this equality of classes in $\mathcal{A}/\mathcal{G}$ means that there exists some $\lambda$ such that $$[\delta,\delta]A_\mu - v^\sigma \partial_\sigma A_\mu = \partial_\mu \lambda,$$ which is precisely what we showed earlier with $\lambda = v^\mu A_\mu$.

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user user81003
answered May 11, 2015 by user81003 (130 points) [ no revision ]
I added another comment above. Could you explain how the gauge-covariant translation is in the graded lie algebra?

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user Peter Anderson
I updated my response to answer your question.

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user user81003
I would also like to add one more note that I found while trying to figure out this question. The gauge transformation that occurs in addition of the translation can be removed if one works in the superspace formalism and retains all the extra auxiliary fields. Another interesting, yet obvious, thing that occurs is if you take the commutator of the susy variations on a gauge invariant object you only get a translation.

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user Peter Anderson
My earlier answer was slightly imprecise, so I clarified it a little bit.

This post imported from StackExchange Physics at 2016-05-17 15:58 (UTC), posted by SE-user user81003

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