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  What is the explicit relationship between supersymmetry and BRST invariance?

+ 5 like - 0 dislike

Looking at the following Lagrangian for a non-Abelian gauge theory with gauge fields $A^a_{\mu}$, scalar matter fields $\phi_i$ with Faddeev-Popov ghosts $c^a$ and anti-ghosts $\bar{c}^a$ included

\[\mathcal L_{FP} = \mathcal L [A^a_{\mu},\phi_i] - \frac{1}{2\xi}(\partial_{\mu}A^a_{\mu})^2 + (\partial_{\mu}\bar{c}^a )(D_{\mu}c^a)\]


\[D_{\mu}c^a = \partial_{\mu}c^a+gf^{abc}A^b_{\mu}c^c\]

the BRST transformation under which this Lagrangian is invariant is given by

$$ \phi_i \rightarrow\phi_i + i\theta c^a T^a_{ij}\phi_j$$

imageimage$$ A^a_{\mu} \rightarrow A^a_{\mu} + \frac{1}{g}\theta D_{\mu}c^a$$

$$ \bar{c}^a \rightarrow \bar{c}^a - \frac{1}{g}\theta\frac{1}{\xi}\partial_{\mu} A^a_{\mu} $$

$$ c^a \rightarrow c^a - \frac{1}{2}\theta f^{abc}c^b c^c $$

where $\theta$ is a Grassmann number.

I have heard that (informally speaking?)  such a BRST transformation can be considered to be half a supersymmetry transformation, probably because the contributions of loops containing the fermionic ghost fields are able to cancel the effects of certain bosonic loops which makes the theory renormalizble (?).

How can I explicitly see that the BRST transformation is half a supersymmetry transformation (if it really is), or what is the exact relationship between the BRST and (for example $\mathcal N = 1$) supersymmetry?

Using the superspace formalism, ($\mathcal N = 1$) supersymmetry can be considered to be a translation in two additional fermionic coordinates, so could a BRST transformation be written down as a translation in a single additional fermionic direction?

asked Dec 2, 2015 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]

I don't know if BRST can be made into some literal supersymmetry, but a simple extension of superspace (in your suggestion N=2 superspace?) doesn't seem to work since normal superfields have all its components satisfy the correct spin-statistics

@JiaYiyang if BRST in the toy-model of my question can really be seen as some kind of supersymmetry, I would expect the fermionic ghosts to be the $s=\frac{1}{2}$ superpartners of the scalar fields $\phi$. I dont understand what would be the problem with spin statistics in this case.

The claim that I have read is that BRST is half a supersymmetry, so I would rather expect it to correspond to some kind of $\mathcal N = \frac{1}{2}$ (if such a thing even exists?) with only one supercharge instead of an extended one such as $\mathcal N = 2$ or even higher.

But ghosts have spin-0.

Just a remark, in general ghosts do not only have 0 spin although indeed in d=4 YM they should have 0 spin.

1 Answer

+ 1 like - 0 dislike

In fact there is a superspace formulation of BRST symmetry, see  section 3 of "Local BRST symmetry and the geometry of gauge fixing".

However, note that the paper used two separate superfields to write the gauge fields ($A_\mu$) and ghosts ($c$), this is necessary because of my earlier comment on spin-statistics. This means one cannot extract from a single superfield all the fields in their fundamental form, e.g. from the first superfield one can at best get a $D_\mu c$ but not $c$ itself.  In my humble opinion such a formulation doesn't quite help with making  "BRST is half a supersymmetry" more literal, at least not any more literal than the usual x-space fomulation.

Of course, the motivation of making such a superspace in this paper was different anyway, that is, to make the study of local BRST invariance easier. But I'm unsure on why local BRST should be an interesting subject in the first place.

answered May 8, 2016 by Jia Yiyang (2,640 points) [ revision history ]

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