Supersymmetry algebras

1. Nov 9, 2006

AlphaNumeric

Not sure if this is something you'd put in the homework help area or not, but I've started learning supersymmetry and seem to have hit a bump within the first few pages.

$$\delta_{\lambda}\varphi = \frac{1}{2}\lambda^{\mu\nu}M_{\mu\nu}.\varphi$$

for some field $$\varphi$$, which is one of $$S$$, $$P$$ or $$\psi$$ and $$\lambda^{\mu\nu}=-\lambda^{\nu\mu}$$

$$M_{\mu\nu}.\psi = -(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu})\psi - \Sigma_{\mu\nu}\psi$$

Where $$\Sigma_{\mu\nu} = \frac{1}{2}\gamma_{\mu\nu}$$, the Dirac matrix thing. I need to show that $$\delta_{\lambda}(\bar{\psi}\gamma^{\rho} \partial_{\rho}\psi) = \partial_{\mu}(\lambda^{\mu\nu}x_{\nu}\bar{\psi}\gamma^{\rho} \partial_{\rho}\psi)$$

The notes issue a warning that the algebra of operators, of which $$M_{\mu\nu}$$ is a part, only asks on fields, so $$M_{\mu\nu}.(x^{\rho}\varphi) = x^{\rho}M_{\mu\nu}.\varphi$$ and $$M_{\mu\nu}.(\partial^{\rho}\varphi) = \partial^{\rho}M_{\mu\nu}.\varphi$$

My problem is that I can't get the $$\psi$$ field to transform in that nice way. The $$\Sigma_{\mu\nu}$$ term screws it up and I end up with something I can't write as a total derivative!

If I've made zero sense here, I'm trying to do Exercise I.2 here.

Thanks for any help :)

2. Nov 10, 2006

dextercioby

I get the desired result. Just pay attention when you commute the gamma^{rho} with the spin matrix.

Daniel.

3. Nov 10, 2006

AlphaNumeric

That's what I was trying to do but I must be doign some stupid slip somewhere. Ignoring all the other parts of the question, I get it down to showing that

$$\frac{1}{4}\lambda^{\mu\nu}\Sigma_{\mu\nu}\bar{\psi}\gamma^{\rho}\partial_{\rho}\psi + \frac{1}{4}\lambda^{\mu\nu}\bar{\psi} \gamma^{\rho}\partial_{\rho} ( \Sigma_{\mu\nu}\psi) = 0$$

Commuting the Sigma and gamma spin matrices using the Dirac algebra turns this into

$$\frac{1}{2}\lambda^{\mu\nu}\Sigma_{\mu\nu}\bar{\psi}\gamma^{\rho}\partial_{\rho}\psi + \frac{1}{2}\lambda^{\mu\nu}\delta_{\mu}^{\rho}\delta_{\nu}^{\sigma}\bar{\psi}\gamma_{\rho}\partial_{\sigma}\psi$$

I don't see how that all turns to zero (since I've collected all the other terms into the required result). Obviously I'm doing which is pretty stupid and knowing me it's probably right infront of my face so feel free to make me look stupid by pointing it out because it begins to get to me, thanks :)

4. Nov 25, 2006

AlphaNumeric

I've now realised the total pig's breakfast I was making of the above question and how much simpler it actually is. Just needed to think about it.

Unfortunately, now I'm stuck on the conformal superalgebra and this time it doesn't even involve matrices.

$$D.\varphi = -x^{\nu}\partial_{\nu}\varphi - \varphi$$

$$P_{\mu}\varphi = -\partial_{\mu}\varphi$$

The algebra has the result $$[P_{\mu},D]\varphi = P_{\mu}\varphi$$ but I get the negative answer and the same happens with other commutators in the algebra, I get the right terms but signs wrong in places.

$$[P_{\mu},D]\varphi = P_{\mu}(D\varphi) - D(P_{\mu}\varphi)$$

$$= -\partial_{\mu}(-x^{\nu}\partial_{\nu}\varphi - \varphi) + D(\partial_{\mu}\varphi)$$

$$= x^{\nu}\partial_{\mu}\partial_{\nu}\varphi + \partial_{\mu}\varphi + \partial_{\mu}\varphi - x^{\nu}\partial_{\nu}\partial_{\mu}\varphi - \partial_{\mu}\varphi$$

$$= \partial_{\mu}\varphi = -P_{\mu}\varphi$$

It's the wrong sign but I can't see where I've made the mistake?!

Last edited: Nov 25, 2006
5. Nov 25, 2006

dextercioby

It could be a mistake in the book.

Daniel.

6. Nov 25, 2006

AlphaNumeric

It's both in those BUSSTEPP lecture notes I linked to before and it's in 'Introduction to Supersymmetry and Supergravity' by West.

West defines $$P_{\mu} = \partial_{\mu}$$ but that should still give the same conformal algebra (-1 factor cancels) just a slightly different representation.

Via similar, simple, computation I also get $$[P_{\mu},K_{\nu}] = -2\eta_{\mu\nu}D - 2M_{\mu\nu}$$ instead of $$[P_{\mu},K_{\nu}] = 2\eta_{\mu\nu}D - 2M_{\mu\nu}$$.

If I just got a totally different answer, I'd know I'm doing something fundamentally wrong, but the fact it's just a sign error here and there and all I'm doing it taking derivatices of fields and $$x^{\mu}$$ it's not complicated algebra. It's more frustrating that just getting it totally wrong!!

This would be the week the entire SUSY group in my department go away for a conference!

7. Dec 14, 2006

arivero

It could be not polite to ask for help in this thread considering I have not got to help to the original poster, but in some sense it is a continuation. What I am banging my head against is at the observation that the supercharges are in some sense square roots of the momentum,
$$\{Q, \bar Q\} = ... P_\mu$$
Because then, if one thinks that minimal coupling amounts to replace the momentum operator by the minimally coupled
$$(P_\mu) \to (P_\mu - i e A_\mu - ... )$$
then it seems painfully obvious that the way to introduce the gauge fields in a supersymmetrical theory is to find a set of supercharges $$Q^N$$ reproducing the new minimally coupled operator. It makes sense that in this way we are really bypassing Coleman-Mandula.

I would hope this to be stated in the first chapters of any supersymmetry book. I can not find it. I am wrong, or it is there and just they speak different language?

Last edited: Dec 14, 2006