# SUSY computational questions:dots!

1. Sep 22, 2009

### haushofer

Hi, I'm reading Bilal's notes on SUSY, hep-th/0101055v1, and have some computational questions.

So I understand that spinors can be seen as objects carrying the basis rep. of SL(2,C), and how SO(3,1) is locally isomorphic to SU(2)XSU(2), giving basically two "sectors". With dots and bars we indicate in which SU(2) algebra the specific spinor is sitting.

We can introduce an inner product between spinors via

$$\epsilon^{12}=\epsilon^{\dot{1}\dot{2}}=-\epsilon^{21}=-\epsilon^{\dot{2}\dot{1}}=1$$

and an opposite sign for the indices down.

Now, some identities are mentioned, such as (eq.2.15)

$$\xi\sigma^{\mu}\bar{\psi} = -\bar{\psi}\bar{\sigma}^{\mu}\xi$$

How can I proof this? What's the origin of that minus-sign? And how do I contract the indices exactly in this equation? I'm a little confused, so to speak :)

2. Sep 22, 2009

### RedX

The indices look like this:

$$\xi^{a}\sigma^{\mu}_{a\dot{b}}\bar{\psi}^{\dot{b}}=-\bar{\psi}^{\dot{b}}\sigma^{\mu}_{a\dot{b}}\xi^{a}=-\bar{\psi}_{\dot{b}}\bar{\sigma}^{\mu}^{\dot{b} a}\xi_{a}$$

The negative sign in the 2nd line comes from swapping two fermion fields which anticommute. In the third line, we lowered and raised some indices with the Levi-Civita an even number of times, so there is no sign change (you can verify that for two spinors, $$\psi^a \xi_a=-\psi_a \xi^a$$, so you have to be careful if you're swapping indices only once). By definition, $$\bar{\sigma}^{\dot{b} a}$$ is what you get when you raise the indices of $$\sigma_{a \dot{b}}$$

3. Sep 22, 2009

### javierR

When you write spinors in terms of the indices, you do not have to include "anti-commuting negative signs", this is encoded in the order of the index contractions. Also, it's good practice to rewrite contractions with epsilons. Therefore, according to the conventions in the article, the steps are:
$$\chi^{\alpha} \sigma_{\alpha\dot{\alpha}} \bar{\psi}^{\dot{\alpha}} =\bar{\psi}^{\dot{\alpha}} \sigma_{\alpha\dot{\alpha}} \chi^{\alpha} =-\bar{\psi}^{\dot{\alpha}} \sigma_{\dot{\alpha}\alpha} \chi^{\alpha} =-\bar{\psi}^{\dot{\alpha}} \epsilon_{\dot{\alpha}\dot{\beta}}\epsilon_{\alpha\beta}\bar{\sigma}^{\dot{\beta}\beta} \chi^{\alpha} =-\bar{\psi}_{\dot{\beta}}\bar{\sigma}^{\dot{\beta}\beta} \chi_{\beta}$$

4. Sep 22, 2009

### Avodyne

This is explained in great detail in the QFT book by Srednicki. You can find a draft copy free at his web page.

5. Sep 22, 2009

### xepma

You basically state the origin of the sign: indices are contracted through an antisymmetric tensor rather than a symmetric one. Therefore the order of upper and lower indices matter! And it also matters which component you use to lower or raise an index using the epsilon tensor. One has to keep careful track of the conventions you use.

$$\chi^\alpha \xi_\alpha = (\chi_\beta \epsilon^{\alpha\beta}) \xi_\alpha = \chi_\beta (\epsilon^{\alpha\beta} \xi_\alpha) = -\chi_\beta (\epsilon^{\beta\alpha} \xi_\alpha) = -\chi_\beta\xi^\beta$$

Here I used the convention for the inner product:
$$\chi \cdot \xi \equiv \chi^\alpha\xi_\alpha$$
and the one for raising:
$$\xi^\beta= \epsilon^{\beta\alpha} \xi_\alpha$$

This leads automatically to the statement that $$\chi \cdot \xi = - \xi \cdot \chi$$, which is a bit unconventional if you're just used to ordinary tensor contractions.

6. Sep 23, 2009

### haushofer

Ok, thanks for the answers! So I don't have to introduce minus signs if I swap spinors, but only if I swap indices which concern contractions involving spinors, right?

Let me take the next example:

$$(\xi \sigma \bar{\psi})^{\dagger} = \psi\sigma\bar{\xi}$$

So in indices the LHS is, I would say,

$$(\xi^a\sigma_{a\dot{b}}\bar{\psi}^{\dot{b}})^{\dagger} = (\bar{\psi}^{\dot{b}})^{\dagger}\sigma^{\dagger}_{a\dot{b}}\xi^{a}^{\dagger}$$

I'm tempted to identify

$$(\bar{\psi}^{\dot{b}})^{\dagger} = \psi^b$$

etc, but what happens with the indices of sigma here?

I'll also look at Srednicki, thanks for the tip!

7. Sep 23, 2009

### Avodyne

This is true only if your treat $\chi$ and $\xi$ as commuting. If they are anitcommuting, then you have an extra minus sign from the anticommutation, leading to an overall plus sign.

8. Sep 23, 2009

### Avodyne

Srednicki explains it.

9. Sep 23, 2009

### RedX

I ought to mention that the way I did the problem is entirely the way that Srednicki teaches it in his excellent textbook. Obviously there seems to be other ways of doing it and I'm going to have a look at all them because I'm sort of confused now about the difference between twistors and spinors, i.e., once we multiply out 4-Dirac spinors, shouldn't we treat the components as twistors and not spinors?