# Order or Grassmann, vector fields and tensors

1. Oct 4, 2015

### marir

Hello.
There is one thing I can not find the answer to, so I try here.
For instance, writing a general superfield on component form, one of the terms appearing is:
$\theta \sigma^\mu \bar{\theta} V_\mu$

My question is if one could have written this as
$\theta \bar{\theta} \sigma^\mu V_\mu$ ?

This might be a silly question, but anyway. Is there in any case a reason why one writes it in the first way? I think it looks cleaner if one collects the Grassmann numbers.

2. Oct 4, 2015

### Orodruin

Staff Emeritus
This depends on what the constituents of your expression are. You should always specify your notation and not just assume that everyone will understand it wothout further explanation.

Regardless, I am going to attempt a guess: $\theta$ and $\bar\theta$ being Weyl fermion fields and $\sigma$ a Pauli matrix, the second structure you propose does not make any sense. The first structure has the structure row vector - square matrix - column vector and is therefore a scalar, while your suggestion is row vector - column vector - square matrix. If you want to change the order, you will have to write out the indices of the matrix multiplication explicitly.

3. Oct 4, 2015

### marir

OK, sorry for the spare comments on my notation. $\theta \bar{\theta}$ are Grassmann numbers, and I think this is where my confusion is. They are anticommuting c-numbers, so in my head I am thinking it is irrelevant where I put them in my expression as long as the order among them is the same. Of course (like everything else), they can be represented by matrices, but people tell me to think of them as numbers, so that is what I try to do, and therefore I am having a hard time justifying why their placement is so important? If I must think the Grassmann numbers are matrices, then I get it. But then it is confusing having people telling me they are "just numbers".
Furter, we have $\sigma^\mu = (1,\vec{\sigma})$, where $\vec{\sigma}$ are the ordinary Pauli matrices. Finally, $V_\mu$ is a Lorentz 4-vector.

4. Oct 4, 2015

### Orodruin

Staff Emeritus
Grassman numbers are not just matrices. But just as you can have a matrix containing ordinary numbers, you can have a matrix containing Grassman numbers. This is the case for Weyl fermion fields, they are not just Grassman numbers, but spinors containing Grassman numbers.

5. Oct 5, 2015

### samalkhaiat

Put the appropriate indices on them first, then you can move them around. $$\theta^{\alpha} \ (\sigma^{\mu})_{\alpha \dot{\alpha}} \ \bar{\theta}^{\dot{\alpha}} \ V_{\mu} = \theta^{\alpha} \ \bar{\theta}^{\dot{\alpha}} \ (\sigma^{\mu}V_{\mu})_{\alpha \dot{\alpha}} = \mathbb{V}_{\alpha \dot{\alpha}} \ \theta^{\alpha}\bar{\theta}^{\dot{\alpha}} .$$ Notice that $\chi \ \theta$ and $\bar{\chi} \ \bar{\theta}$ are convenient ways to write the scalar products $$\chi \ \theta \equiv \chi^{\alpha} \ \theta_{\alpha} ,$$ $$\bar{\chi} \ \bar{\theta} \equiv \bar{\chi}_{\dot{\alpha}} \ \bar{\theta}^{\dot{\alpha}} .$$ However, you cannot pair $\theta$ with $\bar{\chi}$ to form a scalar (they belong to different representation space, i.e., carry different indices) so you need the soldering matrices $\sigma_{\mu}$ and $\bar{\sigma}_{\mu}$ to pair them together. $$\theta \ \sigma^{\mu} \ \bar{\chi} \equiv (\sigma^{\mu})_{\alpha \dot{\alpha}} \ \theta^{\alpha}\ \bar{\chi}^{\dot{\alpha}} ,$$ $$\bar{\chi} \ \bar{\sigma}^{\mu} \ \theta \equiv \bar{\chi}_{\dot{\alpha}} \ (\bar{\sigma})^{\dot{\alpha}\alpha} \ \theta_{\alpha} .$$
Notice the consistency of following identities $$\theta \ \chi = \chi \ \theta , \ \ \ \ \bar{\chi} \ \bar{\theta} = \bar{\theta} \ \bar{\chi} , \ \ \ \ \theta \ \sigma_{\mu} \ \bar{\chi} = - \bar{\chi} \ \bar{\sigma}_{\mu} \ \theta ,$$ with the Grassmannian nature of the spinor components $$\theta_{\alpha} \ \chi_{\beta} = - \chi_{\beta} \ \theta_{\alpha} , \ \ \ \bar{\chi}_{\dot{\alpha}} \ \bar{\theta}_{\dot{\beta}} = - \bar{\theta}_{\dot{\beta}} \ \bar{\chi}_{\dot{\alpha}} , \ \ \ \theta_{\alpha} \ \bar{\chi}_{\dot{\beta}} = - \bar{\chi}_{\dot{\beta}} \ \theta_{\alpha} .$$

6. Oct 6, 2015

### marir

Thank you so much, this explained alot. I have a confusion going on about may things, but at least some pieces fell in place now.