# Spinor dotted and undotted indices

I have had an introduction to QFT following the book of Mandl and Shaw. However, I have been asked to write a report on the CPT theorem. For this, the main reference I'm using is PCT, spin and statistics and all that, by Streater and Wightman.

I find that the introduction of dotted and undotted indices is really unclear in this book. I also found a reference on dotted indices in Local Quantum Physics by Haag, but I do not understand what he is talking about.

Can someone explain, what the indices are, how they are defined and most importantly, how, for a given spinor, do you know what their indices are supposed to be?

For example, what is the difference between $\psi$, $\psi_\alpha$, $\psi_\dot{\alpha}$, $\psi^\alpha$, $\psi^\dot{\alpha}$ and $\psi^{\alpha \beta \dot{\gamma}}_{\delta \dot{\epsilon} \dot{\zeta}}$?

The problem I think, lies with the fact I have no experience with representation theory, nor has the course I had ever mentioned such things. So if you could give or reference an explanation of these things, without the need for representation theory, that would be most helpful. Thanks!

• I don't think you will be able to really understand what's going on without using representation theory. The one and only difference between dotted and undotted indices is that they correspond to a certain representation and its conjugate representation. May 30, 2018 at 15:14
• And how are they implemented? How would you know what representation a standard Dirac spinor with no indices belongs to? May 30, 2018 at 15:18
• A Dirac spinor has both dotted and undotted indices inside (cf. wikipedia). Dotted and undotted spinors refer to Weyl spinors. A Dirac spinor consists of the direct sum of a Weyl spinor with a dotted index, and a Weyl spinor with an undotted index. May 30, 2018 at 15:23
• So $\psi_\alpha^{\dot \alpha}$ is a Dirac spinor? May 30, 2018 at 16:05

Weyl spinors are the two-dimensional irreducible representations of the group $\mathrm{SL}(2,\mathbb{C})$. A representation is an action of the group on a vector space, and for Weyl spinors, this vector space is $\mathbb{C}^2$. The different types correspond to the different ways our group $\mathrm{SL}(2,\mathbb{C})$ can act on $\mathbb{C}^2$.

There is one obvious action, called the fundamental representation. In particular, if $N \in \mathrm{SL}(2,\mathbb{C})$ and $v \in \mathbb{C}^2$, then we can simply act $v \mapsto N v$. From any representation we can always construct the conjugate representation and the dual representation, and also the conjugate dual representation. So we can now consider the following four representations:

\begin{align} \psi_\alpha &\mapsto N_\alpha{}^\beta \psi_\beta & & \text{fundamental}\\ \psi_\dot{\alpha} &\mapsto N^*{}_\dot{\alpha}{}^\dot{\beta} \psi_\dot{\beta} & & \text{conjugate}\\ \psi^\alpha &\mapsto (N^{-1})_\beta{}^\alpha \psi^\beta & & \text{dual} \\ \psi^\dot{\alpha} &\mapsto (N^{*-1})_\dot{\beta}{}^\dot{\alpha} \psi^\dot{\beta} & & \text{conjugate dual} \end{align} The nature of the index on our vector $\psi$ is telling us in which of these four ways our vector transforms. The dual representation may be familiar from general relativity – if we consider a contravariant vector as transforming in the fundamental of $\mathrm{GL}(n,\mathbb{R})$, then a covariant vector transforms in the dual representation. The conjugate representation may be less familiar, since we often work with real vector spaces for which complex conjugation does nothing. Note that if $v$ transforms in the fundamental, then $v^*$ transforms in the conjugate.

It turns out that these are all the ways $\mathrm{SL}(2,\mathbb{C})$ can act on $\mathbb{C}^2$. In fact, it turns out that the fundamental and the dual representation are equivalent1 (the precise sense in which this is meant can be found in any representation theory book), owing to the existence of the invariant tensor $\epsilon_{\alpha \beta}$. This means that the conjugate and conjugate dual are also equivalent, so there are really only two different representations, which we often call left-handed and right-handed Weyl spinors.

1Note that "equivalent" does not mean "identical" here. We must still take care to distinguish upper and lower indices, because the transformation laws are not the same – the two representations merely have the same essence.

• Thanks, this already explains a lot actually. But what about spinors with multiple indices? Does $N$ then act the same number of indices and then in the way corresponding to those indices? So for example, $\psi_\alpha^{\dot{\alpha}} \mapsto N_\alpha^\beta (N^{*-1})_{\dot{\beta}}^{\dot{\alpha}} \psi_\beta^{\dot{\beta}}$? Or is this nonsense? May 30, 2018 at 16:44
• Tensor products.... May 30, 2018 at 17:34
• @user353840 That's roughly the idea, yes, although I expect there are multiple conventions (involving symmetrisation and antisymmetrisation, perhaps). Note that a Dirac spinor is not a two-index object in this sense, however, but rather has two components, one of which is a left-handed spinor and one of which is a right-handed spinor: $\psi_D = (\psi_\alpha, \chi^\dot{\alpha})$. This is a direct sum, whilst your object $\psi^\dot{\alpha}{}_\alpha$ is a tensor product. May 30, 2018 at 18:09
• Shouldn't a Dirac spinor be called a bispinor then? Since, from this, I understand that a spinor has only two components, while a Dirac spinor has four.. May 30, 2018 at 18:33
• @user353840 I think Dirac spinors and bispinors are identical, but it is completely legitimate to call a Dirac spinor a spinor! Only Weyl spinors are two-dimensional. May 30, 2018 at 18:37