Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's have Dirac spinor $\Psi (x)$. It transforms as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$ representation of the Lorentz group: $$ \Psi = \begin{pmatrix} \psi_{a} \\ \kappa^{\dot {a}}\end{pmatrix}, \quad \Psi {'} = \hat {S}\Psi . $$ Let's have spinor $\bar {\Psi} (x)$, which transforms also as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$, but as cospinor: $$ \bar {\Psi} = \begin{pmatrix} \kappa^{a} & \psi_{\dot {a}}\end{pmatrix}, \quad \bar {\Psi}{'} = \bar {\Psi} \hat {S}^{-1}. $$ How to show formally that $$ \bar {\Psi}\Psi = inv? $$ I mean that if $\Psi \bar {\Psi}$ refers to the direct product (correct it please, if I have done the mistake) $$ \left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right]\otimes \left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right], $$ what group operation corresponds to $\bar {\Psi} \Psi$?

This question is strongly connected with this one.

share|cite|improve this question
up vote 2 down vote accepted

You need to work out the tensor product and will find a direct sum of different contributions \begin{multline} [(1/2, 0) \oplus (0, 1/2)] \otimes [(1/2, 0) \oplus (0, 1/2)] =\\ \big((1/2, 0) \otimes (1/2, 0)\big) \oplus \big((1/2, 0) \otimes (0, 1/2) \big)\oplus \quad \\\big((0, 1/2) \otimes (1/2, 0)\big) \oplus \big((0, 1/2) \otimes (0, 1/2)\big) = \\ (0, 0) \oplus (1, 0) \oplus (1/2, 1/2) \oplus (1/2, 1/2) \oplus (0, 1) \oplus (0, 0)\end{multline} The states now can be classified:

  • $(0, 0)$ is a scalar or pseudoscalar, i.e. the $\bar \psi \psi$ you are looking for as well as $\bar \psi \gamma_5 \psi$
  • $(1/2, 1/2)$ is the vector / pseudovector component $\bar \psi \gamma^\mu \psi$ or $\bar \psi \gamma^\mu \gamma_5 \psi$
  • (1, 0) and (0, 1) are the (anti)-self dual parts of the tensor $\bar \psi \sigma^{\mu \nu } \psi$

All these transform well-definedly under Lorenty boosts. The $(0, 0)$ part tells you that this rep will transform neither under the left-chirality nor the right-chirality $sl(2)$ that you classify the reps by.

Edit: Let me add that the distribution law I used above to get from the first to the second line is one of reasons we speak of a "direct sum" vs. "direct product".

share|cite|improve this answer
But how to convolute this expression into $(0, 0)$? And what are the differences between $\bar {\Psi} \Psi$ and $\Psi \bar {\Psi}$ (look here:…)? – Andrew McAddams Mar 23 '14 at 19:29
In $\bar \psi \psi = \sum_\alpha \bar \psi_\alpha \psi_\alpha$ you contract the spinor indices between the two spinors. It is in that sense a bispinor scalar of the Lorentz group. $\psi \bar \psi = \psi_\alpha \bar \psi_\beta$ is a matrix in spinor space. – Neuneck Mar 23 '14 at 19:42
So in both cases I will operate with the direct product of representations and in the group language (which was used in your answer) there aren't differences between these cases? Do I need to write tag $\left( \frac{1}{2} , 0\right)_{c} \oplus \left( 0, \frac{1}{2} \right)_{c} $, which corresponds for $\bar {\Psi}$ and means that its components transforms as cospinors? – Andrew McAddams Mar 23 '14 at 19:50
And another one (which is connected with previous) question: for case $\bar {\Psi}\Psi$ corresponding value is matrix rank zero, but for case $\Psi \bar {\Psi}$ corresponding value refers to the matrix rank two. How the direct product notation considers it in both cases? – Andrew McAddams Mar 23 '14 at 19:51
$\psi \bar \psi$ is not a tensor product of Lorentz representations, there simply is no way to build an object that is not a scalar in sinor coordinates and still a representation of the Lorentz algebra. This is analogous to the realization that the addition of two half-integer angular momenta always yields integer results, i.e. $1/2 \otimes 1/2 = 0 \oplus 1$. There is no way to create some half-integer result. The object $\psi \bar\psi$ exists, but should not transform properly under lorentz transformations. – Neuneck Mar 23 '14 at 20:41

If we assume that

$$\Psi {'} = \hat {S}\Psi$$


$${\bar{\Psi}}{'} = \bar {\Psi} \hat {S}^{-1},$$

it follows that the product of the two transforms as

$$(\bar{\Psi}\Psi)'={\bar{\Psi}}{'}\Psi {'}=\bar {\Psi} \hat {S}^{-1}\hat {S}\Psi=\bar{\Psi}\Psi,$$

which is a consequence of

$$\hat {S}^{-1}\hat {S}=\mathbb{1}.$$

share|cite|improve this answer
My question was following: "...I mean that if $\Psi \bar {\Psi}$ refers to the direct product, what group operation corresponds to $\bar {\Psi}\Psi$?..". So I want to get some group operation (like direct product) which show me that $\bar {\Psi} \Psi \equiv (0, 0)$. Not by the language of transformation matrices. Excuse me for my unintelligible explanation. – Andrew McAddams Mar 23 '14 at 19:04

short answer if $ \hat {S}^{-1} S = \mathbb{I}$

I can give you a general example of $\psi^\dagger\psi$ not being invariant.

because for Dirac spinor $\psi$ whe have the following transformation rules $$\psi(x) \rightarrow S[\Lambda] \psi(\Lambda^{-1}x)=S[\Lambda] \psi(x^\prime) \\ \psi^\dagger(x) \rightarrow \psi^\dagger(\Lambda^{-1}x) S[\Lambda]^\dagger $$ So $\psi^\dagger\psi \rightarrow \psi^\dagger(\Lambda^{-1} x)S[\Lambda]^\dagger S[\Lambda] \psi(\Lambda^{-1} x) $ is invarieant if and only if $S[\Lambda]^\dagger S[\Lambda] = \mathbb{I}$

however for the case where $S[\Lambda]$ are formed by the Clifford algebra it can be shown this is not they case. I do not have the capability to show you that they dirac adjoint does satisfy this condition.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.