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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.

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3 Answers 3

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".

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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

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

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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.

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