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Let $\psi$ be a Dirac spinor. The so-called "Dirac conjugate" of $\psi$ is defined to be $\widetilde{\psi}:=\psi ^*\gamma ^0$, where $^*$ denotes the adjoint and the gamma matrices $\gamma ^\mu$ comprise the essentially unique irreducible representation of $\mathcal{C}\ell (1,3)$. Physicists introduce this, in a relatively ad hoc manner, so that the quantity $$ \widetilde{\psi}\psi $$ is Lorentz invariant. This does the trick, but I have a feeling like there is something deeper going on here.

The quantity $\psi ^*\gamma ^0$ makes sense for an arbitrary Clifford algebra $\mathcal{C}\ell (1,2m-1)$, whereas the notion of Lorentz invariance is specific to the case $m=2$, so the significance of $\widetilde{\psi}$ in other dimensions is not obvious to me. It might be the case that the significance of $\psi ^*\gamma ^0$ is unique to the case $m=2$, but I would be surprised if that were the case.

So then, what is the general mathematical significance of the Dirac conjugate $\psi ^*\gamma ^0$.

(Please let me know if explanation of any physics jargon is needed.)

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I think this issue between adjoints and Dirac conjugates is a whole lot of voodoo associated with the explicit matrix representation. Geometric algebra casts the Dirac equation without using matrices at all--the $\gamma^\mu$ are simply basis vectors in a real clifford algebra, on which the Hermitian adjoint has no meaning. This form is $\nabla \psi = -m \psi \gamma_5 \gamma_3$. No complex imaginaries are needed, which is why the Hermitian adjoint is not needed. I suspect the Dirac conjugate is related to reversion, but I honestly do not know for sure at this time. –  Muphrid Feb 13 '13 at 6:28
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Lorentz invariance in this context means that it transforms like a scalar under the fundamental representation of Spin(3,1). So perhaps in the general case the quantity $\tilde{\psi}\psi$ transforms like a scalar under Spin(n,1)? I confess I haven't tried the calculation though; perhaps I'll have a go once I've slogged through the rest of my examples sheets! –  Edward Hughes Feb 17 '13 at 22:03
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1 Answer 1

The Lorentz group, $SO(1,3)$, is non-compact, thus their representations are not unitary (in general).

Therefore, if you have a spinor, $\psi\in \mathcal{S}$, transforming as $\psi\mapsto S\psi$, it follows that the construction $$ \psi^\dagger \psi \mapsto \psi^\dagger S^\dagger S \psi \neq\psi^\dagger \psi,$$ since $S^\dagger\neq S^{-1}$.

This tells us that $\psi^\dagger$ does not belongs to the dual space of the spinors, $\psi^\dagger\not\in \mathcal{S}^*$.

It this point, you can realize that $$S^\dagger\gamma^0 = \gamma^0 S^{-1},$$ and this allows to define a dual spinor to $\psi$ through the construction $$\mathcal{S}^*\ni\bar{\psi}\equiv \psi^\dagger\gamma^0.$$

Hope it would be helpful.

Summary

The Dirac conjugate serves to define a dual spinor, by giving a spinor in the direct space.

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