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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
up vote 1 down vote accepted

Let $S$ denote the space of Dirac spinors with the fundamental representation of $Cl(1,3)$, so elements of $S$ are complex column 4-vectors, and $Cl(1,3)$ is acting by matrix multiplication on the left. The general significance of $\tilde\psi$ is that it is the image of $\psi$ under a linear transformation which is an equivalence of representations.

To be more specific, suppose $e_i \mapsto \gamma(e_i)=\gamma_i$ denotes the fundamental representation above. We can define a different representation of $Cl(1,3)$ by sending $a\mapsto {\overline{\gamma(\beta(a))}}^t$ where $\beta$ is the fundamental anti-automorphism of the Clifford algebra which exchanges the order of the generators. This defines a representation of the Clifford algebra on the conjugate dual space $\bar{S}^*$. In particular, we see that $e_i\mapsto {\bar\gamma_i}^t$ in this representation.

We can seek a linear transformation $A: S \rightarrow \bar{S}^*$ which commutes with the representations (that is, we want $A$ to be an equivalence) by solving

$A\gamma_i={\bar{\gamma_i}}^tA$

which has a unique solution up to phase (i.e. a complex scale factor). This solution turns out to agree with the matrix $\gamma_0$.

Notice that $\bar{A}:\bar{S}\rightarrow S^*$, so we can define a pairing of elements of $S$:

$(\phi, \psi)=<\bar{A}(\bar{\phi}),\psi>$

where $<,>$ denotes the natural pairing of a vector space with its dual space. In the typical chiral (Weyl) choice for the $\gamma$-matrices, $\gamma_0$ is symmetric and real, so the matrix $\gamma_0$ is also the matrix of this bilinear form.

So, the significance of $\tilde{\psi}$ is that it is the contraction of $\bar{\psi}$ with this bilinear form.

A useful principle I keep in mind when I think about this stuff is that whenever physicists do things in an $\textit{ad hoc}$ manner, guided by Lorentz invariance, mathematically we can achieve the same goal by thinking about equivalence of representations.

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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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