I'm considering solutions to the free Dirac equation $$ (i\gamma^\mu\partial_\mu-m)\psi=0 $$ where $\psi$ is the 4-component bispinor field being solved for, and the $\gamma^\mu$ matrices must satisfy $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$.

Using the plane-wave Ansantz $\psi=u(p)e^{-ip_\mu x^\mu}$ (so-called positive frequency), the equation becomes $$ (p_\mu \gamma^\mu-m)u(p)=0 $$ Additionally left-multiplying by $(p_\nu\gamma^\nu+m)$ leads to $$ (p_\nu\gamma^\nu+m)(p_\mu \gamma^\mu-m)u(p)=(p_{\nu}p_{\mu}\eta^{\nu\mu}-m^2)u(p)=0 $$ which constrains $p_\mu$ to lie on the relativistic mass shell.

Similarly, the Ansatz $\psi=v(p)e^{ip_\mu x^\mu}$ (so-called negative frequency) leads to $$ (p_\nu\gamma^\nu+m)v(p)=0 $$ where again $p_\mu$ must lie on the mass shell.

Looking at the equation for $u(p)$, we see the solutions in fact form the kernel of the matrix $$ p_\mu\gamma^\mu-m $$ At this point, all texts I have read proceed with some concrete basis for the $\gamma^\mu$. In the Weyl basis for example, where $$ \gamma^\mu=\begin{pmatrix}0 & \sigma^\mu \\ \bar\sigma^\mu & 0\end{pmatrix}, \quad \quad\sigma^0\equiv \bar\sigma^0 \equiv I_2, \sigma^k\text{are the Pauli matrices}, \bar\sigma^k\equiv -\sigma^k $$ the solutions can be written as (see e.g. David Tong's lecture notes)

$$ u(p)=\begin{pmatrix}\sqrt{p_\mu\sigma^\mu}\chi \\ \sqrt{p_\mu\bar\sigma^\mu}\chi\end{pmatrix} $$ where $\chi$ is an arbitrary two-entry column matrix. That is to say, for a given $p$, there are essentially two linearly independent solutions. Or said another way, the matrix $p_\mu \gamma^\mu-m$ has rank 2.

Working in other usual basis for the $\gamma^\mu$, one similarly finds two independent solutions for $u(p)$. The equation for $v(p)$ leads to 2 further solutions, this time spanning the kernel of the matrix $p_\mu \gamma^\mu + m$

Physically, this seems to be entirely reasonable: for a give momentum, and choice of frequency sign (i.e. u or v), there are two possible independent spin orientations. This leads me to suspect, that the rank of $p_\mu \gamma^\mu \pm m$ should always be 2, regardless of the particular choice of $\gamma^\mu$.

So here's my question: (how) can I show this to indeed be the case? Is there some bispinor basis-independent calculation possible, using nothing more than the anticommutation constraint on the $\gamma^\mu$?

The best I've come up with thus far, is to note that when $p$ is on-shell, $$ (p_\mu\gamma^\mu-m)(p_\nu\gamma^\nu+m)=p_\mu p^\mu - m^2=0 $$ proves that all the columns of $(p_\nu\gamma^\nu+m)$ belong to $(p_\mu\gamma^\mu-m)$'s kernel and therefore it must have dimension > 0. But I haven't found how to show it equals exactly 2...

  • 1
    $\begingroup$ The usual way to prove that the rank is always 2 is to show that any two representations of the Dirac matrices are related by a similarity transformation, which automatically preserves the rank of something like that. However, that is probably a lot more mathematical machinery than is needed for what you are asking. $\endgroup$
    – Buzz
    Jul 27, 2021 at 16:11

1 Answer 1


Ok, here is my proof:

Firstly, the $\gamma^\mu$ matrices, whatever the basis, must be traceless. Indeed, $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$ implies that 1) $\gamma^\mu\gamma^\nu$=$-\gamma^\nu\gamma^\mu$ when $\mu \neq \nu$, and 2) $\gamma^\nu\gamma^\nu=\pm I_4$ (no summation over $\nu$), and therefore $$ 2 tr(\gamma^\mu)=\pm 2 tr(\gamma^\nu\gamma^\nu\gamma^\mu) = \pm tr[\gamma^\nu(\gamma^\nu\gamma^\mu)-(\gamma^\nu\gamma^\mu)\gamma^\nu]=0 $$ since the trace of a matrix product is independent of the order of factors.

Next, if a matrix $A$ has an Eigenvalue $\lambda$, with Eigenvector $v_\lambda$, then $$ A^2 v_\lambda=A \lambda v_\lambda = \lambda A v_\lambda = \lambda^2 v_\lambda $$ i.e. $\lambda^2$ is an Eigenvalue of $A^2$. By corollary, all Eigenvalues of $A$, must be a square root of some Eigenvalue of $A^2$.

Finally, we have $$ p_\mu\gamma^\mu p_\nu \gamma^\nu = \frac{1}{2}p_\mu p_\nu\{\gamma^\mu,\gamma^\nu\}=p_\mu p_\nu \eta^{\mu\nu}=m^2 I_4 $$

Therefore $p_\mu\gamma^\mu$ can ONLY have Eigenvalues $+m$ or $-m$. Furthermore, as a linear combination of the traceless $\gamma^\mu$, we also have $tr(p_\mu\gamma^\mu)=0$. Since the trace equals the sum of Eigenvalues, they must be exactly $+m$ and $-m$, each with multiplicity 2.

Then, $p_\mu\gamma^\mu+m$ has Eigenvalues $2m$ and $0$, each with multiplicity 2, and $p_\mu\gamma^\mu-m$ has Eigenvalues $0$ and $-2m$, each with multiplicity 2, confirming my suspicion, that they have kernels of dimension 2, regardless of the $\gamma^\mu$ representation basis.


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