# Arbitrary basis solutions to the Dirac equation

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...

• 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.
– Buzz
Jul 27, 2021 at 16:11

## 1 Answer

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.