# Plane wave solutions to massless Dirac equation

I'm revising for my QFT exam and have encountered some issues with something I haven't seen before: finding the plane wave solutions to a massless Dirac equation:

$$i\gamma^\mu\partial_{\mu} \psi = 0$$

I then use the two-component spinor representation, $\psi^T = \left(u,v\right)$ and the equation decouples into two independent equations for $u$ and $v$.

I'm okay with the case when the mass is nonzero and I feel like this should be even easier, but when I try to use the ansatz of the form:

$$u = \lambda(p) e^{-ipx}$$

I just get

$$\bar{\sigma}^{\mu}p_{\mu}\lambda(p) e^{-ipx}$$

and I'm not really sure where to go on from there. How does this vanish? How can I find the functional form of $\lambda(p)$?

• $\lambda(p)$ is a column vector. You solve the matrix equation ${\bar \sigma}^\mu p_\mu \lambda(p) = 0$. – Prahar Apr 19 '17 at 22:37
• So I have $\left(p_0 + \sigma^ip_i\right)\lambda(p)$. I could potentially write out the Pauli matrices explicitly and I end up with: $\begin{bmatrix} p_0+p_3 & p_1-ip_2 \\ p_1+ip_2 & p_0-p_3 \\ \end{bmatrix} \times \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \end{bmatrix} = 0$. But this requires that either the momenta are such that the expressions cancel or that both $\lambda_1$ and $\lambda_2$ are zero. Sorry if I'm being blunt here but I really don't see how this works out to give a nontrivial solution. – Piotr Apr 20 '17 at 8:35
• The former happens. Why don't you actually work through the calculations before jumping to conclusions. You have a matrix equation. Solve it in the usual way. – Prahar Apr 20 '17 at 12:53
• Sorry @Prahar, I didn't explain myself very well. I did do that; it returns the fact I already know: that the particle is mass-less/ultra-relativistic, i.e. $p_0^2 = \vec{p}^2$. This answers the first of my questions - it shows that plane wave is indeed a solution if the particle is massless, which it is. $\lambda$ however remains frustratingly unconstrained and I'm not quite sure how to derive the functional form of it. I really appreciate your help, thanks a lot for your time! – Piotr Apr 20 '17 at 13:05
• I do not understand why you think that. You start out with two equations for two variables. The fact that the square matrix has zero determinant means that the two equations are linearly dependent. So then you are left with one equation for 2 variables. This means you can solve one variable in terms of the other. The one remaining variable cannot be fixed by Dirac equation and is the overall normalization of $\lambda$. You fix this by requiring that $\lambda$ has norm 1 with respect to the Dirac inner product. – Prahar Apr 20 '17 at 13:08