# Understanding the substitution of a solution of Dirac equation in the Dirac equation

Dirac equation is $$(i \gamma^{\mu} \partial_{\mu}) \psi =0.$$

a solution of Dirac equation for massless fermion case is $$\psi (x) =u (\vec{p}) e^{ip^{\mu} x_{\mu}}.$$

substitution should give

$$(\gamma^{0} p_{0}-\vec{\gamma}.\vec{p}) u(\vec{p})=0$$

substituting and taking derivative gives zero with respect to $e^{ip^{\mu} x_{\mu}}$ so we need to take the partial with respect to $ip^{\mu} x_{\mu}$ only. can somebody show the work for how to get to the solution from here?

$0=(i \gamma^{\mu} \partial_{\mu}) \psi=(i \gamma^{\mu} \partial_{\mu}) u (\vec{p}) e^{ip^{\mu} x_{\mu}}=i \gamma^{\mu} u (\vec{p}) \partial_{\mu} e^{ip^{\mu} x_{\mu}}=i \gamma^{\mu} u (\vec{p}) i p_{\mu} e^{ip^{\mu} x_{\mu}}=- \gamma^{\mu}p_{\mu} u (\vec{p}) e^{ip^{\mu} x_{\mu}}$
So divide by $-e^{ip^{\mu} x_{\mu}}$ and get $0=\gamma^{\mu}p_{\mu} u (\vec{p})$ as you wanted.
Let us begin by taking the same Ansatz, so we want to solve for $\psi(t,\vec{x})=\psi(x^\mu)$ in $$i\gamma^\mu\partial_\mu \psi(x^\nu) = 0,$$ in order to get rid of the derivative, apply a Fourier Transform and focus on only one mode, that is for a fixed $p^\mu$, but otherwise arbitrary, thus: $$\psi(x^\mu) \longrightarrow e^{-ip^\nu x_\nu}\,u(p^\mu).$$ Our job is now to solve for the vector $u$ which has four spin components in principle. Writing down both, Lorentzian and Spinorial indices, the equation is now: $$i\gamma^\mu\partial_\mu e^{ip^\nu x_\nu}\,u(p^\mu) = \gamma_{ab}^0 p_0 u_b(p^\mu)-\gamma_{ab}^i p_i u_b(p^\mu) = 0$$ where we have used the $(+,-,-,-)$ signature. Let us now pick a particular representation for the $\gamma$ matrices, the Dirac representation, where: $$\gamma^0 = \begin{pmatrix}\mathbb{I} & 0 \\ 0 & \mathbb{I} \end{pmatrix} \qquad \gamma^1 = \begin{pmatrix}0 & \sigma_x\\ -\sigma_x & 0 \end{pmatrix}$$ $$\gamma^2 = \begin{pmatrix} 0 & \sigma_y\\ -\sigma_y & 0 \end{pmatrix} \qquad \gamma^3 = \begin{pmatrix} 0 & \sigma_z\\ -\sigma_z & 0 \end{pmatrix},$$ with $\sigma_i$ the Pauli matrices. The Dirac equation becomes then (after removing the exponential and the $i$ factors) $$\begin{pmatrix}\mathbb{I} & \frac{\vec{p}\cdot\vec{\sigma}}{p_0}\\ -\frac{\vec{p}\cdot\vec{\sigma}}{p_0} & -\mathbb{I} \end{pmatrix}u(p^\mu) = 0\tag{1}\label{eq:matrixDirac},$$ where $\vec{\sigma} = (\sigma_x,\sigma_y,\sigma_z)$ is a vector made out of the Pauli matrices. One can now exploit the property that $$(\vec{a}\cdot\vec{\sigma})(\vec{b}\cdot\vec{\sigma}) = (\vec{a}\cdot\vec{b})\mathbb{I} + i(\vec{a}\times\vec{b})\cdot\vec{\sigma}$$ Multiplying equation \eqref{eq:matrixDirac} by a suitable matrix we can quickly find the set of solutions for $u$: $$0 = \begin{pmatrix}\mathbb{I} & 0\\ 0 & -\frac{p_0\vec{p}\cdot\vec{\sigma}}{|\vec{p}|^2} \end{pmatrix} \begin{pmatrix}\mathbb{I} & \frac{\vec{p}\cdot\vec{\sigma}}{p_0}\\ -\frac{\vec{p}\cdot\vec{\sigma}}{p_0} & -\mathbb{I} \end{pmatrix}u(p^\mu) = \begin{pmatrix}\mathbb{I} & \frac{\vec{p}\cdot\vec{\sigma}}{p_0}\\ \mathbb{I} & \frac{p_0}{|\vec{p}|^2}\vec{p}\cdot\vec{\sigma}\end{pmatrix}u(p^\mu)$$ Now focus on the matrix only. Non-trivial solutions exist only for $p_0=|\vec{p}|$ in which case both rows are identical and the matrix gets a non-trivial nullspace, this is the correct dispersion relation for a massless particle so we are on the right track. Assuming such condition and taking $u(p^\nu)=(a,b,c,d)$ we can derive the following equations: $$a + \frac{p_3}{p_0}c + \frac{p_1-ip_2}{p_0}d = 0$$ $$b + \frac{p_1+ip_2}{p_0}c - \frac{p_3}{p_0}d = 0$$ Since $c$ and $d$ are free, as expected in terms of dimensions. You can take for example ($c=-p_0$,$d=0$) and $(c=0,d=-p_0)$ to get: $$u^{(1)}_s = \begin{pmatrix}p_3\\ p_1 +ip_2\\ -p_0\\ 0 \end{pmatrix} \qquad u^{(2)}_s = \begin{pmatrix}p_1-ip_2\\ -p_3\\ 0 \\-p_0 \end{pmatrix}$$ This gives you a completely specified solution. You may take more convenient bases for your case or go back to position space.