# Vanishing of a solution of Dirac equation

Let $$\psi(x,t)$$ be a solution of the free Dirac equation. Assume that $$\psi(\vec x,0)=\delta^{(3)}(\vec x) u,$$ where u is a fixed spinor. (In other words $$\psi(\vec x,0)$$ is assumed to be supported at 0.)

Is it true that for $$t>0$$ the wave function $$\psi(\vec x,t)$$ vanishes outside of the light cone?

I'm not sure I can answer your question for the dirac equation since I think I would need a starting condition in $$L^2(\mathbb{R}^3,\mathbb{C}^4)$$ and you gave me a vector in $$L^2(\mathbb{R}^3,\mathbb{C})$$ maybe I misunderstand what you meant. I will therefore assume that we are talking about the positive energy solution to the klein–gordon equation in stead. Since the initial condition is constant in momentum space lets look at the problem there. We have
$$\psi(p,t)=e^{-\frac{i}{\hbar}H t}\psi(p,0)=(2\pi\hbar)^{-3/2}e^{-\frac{it}{\hbar}\sqrt{m^2c^4+p^2c^2}}$$
$$\psi(r,t)=(2\pi\hbar)^{-3/2}\int dp\:\psi(p,t) e^{\frac{i}{\hbar}p\cdot r}=(2\pi\hbar)^{-3}\int dp e^{-\frac{it}{\hbar}\sqrt{m^2c^4+p^2c^2}}e^{\frac{i}{\hbar}p\cdot r}.$$
• Thanks, I corrected my statement. I think for the Klein-Gordon equation the last two formulas you used are not complete since $\psi(r,t)$ is not defined uniquely by the initial value $\psi(r,0)$ since the Klein-Gordon equation is of the second order in $t$. – MKO Mar 17 at 0:44