I am having trouble understanding solutions of the Dirac equation. From what I understand, the probability current four-vector is $J^\mu=\bar\psi\gamma^\mu\psi=\psi^\dagger\gamma^0\gamma^\mu\psi$.

The problem is that the solutions that I have seen of the Dirac equation are said to be

$$\psi^{(1)} = e^{\frac{-imc^2t}{\hbar}}\begin{pmatrix}1\\0\\0\\0\\\end{pmatrix},\psi^{(2)} = e^{\frac{-imc^2t}{\hbar}}\begin{pmatrix}0\\1\\0\\0\\\end{pmatrix},\psi^{(3)} = e^{\frac{imc^2t}{\hbar}}\begin{pmatrix}0\\0\\1\\0\\\end{pmatrix},\psi^{(4)} = e^{\frac{imc^2t}{\hbar}}\begin{pmatrix}0\\0\\0\\1\\\end{pmatrix}.$$

But shouldn't the time component of the probability current be normalized so that

$$\int \,J^0\,dx\,dy\,dz =\int \,\rho \, dx\,dy\,dz = 1~?$$

This isn't the case for the solutions I've listed. What happened?


2 Answers 2


You are interpreting the time component of the probability current as a probability density. But it is a component of a four vector, so is bound to change under Lorentz transformations. Normalization of this quantity thus breaks Lorentz invariance.

However, you are not far off. What we know is that the Dirac current is conserved, i.e. $$ \partial_\mu j^\mu=\partial_\mu \bar{\psi}\gamma^\mu\psi = 0 $$ Writing out components of $j^\mu$, this reads: $$ \partial_tj^0 - \partial_i j^i = 0$$ where the Latin indices run only over spatial directions. Integrate this whole equation over space. We can take the time-derivative outside of this integral: $$ \partial_t \int d^3x j^0 - \int d^3x \partial_ij^i = 0 $$ We can evaluate the second term using the "divergence theorem": $$ \int d^3x \partial_ij^i = \oint (j \cdot \hat{n}) dS $$ where we evaluate the surface integral over a sphere at infinity. Here is when 'normalization' comes in. If we want to have a normalizable solution, surely the solution should be going to zero as we approach infinity. Hence the spatial components of the probability current should vanish infinitely far away. If this is the case, the integral above vanishes.

Letting $Q = \int d^3x j^0$, we are left with the equation $\partial_t Q = 0$. In classical field theory (like electromagnetism) we would interpret this as the "conservation of electric charge". So your integral is not normalized, but it is conserved in time precisely because we require our solutions to be sufficiently localized as to be normalizable.

  • $\begingroup$ I think it is a consequence of uncertainty. The momentum is exactly known so there is infinite uncertainty in position. I'm not sure if normalization breaks Lorentz invariance. I think that there should be wave packet solutions to the Dirac equation in which position and momentum uncertainty are both nonzero. $\endgroup$ Oct 27, 2020 at 9:25
  • $\begingroup$ In the treatment here, nothing is actually quantum. This is just the classical field theory of fermions coupled to EM. $\endgroup$
    – hulsey
    Nov 3, 2020 at 2:02

Your assertion that $$\int \,J^0\,dx\,dy\,dz =\int \,\rho \, dx\,dy\,dz = 1$$ is correct. The solutions you list fullfil this condition.


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