Consider the Klein-Gordon equation $$(\square+m^2)\varphi=0.$$

- People usually claim that $\varphi^* \varphi$ cannot be interpreted as a probability density because $\int d^3\vec{x}\varphi(t,\vec{x})^*\varphi(t,\vec{x})$ is not time independent. I understand that indeed there is no current $j^\mu$ satisfying the continuity equation $\partial_\mu j^\mu=0$ for which $j^0=\varphi^*\varphi$. That would guarantee the time invariance of our integral. However, having not having such a current doesn't guarantee that the integral won't be time independent. Does any one know of a better explanation for the time dependence. A good example would certainly clarify.
- People claim that there is a current, namely $j^\mu:=\varphi^*\partial^\mu\varphi-\varphi\partial^\mu\varphi^*$ which does satisfy the continuity equation. However, $j^0$ is not positive definite (again, an example would be nice). How do we know that there aren't other combinations of $\varphi$ which may lead to conserved currents that have a positive definite zeroth component?
- It is usually said that the above problems are due to the Klein Gordon equation being of second order in time. Why?

Thanks everyone!

## Best Answer

The main problem with the Klein-Gordon equation is that the probability density cannot be made to be positive-defined indefinitely.

The general solution can be written in mode expansions satisfying the mass-shell condition $k_{\mu}k^{\mu}=m^2$: one can show that the $j^0$ component cannot be constrained to always be positive because you have two initial conditions that you must allow for, namely the choice of initial position and velocity, which follows by the equation being of second order. However you want to prepare your initial state, the evolution will always bring it to another state where the density may be negative.

The point is really just that the current oscillates between positive and negative values on the mass-shell $k_{\mu}k^{\mu}=m^2$.

Because there are not. The current is nothing but re-writing the original equation so that the derivative appears in front, namely $$ \partial_{\mu} ( \cdot ) = 0 $$ with the quantity in parentheses needing to vanish if the original equation holds. You can try manipulating the derivatives left and right but there is not much more you will get.

As stated, a second order differential equation leaves you with two arbitrary initial conditions: the consequence of this arbitrarity is that you cannot force them to always combine so that the density is positive everywhere.