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

  1. 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.
    1. 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?
    2. It is usually said that the above problems are due to the Klein Gordon equation being of second order in time. Why?

Thanks everyone!

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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.

However, having not having such a current doesn't guarantee that the integral won't be time independent

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

How do we know that there aren't other combinations

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.

It is usually said that the above problems are due to the Klein Gordon equation being of second order in time. Why

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.

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  • $\begingroup$ Regarding your first answer, there is no current for the "probability density attempt $\varphi^*\varphi$. That is clear since this transforms as a scalar and not like the zeroth component of a four-vector. Why is it time-dependent? An answer mentioning a current is not satisfactory. The answer that most people have said to me is that there is no theorem (such as the one that would come from a conserved current) guaranteeing that the integral of this density is constant. However, I would then like to see an example of a solution to the KG equation in which the time dependence is seen. $\endgroup$ – Iván Mauricio Burbano Mar 19 at 18:13
  • $\begingroup$ Regarding your second answer, how do you know that there isn't a clever way to manipulate the derivatives to get some other conserved corrent? In fact there are at least four more! The spacetime invariance yields that for solutions of the KG equation $\partial_\mu (j^\nu)^\mu=0$ for $(j^\nu)^\mu=T^{\mu\nu}$. Is there something that makes the $U(1)$ symmetry special when trying to describe probability densities? $\endgroup$ – Iván Mauricio Burbano Mar 19 at 18:17
  • $\begingroup$ Regarding the third answer, I see how that may lead to a problem. However, is there an example where we may see this clearly? $\endgroup$ – Iván Mauricio Burbano Mar 19 at 18:20
  • $\begingroup$ "However, I would then like to see an example of a solution to the KG equation in which the time dependence is seen." well, any solution of the KG equation has an explicit time-dependence. You can write down the expansion in Fourier modes and take derivatives w.r.t. the time and see that the result cannot always be constrained to be zero. In the case of the Schrödinger equation the property that helps is the fact that the hamiltonian is Hermitian, whereas here the same equation just doesn't hold. $\endgroup$ – gented Mar 19 at 19:19
  • $\begingroup$ "...Is there something that makes the $U(1)$ symmetry special when trying to describe probability densities?..." that is just the definition of current density. The zero-th component can be interpreted as probability density only in that case: remember that electric charge is associated with $U(1)$ symmetry, anything else just isn't a charge. $\endgroup$ – gented Mar 19 at 19:19

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