# Probability density of Klein-Gordon equation

This may, perhaps, stir some healthy debate; at least I am having some "fun" thinking about it, hopefully I can solicit some outside views too.

It is often regarded that the Klein-Gordon equation does not have a positive definate probability density. In short the argument is that the density in the Schrodinger equation is $\rho=\psi^*\psi$. And the continuity relation is $\nabla\cdot\vec{J}-\partial_t \rho=0$, where $J\equiv$ probabilty current density. Fair enough.

As we go to the relativistic version "it is said" that the $\rho$ in the above must be extended to $\rho\rightarrow \psi^*\partial_t\psi-\psi\partial_t\psi^*$, and is no longer positive definate because we are free to pick initial values for each $\psi$ and $\partial \psi$ in the above.

Here is why the above argument drives me nuts. There is the viewpoint that:

They have simply transformed the expression for the density in error. Sure, it is true that the thing they have decided to label as the new $\rho$ is clearly not positive definate. I feel it is appropriate to entertain the view that this is because the thing they have chosen to reinvent $\rho$ as is simply NOT the probablity density anymore. Who decided to call it hat in the first place?

The continuity equation is really just a statement that ${D\over d{\lambda}}\rho=0$...conservation of probablity density. This is identical to the equation $"v"\cdot\nabla \rho=0$ or if you choose $\nabla\cdot\rho"v"=0$ , where $\nabla$ now pertains to a four-operator. The 'v' in quotation marks corresponds with (though perhaps not literally) proper 4 velocity ala $dx\over d\lambda$.

In short DON'T redefine $\rho$. We continue to have$\rho=\psi*\psi$ to be the case. No need to not call the ford a ferrari when it gets us around just fine. Now consider $"v^0"(\rho)\rightarrow \psi^*v^0\psi+\psi v^0\psi^*\leftrightarrow \psi^*\partial_t\psi-\psi\partial_t\psi^*$. the quantity they were attempting to call $\rho$ earlier.

The continuity equation remains satisfied here. And, there appears to be no obvious reason why $\psi^*\psi$ cannot be considered positive definite under the previous rational. $"v^0"\psi*\psi=J^0$ is not positive definite, but $\rho\neq J^0$ anymore; $\rho=\psi^*\psi$.

Please comment, tell me why you disagree, agree, indifferent, etc. I won't argue much, may add comment, clarify if wanted. I will be interested

• I have changed the title to better reflect your question. If you do not agree, then (of course) you can change it to something else. – Hunter Jun 1 '14 at 21:44
• It's difficult for me to understand what you're trying to say. For example, "The continuity equation is really just a statement that $\frac{D}{d\lambda}ρ=0$..." - what is this? Where did you get that? What is $\lambda$? And in general, as far as I know, $\psi^*\psi$ is not conserved under Klein-Gordon. – akhmeteli Jun 1 '14 at 23:40
• I didn't understand fully your your question, but it seems that you wish to use $\rho=\psi^* \psi$ and try to make sense out of it. Well, the problem with this idea in the case of Klein-Gordan Equation is that you end up with something that is not a continuity equation, so, if you try to use it as a probability density, you end up with something that have non-constant normalization, and, thus, can't be used to define a probability distribution. One way or the other, you end up with problems. – Hydro Guy Jun 2 '14 at 1:24
• $D\over{d\lambda}$ is (proportionally) the full proper time derivative of a quantity in this case $\rho\equiv\psi^*\psi$ is the density. It is invariant. Note the equivalence of the operations $D\over d\lambda$ with $u\cdot\nabla$. $\nabla$ is the 4-space del. There is more discussion in books such as Misner, Thorne, and Wheeler 'Gravitation' in some sections , but the operations are simple. Note: $\dot{x^0}\partial_t+\dot{x^1}\partial x_1+...\rightarrow {d\over d\lambda}$ use chain rule to verify. $\lambda$ may loosely be thought of as proportional to proper time if desired. – IntuitivePhysics Jun 2 '14 at 19:58