My question is on the Klein-Gordon equation and it's relation to the continuity equation, so for a Klein-Gordon equation & continuity equation of the following form, I have attained the following probability density and probability current relations (although not normalised correctly? If that's the correct term):

[tex]-\frac{\partial^2 }{\partial t^2}\psi +\bigtriangledown ^2\psi-m^2\psi=0 \: ,\: \frac{\partial }{\partial t}\rho +\mathbf{\bigtriangledown } \cdot \mathbf{j}=0[/tex]

[tex]\rho=i(\psi^{*}\dot{\psi}-\psi\dot{\psi^{*}})\,,\; \mathbf{j} =i(\psi\bigtriangledown \psi^{*}-\psi^{*}\bigtriangledown\psi)[/tex]

We then impose conservation of probability by making the time derivative of the probability equal to zero, and the following manipulation of this derivative is shown below:

[tex]\frac{\partial P}{\partial t}=\int_{V}\frac{\partial \rho}{\partial t}\, d^3x=-\int_{V}\bigtriangledown\cdot \mathbf{j}\, d^3x=-\oint_{S}\mathbf{j}\cdot d\mathbf{s}=0[/tex]

This above equation makes use of the divergence theorem in the last pair. This is true if j=0, but does j in general need to be equal to zero? I thought 'j' would vary around the surface and so the integration of all 'j' at each part on the surface would equal zero but not necessarily all 'j' =0.

To be honest I'm quite unsure at what 'j' represents in terms of probability!

Can anyone explain this?

Thanks,

SK