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I read somewhere that the Klein-Gordon equation doesn't allow for conservation of probability. Can someone prove this mathematically?

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See phy.ohiou.edu/~elster/lectures/advqm_3.pdf or there are many other easily Googlable explanations. –  John Rennie Sep 20 '12 at 10:02
    
The Klein-Gordon equation is best interpreted as a classical field theory equation, rather than a la Schroedinger equation, and this is one of the reasons why. –  alexarvanitakis Mar 5 '13 at 21:42
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up vote 3 down vote accepted

I can't comment on your question yet but in order to get this moved out of the "unanswered" bin I'm going to type up the proof given in the pdf linked by @JohnRennie. I'm not sure if my permissions are up for tagging users yet, but I'll hope he's notified somehow (I have tried tagging him and it appears not to work).

So the Klein-Gordon equation with proper units is written as: $$ \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \nabla^2 \psi + \frac{m^2 c^2}{\hbar^2} \psi = 0 $$

When we interpret the solutions $\psi(x)$ as probability amplitudes we need to have a probability density, $\rho(x)$, and current, $\vec{j} (x)$, which satisfy the continuity equation (and of course we have normalization etc., etc.)

In this case we define: $$ j^0 (x) \equiv c \rho(x) $$ and $$ j^\mu (x) \equiv ( j^0 (x), \vec{j} (x) ) $$ And as usual the continuity equation can be re-expressed as $$ \partial_\mu j^\mu (x) =0 $$ In a relativistic context like this we have: $$ j^\mu \equiv \frac{i \hbar}{2m} \psi^* \overleftrightarrow{\partial^\mu} \psi $$ where $A \overleftrightarrow{\partial^\mu} B \equiv A(\partial^\mu B) -(\partial^\mu A) B $. Considering the four-divergence of this current, $$ \partial_\mu j^\mu (x) = \frac{i \hbar}{2m} \partial_\mu \left( \psi^* \overleftrightarrow{\partial^\mu} \psi \right) $$ $$ = \frac{i \hbar}{2m} \left[ \psi^* ( \Box \psi ) - ( \Box \psi^* ) \psi \right] $$ for a $\psi$ which satisfies the Klein-Gordon eq it's clear that this gives zero.

The problem with this is current is that the density we used, $j^0$ is not positive definite. $$ \rho = \frac{1}{c} j^0 = \frac{i \hbar}{2mc} \partial_\mu \left( \psi^* \overleftrightarrow{\partial^0} \psi \right) $$ $$ = \frac{i \hbar}{2mc^2} \left[ \psi^* ( \partial_t \psi ) - ( \partial_t \psi^* ) \psi \right] $$ Which is clearly not always non-negative.

Hope this is helpful to anyone who finds this later!

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Congratulations on your first ever best answer! –  Mew Mar 5 '13 at 23:23
    
@Chris: How do you define the positive definiteness of a complex function such as $\rho$ above? Do we want it to be real and positive? –  ramanujan_dirac Dec 3 '13 at 23:18
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