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Physics
Quantum Physics
Dirac's solution to the Klein-Gordon equation
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[QUOTE="stevendaryl, post: 6061622, member: 372855"] The issue that I pointed out is that the quantity ##\psi^\dagger \psi## in the Dirac equation is a nonnegative quantity whose integral is conserved. So it's a candidate for a probability density. In the case of the Klein-Gordon equation, the conserved current is not always positive, so it can't be interpreted as a probability density. I think that this is related to it's being first-order. If you have a first-order equation of the form: ##\dot{\psi} = - i H \psi##, then you can form a positive density by taking ##\rho = \psi^\dagger \psi##. Then you have ##\frac{d \rho}{dt} = -i [(\psi^\dagger H \psi) - (\psi^\dagger H \psi)^*]##. To have a conserved current, you have to be able to write ##-i [(\psi^\dagger H \psi) - (\psi^\dagger H \psi)^*]## as ##-\nabla \cdot \overrightarrow{J}## for some quantity ##\overrightarrow{J}##. You can do that with Schrodinger and with Dirac, but not with Klein-Gordon. [/QUOTE]
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Dirac's solution to the Klein-Gordon equation
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