Deriving the continuity equation from the Dirac equation (Relativistic Quantum)

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toam
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So I am trying to derive the continuity equation:

[tex]\frac{\partial}{\partial x^{\mu}}J^{\mu} = 0[/tex]

From the Dirac equation:

[tex]i\gamma^{\mu} \frac{\partial}{\partial x^{\mu}}\Psi - \mu\Psi = 0[/tex]

And its Hermitian adjoint:

[tex]i\frac{\partial}{\partial x^{\mu}}\overline{\Psi}\gamma^{\mu} - \mu\overline{\Psi} = 0[/tex]

Where:

[tex]\overline{\Psi}=\Psi^{+}\gamma^{0}[/tex] (Dirac conjugate)



The Attempt at a Solution


By multiplying the Dirac equation on the right by [tex]\overline{\Psi}[/tex] and the adjoint on the right by [tex]\Psi[/tex] I get:

[tex]i(\frac{\partial}{\partial x^{\mu}}(\gamma^{\mu}\Psi)\overline{\Psi} + \frac{\partial}{\partial x^{\mu}}(\overline{\Psi})\gamma^{\mu}\Psi) - \mu(\Psi\overline{\Psi} - \overline{\Psi}\Psi)=0[/tex]

The first term is basically what I am after (except I am not 100% sure I can simply apply the product rule - what is the correct order?) which means I shoudl expect the second term to go to zero:

[tex]\Psi\overline{\Psi} - \overline{\Psi}\Psi =0[/tex]

But because [tex]\gamma^{0}[/tex] is a 4x4 matrix, [tex]\Psi[/tex] is a 4x1 and [tex]\overline{\Psi}[/tex] is a 1x4, I should also expect the second term to be multiplied by the 4x4 identity matrix (so that the subtraction makes sense). However the first term is NOT a constant multiplied by the identity so I don't see how this works.



Any help would be greatly appreciated...
 
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toam said:
By multiplying the Dirac equation on the right by [tex]\overline{\Psi}[/tex] and the adjoint on the right by [tex]\Psi[/tex] …

Hi toam! :smile:

Don't you have to multiply one of them on the left? :confused:
 
I tried that and got something else that didn't work. However I will try again because I was surprised that it didn't work so I may have made a mistake or missed something obvious...
 
Ok so it turned out I had multiplied the wrong function on the left. It worked out quite simply when I fixed that. The lecture notes had erroneously shown both functions multiplied on the right.

Thanks, tiny-tim.