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Imposing KleinGordon on Dirac Equation 
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#1
Jan3113, 01:43 PM

P: 209

Hey,
My question is on the Dirac equation, I am having a little confusion with the steps that have been taken to get from this form of the Dirac equation: [tex]i\frac{\partial \psi}{\partial t}=(i\underline{\alpha}\cdot \underline{\nabla}+\beta m)\psi[/tex] to [tex]\frac{\partial^2 \psi}{\partial t^2}=[\alpha^{i}\alpha^{j}\nabla^{i}\nabla^{j}i(\beta\alpha^{i}+\alpha^{i}\beta)m\nabla^{i}+\beta ^{2}m^{2}]\psi[/tex] I believe we are imposing the KleinGordon (maybe not) on the Dirac Equation to determine the conditions required for a free particle description via the Dirac equation, however I cannot see how this is done from those steps above. I'm not exactly sure what these mean ∇^i and alpha's^i... We are told we apply the 'operator' to both sides of the top equation  I'm not sure what operator this is  I'm guessing it's the Klein Gordon operator though. Any help would be appreciated on how to get from equation 1 to equation 2, Thanks, SK 


#2
Jan3113, 02:04 PM

P: 209

Ok I've just realised we must square it to attain the second equation, though I'm still unsure what the ∇^i's represent and ditto for the alpha's.
I'll keep having a look. 


#3
Jan3113, 02:29 PM

P: 209

Actually I've figured it now, it's just an index to sum over, I think!
SK 


#4
Feb113, 12:36 AM

P: 1,020

Imposing KleinGordon on Dirac Equation
https://docs.google.com/viewer?a=v&q...wZSsogrMAyVO0g
it is treated in last chapter of baym'quantum mechanics'. 


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