Alright. So the Dirac Eq is(adsbygoogle = window.adsbygoogle || []).push({});

[tex] (i \gamma^{\mu} \partial_{\mu} - m) \psi = 0 [/tex]

or putting the time part on one side with everything else on the other and multiplying by [itex] \gamma^0 [/itex],

[tex] i \partial_t \psi = (i \gamma^0 \vec{\gamma} \cdot \nabla + \gamma^0 m) \psi [/tex]

I would think that this is the Dirac Hamiltonian, but everywhere (including Peskin/Schroeder, p 52) seems to say that its this

[tex] \hat{H}_D = -i\gamma^0 \vec{\gamma} \cdot \nabla + \gamma^0 m [/tex]

So to be more careful, I tried starting with the Lagrangian density, a la Peskin/Schroeder, and got (using the Lagrangian from Peskin)

[tex] L = \bar{\psi}(i \gamma^{\mu} \partial_{\mu} - m) \psi [/tex]

Then [itex] \pi = i \psi^{\dagger} [/itex] so that the Hamiltonian density is

[tex] H = \pi \dot{\psi} - L = i \psi^{\dagger} \dot{\psi} - ( \psi^{\dagger} \gamma^0 (i \gamma^0 \dot{\psi} - [i\vec{\gamma} \cdot \nabla + m]\psi)) = + \psi^{\dagger} [i \gamma^0 \vec{\gamma} \cdot \nabla + \gamma^0 m]\psi [/tex]

Anyone see where am I screwing up the sign? Is it from having the sign in front of the mass flipped when you do the "other" Dirac eq (from the fact that it satisfies Klein Gordon)?

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# Brain freeze on Dirac EQ v. Dirac Hamiltonian

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