## Brain freeze on Dirac EQ v. Dirac Hamiltonian

Alright. So the Dirac Eq is

$$(i \gamma^{\mu} \partial_{\mu} - m) \psi = 0$$

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

$$i \partial_t \psi = (i \gamma^0 \vec{\gamma} \cdot \nabla + \gamma^0 m) \psi$$

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

$$\hat{H}_D = -i\gamma^0 \vec{\gamma} \cdot \nabla + \gamma^0 m$$

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

$$L = \bar{\psi}(i \gamma^{\mu} \partial_{\mu} - m) \psi$$

Then $\pi = i \psi^{\dagger}$ so that the Hamiltonian density is

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

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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 -+++ metric ? time having a - when seperated out?
 Nah, its the standard +--- metric.

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 I thought I did take that into account. Perhaps I should've include more steps between eq 1 and 2, so here they are: Start w/ dirac eq $$(i \gamma^{\mu} \partial_{\mu} - m) \psi = (i \gamma^0 \partial_t -i \vec{\gamma} \cdot \nabla - m) \psi = 0$$ Keep time derivative on the left, move everything else on the right: $$i \gamma^0 \partial_t \psi = (i \vec{\gamma} \cdot \nabla + m) \psi$$ Multiply by $\gamma^0$ and use $(\gamma^0)^2 = \eta^{00} =1$ to get $$i \partial_t \psi = (i \gamma^0 \vec{\gamma} \cdot \nabla + \gamma^ 0 m) \psi$$ Still don't see my mistake. Its probably going to be one of those things where I slap myself repeatedly for being so silly when it gets sorted out.
 Wait a minute... When the Hamiltonian is written as $$\hat{H}_D = -i\gamma^0 \gamma \cdot \nabla + \gamma^0 m$$ then $\gamma \cdot \nabla$ is a four vector product?? So that $\gamma \cdot \nabla = -\vec{\gamma} \cdot \vec{\nabla}$. I think that's it. Is that what you meant, Bill? So. Much. Rage. Thanks Bill :) I feel so silly now. Always with the notation. Sheesh!
 You should notice that $${\gamma ^\mu }{\partial _\mu } = {\gamma ^0}{\partial _0} + {\bf{\gamma }} \cdot \nabla$$ while $${x^\mu }{y_\mu } = {g_{\mu \nu }}{x^\mu }{y^\nu } = {x^0}{y^0} - {\bf{x}} \cdot {\bf{y}} .$$ There is no metric tensor in the four vector product $\gamma^\mu \partial _\mu$.
 dazhuzai8 is correct. $$\partial_\mu = (\partial_t, \partial_x)$$ whereas for other four-vectors, $$A_\mu=(A^0,-A^i)$$. You can include a metric tensor: $$g_{\mu\nu}\gamma^\mu \partial^\nu$$ but remember now: $$\partial^\nu = (\partial_t, -\partial_x)$$