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The dirac equation;$$(i\gamma^\mu\partial_{\mu} - m)\psi=0 $$ is just; $$(i\gamma^{0}\partial_{0} - i\gamma^{i}\partial_{i} - m)\psi=0 $$ in a (+,---) metric right?

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2 Answers 2

The expression $A^{\mu}B_{\mu}$ simply means that $$A^{\mu}B_{\mu}=A^{0}B_{0}+A^{1}B_{1}+A^{2}B_{2}+A^{3}B_{3}$$

Using the Minkowski metric with signature $(+---)$ you write this as

$$A^{\mu}B_{\mu}=A^{\mu}\eta_{\mu\nu}B^{\nu}=A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}$$

The metric simply tells you have how the components of a vector and its dual vector (covariant vector) are associated. With this specific metric and signature we have $B_0=B^0$ and $B_i=-B^i$, $i=1,2,3$.

In Dirac equation we have

$$\gamma^{\mu}\partial_{\mu}=\gamma^{0}\partial_{0}+\gamma^{1}\partial_{1}+\gamma^{2}\partial_{2}+\gamma^{3}\partial_{3}=\gamma^{0}\partial_{0}+\gamma^{i}\partial_{i}=\\ =\gamma^{\mu}\eta_{\mu\nu}\partial^{\nu}=\gamma^{0}\partial^{0}-\sum_{i=1}^{3}\gamma^{i}\partial^{i}$$

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

$$%(i\gamma^{0}\partial^{0} - i\sum_{i=1}^{3}\gamma^{i}\partial^{i} - m)\psi=0 $$

or

$$(i\gamma^{0}\partial^{0} - i\sum_{i=1}^{3}\gamma^{i}\partial^{i} - m)\psi=0 $$

$$\partial_0=\partial^0=\frac{\partial}{\partial t},\hspace{0.2cm}\partial_i=-\partial^i=\vec{\nabla}_i$$

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Thx guys, is it fair when deriving $\mathcal{H}$ to set it equal to $+\mathcal{L}-\frac{\partial \mathcal{L}}{\partial\dot{\psi}}\dot{\psi}$ to say that the lag density $\rightarrow 0$ so $\mathcal{H}= i\psi^{\dagger}\dot{\psi}$? –  user21119 Mar 19 '13 at 19:54
    
what is the significance of rising and lowering the indices of gamma matrices? isn't it $\gamma_i=\gamma^i$; $i=1,2,3$ –  Ome Nov 6 '13 at 13:57

No, it's with a plus. You don't need to change the sign unless both indices are upper (or lower).

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