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Following the notation used in David J. Griffiths Introduction to Elementary Particles, recall that $a^{\mu}b_{\mu}:=a^{0}b^{0}-a^{1}b^{1}-a^{2}b^{2} - a^{3}b^{3}$. Recall the relativistic Dirac equation $$ i\hbar\gamma^{\mu}\partial_{\mu}\psi = mc \psi $$ In this case, our notational definition above does not hold, in that $$ \gamma^{\mu}\partial_{\mu} = \gamma^{0}\partial_{0} + \gamma^{1}\partial_{1} + \gamma^{2}\partial_{2} + \gamma^{3}\partial_{3}. $$ Why is this?

Note that the Dirac equation as given above derives from $$ \gamma^{\mu}p_{\mu}\psi = mc \psi $$ where the convention above does hold, in that $ \gamma^{\mu}p_{\mu} = \gamma^{0}p^{0} - \gamma^{1}p^{1}-\gamma^{2}p^{2}-\gamma^{3}p^{3} $ then the quantum substitution $p_{\mu}\to i\hbar\partial_{\mu}$ (is this right?) is given.

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2 Answers 2

up vote 1 down vote accepted

That is correct if you remember that

$${\bf p}=-i\nabla$$

and

$$p_0=i\partial_t.$$

Then, if you put this in $\gamma^{0}p_{0} - \gamma^{1}p_{1} - \gamma^{2}p_{2} - \gamma^{3}p_{3}$ you will get immediately the result that is also in agreement with the definition of $\partial_\mu$, that is $\partial_\mu=(\partial_t,-\nabla)$.

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Thank you so much! –  Nick Thompson Feb 4 '12 at 20:11

This is all explained in chapter 3.2, notably equation 3.16 $x_{\mu}=g_{\mu\nu} x^{\nu}$, where $g_{\mu\nu}$ is the Minkowski metric.

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Ah, thank you so much! That's exactly what I had missed. If you don't mind me asking, is $p_{\mu} \to i\hbar\partial_{\mu}$ (eq. 7.5) a typo? I was under the impression that (say) $p_{1} \to -i\hbar\partial_{1}$. –  Nick Thompson Feb 4 '12 at 19:53

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