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I'm stuck in an equation derivation of Ryder's QFT book.

Starting with Dirac's equation:

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

If I multiply by $i\gamma^\nu\partial_\nu$, I get:

$$((\gamma^\nu\partial_\nu)(\gamma^\mu\partial_\mu)+i\gamma^\nu\partial_\nu m)\psi=0$$

I should get:

$$(\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu+m^2)\psi=0$$

I suppose that this means:

$$i\gamma^\nu\partial_\nu=m$$

But I don't know why. ¿Could anyone show me the way to prove this property?

Thanks.

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Multiply by $i \gamma^\nu \partial_\nu + m$. –  user1504 May 2 '13 at 23:59

3 Answers 3

up vote 4 down vote accepted

First, to get the equation you want apply $(i\gamma^\nu\partial_\nu + m)$ to both sides, then on the left hand side you'll get \begin{align} (i\gamma^\nu\partial_\nu + m)(i\gamma^\mu\partial_\mu - m)\psi &= (-\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu-m^2)\psi \end{align} which, when set to zero, gives $$ (\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu+m^2)\psi = 0 $$ as desired. The expression you wrote down is missing an $i$; if you apply $i\gamma^\nu\partial_\nu$ to both sides, then you get $$ (\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu + im\gamma^\nu\partial_\nu)\psi = 0 $$ which combined with the equation you wanted to get gives $$ i\gamma^\nu\partial_\nu\psi = m\psi $$ which is fine because this is just the Dirac equation again. You cannot conclude from this that $i\gamma^\nu\partial_\nu = m$; this is only true when acting on solutions to the Dirac equation, not as a statement about differential operators.

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Hey man, hope you don't mind me sticking the Clifford algebra bit on the end! I'm sure it must complete the exercise in Ryder? –  innisfree May 3 '13 at 0:13
    
@innisfree I feel like that's a bit too unrelated to the original question, but it's a cool observation; maybe you should add a response with that in it? –  joshphysics May 3 '13 at 0:18
    
sure, i guess it is unrelated to the question, though i'm pretty sure (without having read the book!) that that is where the calculation is going... –  innisfree May 3 '13 at 0:23
    
@innisfree You're right, but I didn't find it necessary to add it to the question. –  jinawee May 3 '13 at 0:38
    

Once you have shown, $$ (\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu + m^2)\psi = 0 $$ You can replace $\gamma^\nu\gamma^\mu$ with its symmetric part - because $\partial_\nu\partial_\mu$ is symmetric in $\mu\nu$, the antisymmetric part of $\gamma^\nu\gamma^\mu$ does not contribute. $$ (1/2\{\gamma^\nu,\gamma^\mu\}\partial_\nu\partial_\mu + m^2)\psi = 0 $$ You can then recognize by comparison with the KG equation that $$ \{\gamma^\nu,\gamma^\mu\} = 2 g^{\mu\nu} $$ i.e. you have shown that the gamma matrices satisfy a Clifford algebra.

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You cannot write

$$i\gamma^\nu\partial_\nu=m$$

in general. It is only when they are fed a solution $\psi$ that they have equal outputs.

$$i\gamma^\nu\partial_\nu\psi=m\psi$$

What would happen if for $\psi$ you jammed in some arbitrary Dirac spinor-valued function? You do not get equality except by luck or when $\psi$ describes an electron or positron in free space with the correct relations among the four complex components.

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