Problem involving Dirac's equation 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.
 A: 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.
A: 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.
A: 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.
