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## Main Question or Discussion Point

I saw this somewhere but I think it is wrong...

$$ ( i \gamma \cdot \mathbf{p} + m ) \psi ( \mathbf{p} ) = 0. $$

And what is the C matrix?

Can anyone help?

I already read Griffiths' "Introduction to Particle Physics" (the 1st edition) from the page 216 to the page 222 (chapter of Quantum Electrodynamics - section "Solution to the Dirac Equation") and I didn't understood why was there the imaginary number in the equation:Free fermions are solutions of Dirac's equation:

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

where $$ \psi $$ is a four vector. Working the equation in terms of $$p_ \mu : $$

$$ ( i \gamma \cdot \mathbf{p} + m ) \psi ( \mathbf{p} ) = 0, $$

because $$p_\mu \rightarrow i \hbar \partial_\mu $$.

There is a C matrix that when applied to a state $$ \psi (\mathbf{p} ), $$ we have:

$$C \langle \psi ( \mathbf{p} ) \mid = \mid \psi ( \bar{ \mathbf{p} } ) \rangle ,$$

i.e., the particle turns into the antiparticle.

If we apply C matrix to the field we obtain:

$$ C ( i \gamma \cdot \mathbf{p} + m )^{T} \psi^{T} ( \mathbf{p} ) = ( -i \gamma \cdot \mathbf{p} + m )C \psi^{T} ( \mathbf{p} ) . $$

Introducing a field $$ \psi^c ( \mathbf{p} )$$, and applying it in Dirac's equation we have:

$$ ( i \gamma \cdot \mathbf{p} + m ) \psi^c ( \mathbf{p} ) = 0. $$

$$ ( i \gamma \cdot \mathbf{p} + m ) \psi ( \mathbf{p} ) = 0. $$

And what is the C matrix?

Can anyone help?