Charge conjugation matrix and Dirac equation's solutions

[StephvsEinst]

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

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 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:

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

And what is the C matrix?

Can anyone help?

 

[StephvsEinst]

EDIT: In the quote I meant:

$$ C \mid \psi \left( \mathbf{p} \right) \rangle = \mid \psi \left( \bar{ \mathbf{p} } \right) \rangle . $$

 

[HomogenousCow]

He basically just Fourier transformed the dirac equation, try it yourself and see what you get.

 

[phyzguy]

In the Dirac equation, psi is not a four-vector. It is a spinor, and they are very different physical objects.

 

[StephvsEinst]

In the Dirac equation, psi is not a four-vector. It is a spinor, and they are very different physical objects.

I was thinking that psi could be described by two different relativistic fields, with each one having a (1 0) or (0 1) (those two states must be read in a COLUMN vector). Wouldn't this give us the four states: 2 different states of spin (up and down) of the particle and 2 different states of spin of the ANTIparticle?

 

[StephvsEinst]

He basically just Fourier transformed the dirac equation, try it yourself and see what you get.

How the hell did I not think for once of doing the Fourier transform...
Thank you, I will work on that!

 

[phyzguy]

I was thinking that psi could be described by two different relativistic fields, with each one having a (1 0) or (0 1) (those two states must be read in a COLUMN vector). Wouldn't this give us the four states: 2 different states of spin (up and down) of the particle and 2 different states of spin of the ANTIparticle?

The Dirac spin ψ does in fact have four components, but that does not make it a four-vector. A four-vector and a Dirac spinor transform very differently under coordinate transformations.