How Does the Charge Conjugate Dirac Field Transform in Quantum Field Theory?

[maverick280857]

Hi,

I'm trying to work my way through Halzen and Martin's section 5.4. I'd appreciate if someone could answer the following question:

How does

$$j^{\mu}_{C} = -e\psi^{T}(\gamma^{\mu})^{T}\overline{\psi}^{T}$$

become

$$j^{\mu}_{C} = -(-)e\overline{\psi}\gamma^{\mu}\psi$$

? Is there some identity I'm missing?

Thanks in advance.

-Vivek

 

[maverick280857]

Anyone?

 

[ansgar]

transpose the entire thing, then use the fact that the psi-bar contains a gamma_0 matrix

 

[maverick280857]

transpose the entire thing, then use the fact that the psi-bar contains a gamma_0 matrix

Why? If I transpose the entire thing, I get the next line without a minus sign. But why do I transpose? Not sure I follow you..

 

[ansgar]

ok I can get this:



$$-e\psi^T\gamma^0{\gamma^{\mu}}^*\psi^*$$

so performing a complex conjugation one gets

$${j_C^{\mu}}^* -e\overline{\psi}{\gamma^{\mu}}\psi$$

 

[maverick280857]

ok I can get this:



$$-e\psi^T\gamma^0{\gamma^{\mu}}^*\psi^*$$

so performing a complex conjugation one gets

$${j_C^{\mu}}^* -e\overline{\psi}{\gamma^{\mu}}\psi$$

I don't think you understand my question here. The two expressions are equal. But are you asking me to transform one to the other by performing a transpose followed by a complex conjugation (in other words asking me to take the Hermitian adjoint)? That is, to prove A = B, I should take the Hermitian adjoint of A and find it to be equal to B. Is that what you're saying?

(Do you intend to utilize the fact that the current density 4 vector is real? If so, we should merely be taking the complex conjugate.)

 

[ansgar]

can you just for completeness write down the four current and the C- transformation?

there are a couple of conventions out there you know..

 

[ansgar]

ok if you go to peskin page 70, if you have it then you can work it you I think, with Halzens definitions I have no clue sorry

 

[ansgar]

the current is a spinor scalar, it has no spinor indecies, so do a transpose in spinor space and use that the \psi's anticommute.