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

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SUMMARY

The transformation of the charge conjugate Dirac field in Quantum Field Theory is addressed through the equation j^{\mu}_{C} = -e\psi^{T}(\gamma^{\mu})^{T}\overline{\psi}^{T}, which simplifies to j^{\mu}_{C} = -(-)e\overline{\psi}\gamma^{\mu}\psi. The key steps involve transposing the entire expression and applying the properties of the gamma matrices, particularly the gamma_0 matrix. The discussion emphasizes the importance of recognizing that the current density 4-vector is real, necessitating a complex conjugation to establish the equivalence of the two expressions.

PREREQUISITES
  • Understanding of Dirac spinors and their properties
  • Familiarity with gamma matrices in Quantum Field Theory
  • Knowledge of Hermitian adjoints and complex conjugation
  • Basic concepts of charge conjugation in particle physics
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  • Study the properties of gamma matrices in detail, particularly their transposition and complex conjugation
  • Learn about the implications of charge conjugation on Dirac fields
  • Explore the derivation of the four-current in Quantum Field Theory
  • Review Halzen and Martin's Quantum Field Theory text, focusing on section 5.4 for deeper insights
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Quantum physicists, particle physicists, and students studying Quantum Field Theory who are looking to understand the transformation properties of Dirac fields and charge conjugation.

maverick280857
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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
 
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Anyone?
 
transpose the entire thing, then use the fact that the psi-bar contains a gamma_0 matrix
 
ansgar said:
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..
 
ok I can get this:



<br /> -e\psi^T\gamma^0{\gamma^{\mu}}^*\psi^*<br />

so performing a complex conjugation one gets

<br /> {j_C^{\mu}}^* -e\overline{\psi}{\gamma^{\mu}}\psi<br />
 
Last edited:
ansgar said:
ok I can get this:



<br /> -e\psi^T\gamma^0{\gamma^{\mu}}^*\psi^*<br />

so performing a complex conjugation one gets

<br /> {j_C^{\mu}}^* -e\overline{\psi}{\gamma^{\mu}}\psi<br />

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

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

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