How Does the Dirac Conjugation Operator Affect Majorana Neutrino Mass Terms?

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SUMMARY

The Dirac conjugation operator, denoted as C, is represented by the matrix \( C = (i\sigma^2) = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \) when applied to the left-handed Majorana neutrino fields \( \nu_L \). The Majorana mass term is expressed as \( m_L \nu_L^T C^\dagger \nu \). A common point of confusion arises from the anticommutation properties of fermionic fields, which leads to the realization that terms like \( \nu_1 \nu_2 - \nu_2 \nu_1 \) do not equal zero due to these properties. This discussion references the Dirac chapter in Peskin's text for further clarification.

PREREQUISITES
  • Understanding of Majorana neutrinos and their mass terms
  • Familiarity with the Dirac conjugation operator and its representation
  • Knowledge of fermionic field theory and anticommutation relations
  • Basic grasp of matrix representations in quantum field theory
NEXT STEPS
  • Study the Dirac chapter in Peskin's "An Introduction to Quantum Field Theory"
  • Explore the mathematical properties of the Dirac conjugation operator
  • Research the implications of fermionic anticommutation in quantum field theory
  • Learn about Majorana mass terms in the context of neutrino physics
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Physicists, particularly those specializing in particle physics, quantum field theory, and neutrino research, will benefit from this discussion.

rkrsnan
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In books I find that the Majorana mass term for the neutrinos is given by m_L \nu_L^T C^\dagger \nu where C is Dirac Conjugation operator. How does C look like if I write \nu_L as in terms of its two components \left(\begin{array} (\nu_{L1} \\ \nu_{L2} \end{array} \right)?

Is C= (i\sigma^2) =\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)?

Thanks for your help!
 
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Yes, that looks ok to me. You can read about this in the Dirac chapter of Peskin, there is even a problem about it as I recall.
 
Thanks, the expression is correct. I was confused earlier because when I expand it I get \nu_1 \nu_2 - \nu_2 \nu_1 which I thought is zero. Then it didn't occur to me that the fields are fermionic and they anticommute.
 

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