In the context of seesaw mechanism or Dirac and Majorana mass terms, one often see the following identity $$ \overline{\psi_{L}^{c}}\psi_{R}^{c}=\overline{\psi_{R}}\psi_{L}. $$

Here, I am using 4 components notation in the chiral basis. The convention for the charge conjugation is $\psi^{c}=-i\gamma^{2}\psi^{*}$, and $\psi_{L}^{c}=\left(\psi_{L}\right)^{c}$. The following is my effort of proving it. $$ \overline{\psi_{L}^{c}}\psi_{R}^{c}=\overline{i\gamma^{2}\psi_{L}^{*}}i\gamma^{2}\psi_{R}^{*}=\left(i\gamma^{2}\psi_{L}^{*}\right)^{+}\gamma^{0}i\gamma^{2}\psi_{R}^{*}=\psi_{L}^{T}i\gamma^{2}\gamma^{0}i\gamma^{2}\psi_{R}^{*} $$ $$ =-\psi_{L}^{T}\gamma^{2}\gamma^{0}\gamma^{2}\psi_{R}^{*}=\psi_{L}^{T}\gamma^{2}\gamma^{2}\gamma^{0}\psi_{R}^{*}=-\psi_{L}^{T}\gamma^{0}\psi_{R}^{*}=-\psi_{L,i}\gamma_{ij}^{0}\psi_{R,j}^{*}. $$ Now if $\psi_{L,i}$ and $\psi_{R,i}$ are anticommuting, then one have $$ -\psi_{L,i}\gamma_{ij}^{0}\psi_{R,j}^{*}=\psi_{R,j}^{*}\gamma_{ji}^{0}\psi_{L,i}=\overline{\psi_{R}}\psi_{L}. $$ Question:

Is the anticommuting assumption still true if $\psi_{R}$ and $\psi_{L}$ are two different species of fermion? (For example, $\psi_{L}=\chi_{L}$)

Do we assume any two fermions are anticommuting even if they are two different fields in QFT?

  • $\begingroup$ What do you mean by $\psi_L=\chi_L$? Two different fermionic fields always commute. Here, $\psi_R$ and $\psi_L$ are just different projections of the same field. $\endgroup$ Apr 19, 2014 at 20:22
  • $\begingroup$ By $\psi_{L}=\chi_{L}$, I mean to replace all the $\psi_{L}$ by $\chi_{L}$ in the above derivation. So now $\chi_{L}$ and $\psi_{R}$ are two different Weyl spinors. They are not just the projection of a single Dirac fermion. I guess your answer is that if $\chi_{L}$ and $\psi_{R}$ are two different fields, then they commute. So we will have $\overline{\chi_{L}^{c}}\psi_{R}^{c}=-\overline{\psi_{R}}\chi_{L}$ instead of $\overline{\chi_{L}^{c}}\psi_{R}^{c}=\overline{\psi_{R}}\chi_{L}$. $\endgroup$
    – Louis Yang
    Apr 19, 2014 at 22:06
  • $\begingroup$ Yes, that is exactly what I thought. $\endgroup$ Apr 19, 2014 at 23:33


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