# Is $\overline{\psi_{L}^{c}}\psi_{R}^{c}=\overline{\psi_{R}}\psi_{L}$ true for two different spin 1/2 fermions?

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?

• 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. – Melquíades Apr 19 '14 at 20:22
• 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 $\overline{\chi_{L}^{c}}\psi_{R}^{c}=\overline{\psi_{R}}\chi_{L}$. – Louis Yang Apr 19 '14 at 22:06
• Yes, that is exactly what I thought. – Melquíades Apr 19 '14 at 23:33