Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.