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In chiral basis, $\psi=\begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix}$ and therefore, $\overline\psi=\psi^\dagger\gamma^0=\begin{pmatrix} \psi^\dagger_L & \psi^\dagger_R \end{pmatrix}\gamma^0=\begin{pmatrix} \psi^\dagger_R & \psi^\dagger_L \end{pmatrix}$. Hence, by matrix multiplication, $\overline \psi \psi=\psi^\dagger_R \psi_L+\psi^\dagger_L \psi_R$.

Again using chiral projection operators it can be shown that, $\overline\psi\psi=\bar\psi_R \psi_L+\bar\psi_L \psi_R$.

Therefore, these two relations suggest that, we can also write, $\overline\psi$ as $\overline\psi=\begin{pmatrix} \bar\psi_R & \bar\psi_L \end{pmatrix}$.

  1. Am I right? If yes, how can I show $\overline\psi=\begin{pmatrix} \bar\psi_R & \bar\psi_L \end{pmatrix}$starting from the $\overline\psi=\psi^\dagger\gamma^0=\begin{pmatrix} \psi^\dagger_R & \psi^\dagger_L \end{pmatrix}$? Moreover, this implies:

    $$\psi^\dagger_R \psi_L+\psi^\dagger_L \psi_R=\bar\psi_R \psi_L+\bar\psi_L \psi_R$$

  2. Am I right? If yes, how can I prove the last relation starting from either side of it?

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  • $\begingroup$ On the chance to sound ignorant, but what does $\bar{\psi}_R$ mean? Also, can you give a reference where the claim is proven? $\endgroup$
    – Nephente
    Mar 8, 2014 at 9:59
  • $\begingroup$ @nephente $\bar\psi_R=\frac{1+\gamma^5}{2}\psi$ and $\bar\psi_L=\frac{1-\gamma^5}{2}\psi$. I didn't get which claim are you talking about. Can you please mention it specifically? $\endgroup$
    – SRS
    Mar 8, 2014 at 10:04
  • $\begingroup$ Ok. I meant $\bar{\Psi}\Psi=\bar{\Psi}_R\Psi_L+\bar{\Psi}_L\Psi_R$ $\endgroup$
    – Nephente
    Mar 8, 2014 at 10:19
  • $\begingroup$ @nephente- the proof goes as follows: $\bar\psi_L\psi_R+\bar\psi_R\psi_L=\bar\psi(\frac{1+\gamma^5}{2})(\frac{1+\gamma^5}{2})\psi+\bar\psi(\frac{1-\gamma^5}{2})(\frac{1-\gamma^5}{2})\psi=\bar\psi\psi $. Intermediate steps are trivial to work out. $\endgroup$
    – SRS
    Mar 8, 2014 at 10:30

1 Answer 1

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I think you're confusion is with notation since, unfortunately, there are two common ways to denote projected spinors. One form is to write: \begin{equation} \psi \equiv \left( \begin{array}{c} \psi _L \\ \psi _R \end{array} \right) \end{equation} In this notation $ \psi _L $ and $ \psi _R $ are two component Weyl spinors. However, a second notation is also used where \begin{equation} \psi _L \equiv P _L \psi , \quad \psi _R \equiv P _R \psi \end{equation} and $ P _{L/R } $ are the projection operators. Now $ \psi _L $ and $ \psi _R $ are four component spinors with a zero value for two the components.

In the first notation (where $ \psi _{ L/ R } $ are Weyl spinors) the Dirac term takes the form, \begin{equation} m \bar{\psi} \psi = m \left( \psi _R ^\dagger \psi _L + h.c. \right) \end{equation} and in the second (where $ \psi _{ L/R} $ are four component objects) it takes the form, \begin{equation} m \bar{\psi} \psi = m \left( \overline{ \psi _R} \psi _L + h.c. \right) \end{equation} Its just a matter of convention and the end result is the same.

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  • $\begingroup$ @ JeffDror -Therefore, I think I can also write $\bar\psi=\begin{pmatrix} \bar \psi_R & \bar\psi_L\\ \end{pmatrix}$, if I consider all $\bar\psi_{L,R}$ as 4-component rows. Am I right? $\endgroup$
    – SRS
    Mar 8, 2014 at 13:35
  • $\begingroup$ @Roopam, yes that is one way to write it. Though its a bit sloppy. What we really mean is in four-component notation: $\overline{\psi}=\overline{\psi_L}+\overline{\psi_R}$ $\endgroup$
    – JeffDror
    Mar 8, 2014 at 13:51
  • $\begingroup$ @ JeffDror - Okay. I understand. Your answers are very helpful. I was indeed confused by the notations. $\endgroup$
    – SRS
    Mar 8, 2014 at 13:58
  • $\begingroup$ No problem. I'm glad you like them. $\endgroup$
    – JeffDror
    Mar 10, 2014 at 13:37
  • $\begingroup$ The notation in which $\psi_L$ and $\psi_R$ denote 2-component spinors suggest that there is a $2\times 2$chirality projection operator. However, as far as I know, there is no such thing and the chirality projection operators are always $4\times 4$ matrices $\frac{1}{2}(1\pm\gamma^5)$ @JeffDror $\endgroup$
    – SRS
    Nov 12, 2018 at 13:19

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