This is a two part question.

  1. What is the definition of $\bar\psi$ in QCD?

In QED I know that $\bar\psi=\psi^\dagger\gamma^0$, but in QCD we also have flavor and/or color space. In particular, I'm reading Klevansky's review of the Nambu–Jona Lasinio model of QCD, in which the Lagrangian and several symmetry transformations seem to invole the Pauli isospin matrices $\tau$.

  1. How can I prove the following symmetry transformation, given in Eq. (2.8) of the source cited above?

$$\psi\to e^{-i(\tau\cdot\theta)\gamma_5/2}\psi \Rightarrow (\bar\psi\psi)\to (\bar\psi\psi)\cos\theta-(\bar\psi i\gamma_5\tau\cdot\hat\theta)\sin\theta.$$

Obviously (1) is necessary in order to answer (2), but at any rate I'm confused as to how this works.


In a NJL model with two flavors $(u,d)$, the field $\psi$ is defined as $$\psi=\begin{pmatrix} u \\ d \end{pmatrix},$$ being $u$ and $d$ ordinary Dirac spinors. This means that, in your transformation, the SU(2) part applies to $\psi$ while $\gamma_5$ goes on the single Dirac spinors. Remebering that $\gamma_5^2=I$, you have $$ e^{-i{\mathbf\tau}\cdot{\mathbf\theta}\gamma_5/2}=\cos\left(\frac{|{\mathbf\theta}|}{2}\right)-i\frac{{\mathbf\tau}\cdot{\mathbf\theta}}{|{\mathbf\theta}|}\gamma_5\sin\left(\frac{|{\mathbf\theta}|}{2}\right) $$ that yields $$ e^{-i{\mathbf\tau}\cdot{\mathbf\theta}\gamma_5/2}\psi=\cos\left(\frac{|{\mathbf\theta}|}{2}\right)\begin{pmatrix} u \\ d \end{pmatrix}-i\frac{{\mathbf\tau}\cdot{\mathbf\theta}}{|{\mathbf\theta}|}\sin\left(\frac{|{\mathbf\theta}|}{2}\right)\begin{pmatrix} \gamma_5u \\ \gamma_5d \end{pmatrix}. $$

  • $\begingroup$ As for my question 1, is $\bar\psi$ then the transpose of $\psi$ in flavor space, with $u$ and $d$ also becoming $u^\dagger\gamma^0$ and $d^\dagger\gamma^0$? The multiplication by $\gamma^0$ for the spinor part makes me wonder whether there is also a multiplication by some $\tau$ matrix for the flavor part. $\endgroup$ – WillG Jan 10 '20 at 17:27
  • $\begingroup$ You are right and this is the interesting part as $\gamma_5$ anticommutes with $\gamma_0$ and this gives exactly the transformation you get for $\bar\psi\psi$. $\endgroup$ – Jon Jan 10 '20 at 21:01
  • $\begingroup$ Coming back to this (weeks later), I'm still confused about how the $\tau\cdot\theta$ turns into $|\theta|/2$. I have been assuming $\tau\cdot\theta=\theta_1\tau_1+\theta_2\tau_2+\theta_3\tau_3$ for arbitrary real parameters $\theta_i$. Is this true? If so, how does the $\tau\cdot\theta$ disappear from the $\cos$ term? $\endgroup$ – WillG Feb 5 '20 at 20:47
  • $\begingroup$ Never mind, now I see it: commutation relations for Pauli matrices show that $(\tau\cdot\theta)^2 = |\theta|^2$, and that does the trick. $\endgroup$ – WillG Feb 5 '20 at 21:10

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