The pseudo-scalar Yukawa theory Lagrangian is

$$\mathcal{L}=\bar{\psi}(i\gamma ^\mu \partial_\mu - m)\psi -g\bar{\psi}i\gamma^5\phi\psi,$$ where $g$ is a coupling constant. How can I show it is invariant under a chiral transformation, $\psi\to e^{i\lambda \gamma_5}\psi$?

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    $\begingroup$ Um...plug the transformation in and see what happens? $\endgroup$
    – ACuriousMind
    Nov 2, 2014 at 16:08

1 Answer 1


(This is a largely a response prompted by your comment.)

You can get the answer by just remembering the commutation/anti-commutation properties of the $\gamma$ matrices, and the fact that ${\bar \psi} = \psi^{\dagger} \gamma^0$. To see the following, you would have to expand the exponential factor, up to linear order $e^{M} = I + M + \ldots$.

(I'm not going to do your homework, this is just a guide!)

1) The kinematic term $i \bar{\psi}\gamma ^\mu \partial_\mu\psi$ goes into itself, using $ \{\gamma^{\mu}, \gamma^5\} = 0$.

2) The Yukawa coupling term follows suite.

3) There is no such cancellation in the mass term $m \bar \psi \psi$, but the two factors reinforce each other. This term picks up an overall factor of $e^{2i\lambda \gamma_5}\psi$, i.e. two times either factor. Thus, the mass term is not invariant under this transformation, and breaks chiral symmetry.

B.T.W. This transformation is called the axial-vector transformation, since the corresponding conserved (in the m=0 limit) Noether current transforms like an axialvector $\bar \psi \gamma^{\mu} \gamma^5 \psi$.

Resolving into Weyl spinors $\psi_{L,R} = (1\mp \gamma^5)\psi/2$ is an alternative way of seeing this. With this, you will again have to use the $\gamma$ matrices' properties, and you will arrive at the result that only the mass term mixes up the two chiralities, i.e. becomes $m (\bar \psi_L \psi_R + \bar \psi_R \psi_L)$. The kinematic term would transform into $i \bar{\psi_L}\gamma ^\mu \partial_\mu\psi_L + i \bar{\psi_R}\gamma ^\mu \partial_\mu\psi_R$ and hence, it is like the the kinematic terms of two independent Lagrangians added up. No mixing. The two formulations are absolutely equivalent.

  • $\begingroup$ Thanks for your very clear guidance. I have a question though, while I am performing this transformation on Yukawa coupling term: I get the following:$$L'=-g\bar{\psi}i\gamma^5\phi\psi = -gi\bar{\psi}e^{i\lambda \gamma^5}\gamma^5 \phi e^{i\lambda\gamma^5}\psi$$ Why aren't the exponentials cancelling so it would be invariant.. $\endgroup$ Nov 2, 2014 at 19:09
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    $\begingroup$ In one of your exponentials there should be a minus sign, because $\bar{\psi}$ is a conjugate of $\psi$. Also, $\gamma^5$ commutes with its exponential (because it commutes with all terms in its Tailor expansion). So the two exponentials cancel each other. $\endgroup$ Nov 2, 2014 at 20:31
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    $\begingroup$ @Fluctuations no, it is not true. But for every matrix $x$ (in your case, $x=i\gamma^5$), the following holds: $$ \left[ x, \exp x \right] = 0 $$ $\endgroup$ Nov 2, 2014 at 21:38
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    $\begingroup$ Hindsight did most of the follow-up job for me, so thanks. Regarding the last point, @Fluctuations, as Hindsight already mentioned, they don't anticommute, they commute! To see this explicitly, expand the exponential in $$ [ x, \exp x ]$$ and exploit the linearity $$ [ x, (y+z) ] = [ x, y ] + [ x, z ]$$. Clearly, $x$ commutes with identity, with $x$, with $x^2$, and so on. :) $\endgroup$
    – 299792458
    Nov 3, 2014 at 5:10
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    $\begingroup$ And @Hindsight, thanks for the follow-up job. Yes, chiral symmetry holds only in the $m=0$ limit. Finite mass explicitly breaks chiral symmetry. But if these mass terms are small, like $u$ and $d$ quarks (the $SU(2)$ case), chiral symmetry can be considered an approximate symmetry of the strong interactions. :) $\endgroup$
    – 299792458
    Nov 3, 2014 at 5:16

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