In Srednicki's text book the axial symmetry is described as invariance under,

$$\Psi (x) \rightarrow e^{-i\alpha (x) \gamma_5}\Psi(x)$$ and $$\bar \Psi (x) \rightarrow \bar \Psi e^{-i\alpha (x) \gamma_5}$$

Where as wikipedia describes it as,

$$\psi_L\rightarrow e^{i\theta_L}\psi_L$$ $$\psi_R\rightarrow \psi_R$$ or $$\psi_L\rightarrow \psi_L$$ $$\psi_R\rightarrow e^{i\theta_R}\psi_R$$

To begin with one of these seems to be local and the other global, but would the local and the global symmetry behave differently? Do they both have an axial anomaly?

I have some vague notion that particles and antiparticles can be composed of left and right handed fields, something like; $$ \Phi(x) = \begin{pmatrix} \phi_R\\ \phi_L \end{pmatrix} $$ But I cannot actually find it written like this. If that were true it might be possible to find the two definitions somewhat equivalent. Are they equivalent? Am I working in the right direction?

Edit; my confusion mostly stems from having confused adjoint representation $\bar \psi$ with some sort of antiparticle state. I guess this was such an odd mistake to make that most people answering assumed I couldn't have gotten that wrong.


1 Answer 1


The axial symmetry is usually taken to be something global. Since $\gamma_5 \psi_R=\psi_R$ and $\gamma_5 \psi_L=-\psi_L$, the expression $\Psi \rightarrow e^{-i\alpha \gamma_5}\Psi$ means $$\psi_R\rightarrow e^{-i\alpha}\psi_R$$ $$\psi_L\rightarrow e^{+i\alpha}\psi_L$$ Since we also have vector symmetry where both L and R transform the same way, we can apply a global phase rotation to both to get $$\psi_R\rightarrow \psi_R$$ $$\psi_L\rightarrow e^{+2i\alpha}\psi_L$$ Since $\alpha$ is arbitrary, and we could equally well have isolated $\psi_R$ instead, this is equivalent to the wikipedia description.

  • $\begingroup$ Fab, do you know why Srednicki is defining it in terms of particles and antiparticles? $\endgroup$
    – Clumsy cat
    Commented Mar 20, 2017 at 18:20
  • $\begingroup$ If you mean the relation $\bar{\Psi}\rightarrow\bar{\Psi}e^{-i\alpha\gamma_5}$ that's just a consequence of your first line, it's nothing new. Taking the complex conjugate flips the sign in the exponent, but then $\gamma_5$ anticommutes with the $\gamma^0$ in the barred field so the sign flips back. $\endgroup$
    – octonion
    Commented Mar 20, 2017 at 18:25
  • $\begingroup$ but are the bared and unbared fields not fields and anti fields, I know I should know this but can I relate them to left and right handed fields? $\endgroup$
    – Clumsy cat
    Commented Mar 20, 2017 at 20:00
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    $\begingroup$ @TheoreticalPerson No, if you write out the usual expansion for a Dirac field, it's something like $$\psi = \int d^3 \mathbf p \sum_r \, C_\mathbf{k} \big(c_r(\mathbf p) u_r(\mathbf p) \exp(-ipx) + d^\dagger_r (\mathbf p) v_r(\mathbf p) \exp(ipx) \big) $$ where $c$ annihilates particles and $d^\dagger$ creates particles. The expansion of $\overline{\psi}$ has $c^\dagger$ that creates particles and $d$ that annihilates anti-particles. You can realize that it has to be like this because if particles have charge $q$, anti-particles have charge $-q$, so the operator that absorbs a particle must $\endgroup$ Commented Mar 20, 2017 at 21:16
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    $\begingroup$ have the same charge as the operator that creates an anti-particle. (Subtracting one $q$ unit vs. adding one $-q$ unit.) By charge, I mean the thing that appears in gauge transformations: $\psi \mapsto \exp(iq\varphi) \psi$. So if you were to try to construct the field out of $c$ and $c^\dagger$, it wouldn't have consistent a transformation law under gauge transformations, or put another way, a well-defined charge. Any field with electric charge must contain both particle and anti-particle operators by this argument. $\endgroup$ Commented Mar 20, 2017 at 21:18

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