# Gauging the chiral symmetry: Is there a vector field that couples to chiral current?

I'm trying to understand the consequences of massless Dirac field $$\mathcal{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi\tag{1}$$ when the chiral symmetry is made local i.e., $$\psi\rightarrow\psi^\prime= \exp[i\alpha(x)\gamma_5]\psi.\tag{2}$$ It turns out that after chiral transformation the massless Dirac Lagrangian becomes $$\mathcal{L}^\prime=i\bar{\psi}^\prime\gamma^\mu\partial_\mu\psi^\prime=i\bar{\psi}\gamma^\mu\partial_\mu\psi-(\partial_\mu\alpha)\bar{\psi}\gamma^\mu\gamma_5\psi$$

In case of $U(1)$ vector symmetry in QED, $$\psi\rightarrow\psi^\prime= \exp[i\theta(x)]\psi,$$ one introduces a field $A_\mu$ that transforms under gauge transformation as $$A_\mu^\prime=A_\mu-\partial_\mu\theta$$ so that a term proportional to $\bar{\psi}\gamma^\mu\psi A_\mu$ when added to the Lagrangian (1) kills the extra term and makes the theory invariant. Similarly, if we postulate a field $B_\mu$, which when introduced via a term proportional to $$\bar{\psi}\gamma^\mu\gamma_5\psi B_\mu$$ make the Lagrangian (1) invariant under (2) if $$B_\mu\rightarrow B_\mu-\partial_\mu\alpha$$ under (2). Is this criterion justified?

If yes, the chiral symmetry can be made local. But is this physically meaningful?

• I'm not sure what exactly you mean by "physically meaningful". If the question is whether you can write down this coupling to $B_\mu$ and make the symmetry local, the answer is yes. Note however that you cannot introduce both $A$ and $B$: there is a mixed anomaly if you try to gauge both symmetries. Is this what you are interested in? – user121664 Dec 28 '16 at 19:30
• @user121664 Yes. It is only the chiral current which is anomalous. Right? What do you mean by mixed anomaly? – SRS Dec 28 '16 at 19:39
• If you gauge the vector symmetry only, the chiral one is anomalous. If you only want to gauge the chiral symmetry, that also gives you a consistent theory, but the vector (global) symmetry is now anomalous. If you try to gauge both, you get an anomalous gauge symmetry. This set of facts comes about because neither symmetry is anomalous by itself, but there is a mixed anomaly, so you cannot gauge both at the same time. – user121664 Dec 28 '16 at 19:41
• @user121664 Ok. If we gauge the chiral symmetry and not the vector symmetry, will it be phenomenologically meaningful? I mean, usually it is the chiral symmetry which is left ungauged. Also, how can I show that there is a "mixed anomaly" and none of the current is individually anomalous? – SRS Dec 28 '16 at 19:45
• Actually I think what I just said is true in 2d (where anomalies are proportional to $Q_L^2-Q_R^2$) but not in 4d, where they go as the cube of the charges. It might be that the theory with only the chiral symmetry is also anomalous in 4d. I'll try to check this. – user121664 Dec 28 '16 at 21:48

1) There indeed exist theories in which the gauge vector-like fields coupled to the axial fermion current enter the game. The most familiar example is, of course, the Standard model, in which there is the local $SU_{L}(2)$ symmetry (left fermion doublets interact with 3 gauge fields $W_{\mu}^{a}$). It is possible to rewrite the theory in terms of vector-axial basis, $$\psi_{L} \equiv \frac{1}{2}\psi - \frac{\gamma_{5}}{2}\psi,$$ and corresponding vector-axial gauge fields are $$V_{\mu}^{a} = \frac{W_{\mu}^{a}}{2}, \quad A_{\mu}^{a} = -\frac{W_{\mu}^{a}}{2}$$ 2) There also exist many realistic field theories in which the axial-vector fields exist, but where they are not the gauge fields; typically there are fields representing some particles or the symmetries of the theory.
The familiar example is axial-vector mesons in the QCD near and below the global chiral symmetry $SU_{L}(3)\times SU_{R}(3)$ breaking scale. An approach to introduce these mesons is following: one may gauge this symmetry by introducing massless axial-vector fields, and after that to break it explicitly by adding the mass terms. Although this approach looks unnatural, in fact it has some theoretical origin (the action of chiral perturbation theory has hidden local gauge $SU_{L}(3)\times SU_{R}(3)$ symmetry), phenomenological origin (axial-vector mesons is of course the part of the QCD which respects the approximate chiral symmetry) deep historical roots (the so-called vector meson dominance model) and more or less succesfully describes the data.
Another example of such effective theory is the Weyl semimetal near the bands crossing point. It is given by the theory of massless chiral fermions with non-zero distance in momentum and energy space between their spectrum (being the Dirac cones), parametrized by $b_{\mu}$. It is local because of tensions and dislocations into the semimetal. The lagrangian of such model effectively coincides with $$L = \bar{\psi}(i\gamma_{\mu}\partial^{\mu} -\gamma^{\mu}\gamma_{5}b_{\mu})\psi$$ 3) Also, background axial gauge fields are sometime introduced when we need to define the axial current through the action. Precisely, if we introduce such field $A_{\mu}$, then the corresponding current is given by $$J_{\mu}^{A}(x) = \frac{\delta \Gamma[A]}{\delta A^{\mu}(x)}$$ After calculating the current and its properties (often related to the chiral anomaly) the coupling to $A$ is set to zero. This trick was used by Bardeen when he calculated the anomaly in $SU_{L}(3)\times SU_{R}(3)$ theory similar to the QCD.
Of course, if You have both vector and axial-vector gauged symmetries, then the gauge anomaly appears. If You have only one fermion specie, then in general the theory will be inconsistent. Other fermions required in order to cancel this anomaly. However, You see from the text written above that often axial (or vector) gauge fields are in fact fictive (or may even correspond to physical particle), and in these cases You have not to worry about the anomalies. Indeed, suppose the theory in which both vector $V_{\mu}$ and axial-vector $A_{\mu}$ fields are present; however, the axial field is physical. Then the (consistent) anomaly for vector and axial currents is $$\tag 1 \partial_{\mu}J^{\mu}_{V} = \frac{1}{48\pi^{2}}\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}^{A}F_{\alpha\beta}^{V},$$ $$\tag 2 \partial_{\mu}J^{\mu}_{A} = \frac{1}{96\pi^{2}}\epsilon^{\mu\nu\alpha\beta}(F_{\mu\nu}^{A}F_{\alpha\beta}^{A} + F_{\mu\nu}^{V}F_{\alpha\beta}^{V})$$ In order to remove the anomaly from the vector current conservation (and corresponding Ward identities, of course) You have to add the local counter-term (in literature it is called the Bardeen counter-term), which cancels the right hand-side of $(1)$. The expression $(2)$ instead takes the form $$\partial_{\mu}J^{\mu}_{A} = \frac{1}{96\pi^{2}}\epsilon^{\mu\nu\alpha\beta}(F_{\mu\nu}^{A}F_{\alpha\beta}^{A} + 3F_{\mu\nu}^{V}F_{\alpha\beta}^{V})$$ This counter-term even generates the additional physical part of the vector current $J_{\mu}$, namely $$\Delta J_{\mu}^{V} \sim \epsilon_{\mu\nu\alpha\beta}A^{\nu}F^{\alpha\beta}_{V}$$