AAnomalies and Bose symmetry
Daniel Kabat
Department of Physics and AstronomyLehman College, City University of New York, Bronx NY 10468, USA [email protected]
We point out a feature of the triangle diagram for three chiral currents whichis perhaps not widely appreciated: Bose symmetry is not manifest and suffersfrom a momentum-routing ambiguity. Imposing Bose symmetry fixes theambiguity and leads to the famous Adler - Bell - Jackiw anomaly.
Dedicated to Roman Jackiw on the occasion of his 80 th birthday a r X i v : . [ h e p - t h ] D ec Introduction
It was a privilege studying under Roman Jackiw at MIT in the early 90’s. Bythat time the threat of midnight phone calls inquiring after the status of acalculation had somewhat subsided, but the benefits of Roman as an advisorremained. Never one to waste time with trivial discussion or unfoundedspeculation, he instead provided an unparalleled source of clear guidanceand direction for his students.Roman had an uncanny ability to formulate and solve mathematical sys-tems of physical relevance. His work often laid the foundation for futuredevelopments, with an impact far beyond its original scope. Witness theresurgence of interest in Jackiw - Teitelboim gravity [1, 2] as a holographicsystem [3, 4, 5], or the Adler - Bell - Jackiw anomaly [6, 7] which had itsorigins in current algebra but then did so much to initiate the use of topo-logical methods in gauge theories [8]. Another example may be his work onnon-associative structures and 3-cocycles [9], a topic I believe will ultimatelyplay an important role in quantizing gravity [10].In these notes I’ll consider a two-component fermion of definite chirality.Thus in section 2 I’ll work with a single chiral current, instead of the vectorand axial combinations which can be built from a pair of chiral fermions,and I’ll discuss a feature of the triangle diagram for three chiral currentswhich, although known, may not be widely appreciated: due to a momentum-routing ambiguity the diagram does not have manifest Bose symmetry. Bosesymmetry may be imposed on the diagram by hand; this fixes the ambiguityand leads to a unique expression for the divergence of the chiral current. Insection 3 I’ll make contact with the usual axial anomaly. In section 4 I’llshow that Bose symmetry can be restored with a local counterterm, while adifferent counterterm gives a covariant expression for the anomaly.
Consider a massless chiral fermion coupled to an external vector field. We’lldescribe the fermion using a Dirac spinor ψ , but with a projection condition1o that the fermion is either right- or left-handed. That is, we’ll assume ψ = 12 (cid:0) ± γ (cid:1) ψ (1) In what follows the upper sign will correspond to a right-handed spinor, thelower sign to left-handed.
The coupling of ψ to the external (non-dynamical)vector field A µ is described by the Lagrangian L = ¯ ψiγ µ ( ∂ µ + iqA µ ) ψ (2)With these ingredients the diagram for scattering three vector fields µ k kp + k λν kpp − k + crossed diagram( k , ν ) ↔ ( k , λ )leads to the scattering amplitude − i M µνλ = ( − (cid:90) d p (2 π ) Tr (cid:40) − iqγ µ ip (cid:14) + k (cid:14) (cid:0) − iqγ ν (cid:1) ip (cid:14)(cid:0) − iqγ λ (cid:1) ip (cid:14) − k (cid:14) (cid:0) ± γ (cid:1)(cid:41) +Tr (cid:40) − iqγ µ ip (cid:14) + k (cid:14) (cid:0) − iqγ λ (cid:1) ip (cid:14)(cid:0) − iqγ ν (cid:1) ip (cid:14) − k (cid:14) (cid:0) ± γ (cid:1)(cid:41) (3) Conventions: the metric g µν = diag(+ − −− ), the antisymmetric tensor (cid:15) = +1,and the Dirac matrices are γ = (cid:18) (cid:19) γ i = (cid:18) σ i − σ i (cid:19) γ ≡ iγ γ γ γ = (cid:18) − (cid:19) In this basis a Dirac spinor has left- and right-handed chiral components ψ = (cid:16) ψ L ψ R (cid:17) . k + k + k = 0. Theprojection operator (cid:0) ± γ (cid:1) ensures that only a single chirality circulatesin the loop.We might expect the (necessarily chiral) currents j µ = ¯ ψγ µ ψ to obeyBose statistics, so we might expect the amplitude to be invariant under per-mutations of the external lines. Indeed there’s a simple argument whichseems to show that Bose symmetry is satisfied. Invariance under exchange( k , ν ) ↔ ( k , λ ) is manifest; given our labellings it just corresponds to ex-changing the two diagrams. However we should check invariance under ex-change of say ( k , µ ) with ( k , ν ). Making this exchange in (3) we get − i M νµλ = ( − (cid:90) d p (2 π ) Tr (cid:40) − iqγ ν ip (cid:14) + k (cid:14) (cid:0) − iqγ µ (cid:1) ip (cid:14)(cid:0) − iqγ λ (cid:1) ip (cid:14) − k (cid:14) (cid:0) ± γ (cid:1)(cid:41) +Tr (cid:40) − iqγ ν ip (cid:14) + k (cid:14) (cid:0) − iqγ λ (cid:1) ip (cid:14)(cid:0) − iqγ µ (cid:1) ip (cid:14) − k (cid:14) (cid:0) ± γ (cid:1)(cid:41) Shifting the integration variable p µ → p µ + k µ in the first line, p µ → p µ − k µ inthe second, and making some cyclic permutations inside the trace, we seemto recover the previous expression (3).Famously, though, this argument for Bose symmetry is invalid. Instead alinearly divergent integral picks up a finite surface term when the integrationvariable is shifted. (cid:90) d p (2 π ) f ( p + a ) = (cid:90) d p (2 π ) (cid:0) f ( p ) + a µ ∂ µ f ( p ) + · · · (cid:1) = (cid:90) d p (2 π ) f ( p ) − ia µ π lim p →∞ (cid:104) p p µ f ( p ) (cid:105) (4)Here angle brackets denote an average over the Lorentz group, and the limitis understood to mean large spacelike momentum.We’ll return below to see that the amplitude (3) indeed violates Bosesymmetry. But for now, rather than study the violation of Bose symmetryin detail, we’re simply going to demand that the scattering amplitude besymmetric. The most straightforward way to do this is to define the scatter-ing amplitude to be given by averaging over all permutations of the external3ines. Equivalently, we average over cyclic permutations of the internal mo-mentum routing. That is, we define the Bose-symmetrized amplitude − i M symm µνλ = 13 (cid:34) +p µ νλ µ νλ pp µ νλ + + crossed diagrams( k , ν ) ↔ ( k , λ ) (cid:35) Explicitly this gives − i M symm µνλ = ∓ q (cid:90) d p (2 π ) Tr (cid:26) γ µ p (cid:14) γ ν p (cid:14) − k (cid:14) γ λ p (cid:14) + k (cid:14) γ + γ µ p (cid:14) + k (cid:14) γ ν p (cid:14) γ λ p (cid:14) − k (cid:14) γ + γ µ p (cid:14) − k (cid:14) γ ν p (cid:14) + k (cid:14) γ λ p (cid:14) γ + γ µ p (cid:14) γ λ p (cid:14) − k (cid:14) γ ν p (cid:14) + k (cid:14) γ + γ µ p (cid:14) + k (cid:14) γ λ p (cid:14) γ ν p (cid:14) − k (cid:14) γ + γ µ p (cid:14) − k (cid:14) γ λ p (cid:14) + k (cid:14) γ ν p (cid:14) γ (cid:27) Here we’ve used the fact that only terms involving γ contribute to the scat-tering amplitude (Furry’s theorem).Having enforced Bose symmetry, let’s check current conservation by dot-ting this amplitude into k µ . Using trivial identities such as k (cid:14) = (cid:0) p (cid:14) + k (cid:14) (cid:1) − p (cid:14) (5)to cancel the propagators adjacent to k (cid:14) , it turns out that most terms cancel,leaving only − ik µ M symm µνλ = ± q (cid:90) d p (2 π ) Tr (cid:26) p (cid:14) + k (cid:14) γ ν p (cid:14) − k (cid:14) γ λ γ − p (cid:14) + k (cid:14) γ ν p (cid:14) − k (cid:14) γ λ γ + 1 p (cid:14) − k (cid:14) γ ν p (cid:14) + k (cid:14) γ λ γ − p (cid:14) − k (cid:14) γ ν p (cid:14) + k (cid:14) γ λ γ (cid:27) After shifting p → p + k − k the second term seems to cancel the first,and after shifting p → p + k − k the fourth term seems to cancel the third.4his naive cancellation means the whole expression is given just by a surfaceterm. − ik µ M symm µνλ = ± q (cid:90) d p (2 π ) ( k α − k α ) ∂∂p α Tr (cid:40) p (cid:14) γ ν p (cid:14) + k (cid:14) γ λ γ (cid:41) + ( k α − k α ) ∂∂p α Tr (cid:40) p (cid:14) + k (cid:14) γ ν p (cid:14) γ λ γ (cid:41) Using the expression for the surface term (4), evaluating the Dirac traceswith Tr (cid:0) γ α γ β γ γ γ δ γ (cid:1) = 4 i(cid:15) αβγδ , and averaging over the Lorentz group with (cid:104) p α p β (cid:105) = g αβ p we see that the amplitude satisfies − ik µ M symm µνλ = ∓ q π (cid:15) νλαβ k α k β . (6)Famously current conservation is violated by the triangle diagram [6, 7].This consistent anomaly [11] can be encapsulated by writing down aneffective action for the vector field Γ[ A ] which incorporates the effect of thefermion triangle. The amplitude we’ve computed corresponds to the followingnon-local term in the effective action. Γ[ A ] = · · · ± q π (cid:90) d xd y ∂ µ A µ ( x ) (cid:3) − ( x − y ) (cid:15) αβγδ F αβ F γδ ( y ) (7)In this expression (cid:3) − ( x − y ) should be thought of as a Green’s function, theinverse of the operator ∂ µ ∂ µ , and F αβ is the field strength of A µ . The factthat the current j µ ≡ − q δ Γ δA µ is not conserved, ∂ µ j µ = − q ∂ µ δ Γ δA µ = ± q π (cid:15) αβγδ F αβ F γδ , (8) To verify (7) note that the term we’ve written down in Γ[ A ] corresponds to a vertex k νλµ k k ∓ q π (cid:18) k k µ (cid:15) νλαβ k α k β + cyclic perms (cid:19) When dotted into one of the external momenta this reproduces (6). The non-locality of Γis crucial, as otherwise the anomaly could be canceled with a local counterterm.
Suppose we have two spinors, one right-handed and one left-handed. Assem-bling them into a Dirac spinor ψ , the currents j µR = ¯ ψγ µ (cid:0) γ (cid:1) ψ j µL = ¯ ψγ µ (cid:0) − γ (cid:1) ψ have anomalous divergences ∂ µ j µR = q π (cid:15) αβγδ R αβ R γδ ∂ µ j µL = − q π (cid:15) αβγδ L αβ L γδ . (9)Here R µ and L µ are background vector fields which couple to the chiralcomponents of ψ , and quantities with two indices are the corresponding fieldstrengths. Note that we’ve taken the right- and left-handed components of ψ to have the same charge. The vector and axial currents j µ = j µR + j µL = ¯ ψγ µ ψ j µ = j µR − j µL = ¯ ψγ µ γ ψ couple to the linear combinations V µ = 12 (cid:0) R µ + L µ (cid:1) A µ = 12 (cid:0) R µ − L µ (cid:1) . As a consequence of (9) these currents have divergences ∂ µ j µ = q π (cid:15) αβγδ V αβ A γδ ∂ µ j µ = q π (cid:15) αβγδ ( V αβ V γδ + A αβ A γδ ) . At first sight this seems no better than having a single chiral spinor. Butconsider adding the following local term to the effective action for V µ and A µ . ∆Γ = cq π (cid:90) d x (cid:15) αβγδ ∂ α V β V γ A δ c is an arbitrary constant. This term violates both vector and axialgauge invariance, so it contributes to the divergences of the correspondingcurrents. ∆ (cid:0) ∂ µ j µ (cid:1) = − q ∂ µ δ (∆Γ) δV µ = − cq π (cid:15) αβγδ V αβ A γδ ∆ (cid:0) ∂ µ j µ (cid:1) = − q ∂ µ δ (∆Γ) δA µ = + cq π (cid:15) αβγδ V αβ V γδ If we add this term to the effective action and set c = 1, we have a conservedvector current but an anomalous axial current. ∂ µ j µ = 0 ∂ µ j µ = q π (cid:15) αβγδ (cid:18) V αβ V γδ + 13 A αβ A γδ (cid:19) (10)Given a conserved vector current we can promote V µ to a dynamical gaugefield – usually the desired state of affairs. If we aren’t interested in making V µ dynamical then other choices for c are possible. It’s no surprise that by imposing Bose symmetry we’ve recovered the standardexpression for the anomaly, as the importance of Bose symmetry for the resultwas emphasized from the very beginning [6, 7]. But still, let’s return to themomentum routing given in (3) and see how Bose symmetry and currentconservation play out. Using identities similar to (5) and the expression forthe surface term (4) we find that − ik µ M µνλ = 0 − ik ν M µνλ = ∓ q π (cid:15) µλαβ k α k β (11) − ik λ M µνλ = ∓ q π (cid:15) µναβ k α k β Thus with the momentum routing (3) the current is conserved at one vertexbut not at the other two, a peculiar state of affairs which shows that Bosesymmetry is violated. To capture this in an effective action we introduce7hree distinct vector fields
A, B, C , with field strengths denoted by the sameletter, and takeΓ[
A, B, C ] = ± q π (cid:90) d xd y (cid:16) ∂ µ B µ ( x ) (cid:3) − ( x − y ) (cid:15) αβγδ A αβ C γδ ( y )+ ∂ µ C µ ( x ) (cid:3) − ( x − y ) (cid:15) αβγδ A αβ B γδ ( y ) (cid:17) (12)Gauge invariance is respected for A but violated for B and C . The breakdownof Bose symmetry is manifest.Since Bose symmetry could be restored by symmetrizing over momentumroutings, it should also be possible to restore it with a local counterterm.Consider adding the following local term to the effective action.∆Γ = ∓ q π (cid:90) d x (cid:15) αβγδ A α ( B βγ C δ + B β C γδ ) (13)When added to (12) the anomalous divergences become symmetric, ∂ µ j µA = − q ∂ µ δ (Γ + ∆Γ) δA µ = ± q π (cid:15) αβγδ B αβ C γδ ∂ µ j µB = − q ∂ µ δ (Γ + ∆Γ) δB µ = ± q π (cid:15) αβγδ A αβ C γδ (14) ∂ µ j µC = − q ∂ µ δ (Γ + ∆Γ) δC µ = ± q π (cid:15) αβγδ A αβ B γδ and can be captured by an effective actionΓ[ A, B, C ] = ± q π (cid:90) d xd y (cid:16) ∂ µ A µ ( x ) (cid:3) − (cid:15) αβγδ B αβ C γδ ( y ) + cyclic (cid:17) (15)Then we’re free to identify the three vector fields and, with a 1 /
3! for Bosesymmetry, describe the anomaly with the effective action (7).The procedure above gives a consistent anomaly. On the other handconsider adding to (12) the counterterm ∆Γ = ∓ q π (cid:90) d x (cid:15) αβγδ A α ( B βγ C δ + B β C γδ ) (16) Related counterterms appear in [12]. ∂ µ j µA = ± q π (cid:15) αβγδ B αβ C γδ ∂ µ j µB = ∂ µ j µC = 0 (17)An effective action which captures this isΓ[ A, B, C ] = ± q π (cid:90) d xd y ∂ µ A µ ( x ) (cid:3) − (cid:15) αβγδ B αβ C γδ ( y ) (18)Since j B and j C are conserved, (18) respects gauge invariance for B and C , which means (17) can be identified as the covariant anomaly for A . Evidently the covariant anomaly can be obtained by varying an effectiveaction, despite the Wess-Zumino consistency condition [13], at the price ofviolating Bose symmetry. It would be interesting to see if a similar resultholds in non-abelian theories.
Happy birthday, Roman!
Acknowledgements
I’m grateful to V. Parameswaran Nair for comments on the manuscript andespecially for suggesting the connection to covariant anomalies. DK is sup-ported by U.S. National Science Foundation grant PHY-1820734.
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