I have seen the following matrix element quoted millions of times, often called the Partially Conserved Axial Current (PCAC) relation:

$$\langle 0|J^5_{\mu,a}(x)|\pi_b(p)\rangle=-if_{\pi}\delta_{ab}e^{-ipx}p_{\mu}\tag{1}$$

where $J^5_{\mu,a}(x)=i\bar q(x)\tau_a\gamma_{\mu}\gamma_5 q(x)$ is the Noether current for $SU(N_f)$ chiral rotations in the quark sector, which is spontaneously broken by the QCD vacuum and henceforth gives rise to the $\pi_b(p)$ pion fields.

From this relation we may motivate the important relation $J^5_{\mu,a}=f_{\pi}\partial_{\mu}\pi_a(x)$. I have similarly seen the following (equivalent?) identity:

$$\langle 0 | \partial^{\mu}J^{5}_{\mu,a}|\pi_b(p)\rangle=-m_{\pi}^2f_{\pi}\delta_{ab} \tag{2}$$

My question is, how do we derive relation (1)? I think historically, in the development of current algebra and $\chi$-PT, this was probably just an assumption/postulate, but how can I see/derive this from a modern perspective?

Does the proper explanation lie in the CCWZ construction (of the chiral Lagrangian through the pions being a nonlinear realization of the broken symmetry)? I have always put off reading about that in detail.

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    $\begingroup$ No, I'd lay off the CCWZ construction. You may expand the axial current of the chiral lagrangian to identify the linear part of the current--if your QFT course skimped on that, shame on it. PCAC was motivated and illustrated by the σ model, which defines the Nambu-Goldstone realization of symmetry ("broken"). It is masterfully detailed in pp 147-152 of T-P Cheng & L-F Li's text, leading up to your (5.181); and no glib elliptical answers would do justice to the question! $\endgroup$ Commented Apr 28, 2020 at 23:18
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    $\begingroup$ ...some people would prefer Peskin & Schroeder pp 669 et seq., to (19.88), (19.92), so I don't see why. But if it worked for you... $\endgroup$ Commented Apr 28, 2020 at 23:26

1 Answer 1


Two more key references in this answer: the Itzykson & Zuber discussion for the σ-model alluded to is peerless, if one really wishes to get to the bottom of it; summarized here. The chiral model calculation is trivial and I am illustrating it below in a thus retargetted question. (The subtler point in your question, however, is connecting quark bilinear currents to pseudoscalar hadrons, a crepuscular art, and the sheer genius brainchild of Murray Gell-Mann, long before the discovery of confinement! His Escoffier pheasant baked between veal slices picture, glossed over here. This is a hard question, split off from the body of yours, requiring subtle verbiage: the core of M G-M's historic tie-breaker to Renner, above.)

Here, then, is the trivial answer of the retargeted question merely illustrating the operator PCAC statement with Gürsey's 1960 spectacular chiral model for $SU(2)_L\times SU(2)_R$ for you to get the basic idea; whose 3-flavor extension is alarmingly called the "Goldstone matrix" at the end of the talk you linked. I do this since the Pauli matrices for SU(2) are better understood intuitively, and I am terminally cavalier with normalizations, using ~ signs instead of =, since you should appreciate the structure of the representations and how they fit, and not the details utilized in monkey-see-monkey-do calculations.

Again, I am eschewing the $64 question of hadronizing the quark operators for which subtler current algebra "hypothesis" arguments were formulated by Gell-Mann--Oakes--Renner, as alluded to in the comments. So everything I'll cover will be solid operator statements in the context of that model, strict equalities, without any hypotheses and pheasants -- the fast talking involved bridging to QCD. It is important below, to illustrate all current statements with how the currents also present in the linear σ model, covered brilliantly in the textbooks cited; and also the quark bilinears of QCD. Both these are alternate realizations of the same group structure illustrated here.

Start with CVC and CAC, for the ideal, merely SSBroken chiral model of Goldstone mesons, without explicit breaking (which does not affect the currents!), yet, $$ {\cal L} \sim \frac{f_\pi^2}{4} \operatorname{Tr} ~ \partial^\mu U^\dagger \partial_\mu U , \\ U= e^{i \vec \pi\cdot \vec \tau /f_\pi} = 1\!\! 1 \cos (|\pi|/f_\pi) + i \frac{\vec \pi \cdot \vec \tau}{|\pi|}\sin (|\pi|/f_\pi),\\ |\pi|\equiv \sqrt{\vec \pi \cdot \vec \pi} . $$ It is then evident all four currents are conserved, $$ {\vec L}_\mu = if_\pi^2 \operatorname {Tr}~ \vec\tau ~ U\partial_\mu U^\dagger/2 \sim f_\pi \partial_\mu \vec \pi +..., \\ \vec R_\mu = if_\pi^2 \operatorname {Tr}~ \vec\tau ~ U^\dagger \partial_\mu U /2 \sim -f_\pi \partial_\mu \vec \pi +..., \\ \vec V_\mu =\vec L_\mu+\vec R_\mu= if_\pi^2 \operatorname {Tr}~ \vec\tau ~ (U\partial_\mu U^\dagger +U^\dagger \partial_\mu U )/2 \sim -2f_\pi \vec \pi \times \partial_\mu \vec \pi +... \\ \vec A_\mu = \vec R_\mu-\vec L_\mu= if_\pi^2 \operatorname {Tr}~ \vec\tau ~ (U\partial_\mu U^\dagger -U^\dagger \partial_\mu U )/2 \sim -2f_\pi \partial_\mu \vec \pi +... , $$ that is, L is the right-invariant left SU(2) current, R the left-invariant right SU(2) current, V the vector/isospin current, and A the axial current your question asks about. The key point here is that only V escapes SSB and acts linearly, in the Wigner-Weyl mode: its charge isorotates Goldstone mesons.

The remaining three, L,R,A are in the Nambu-Goldstone nonlinear mode: their charges shift goldstons into and out of the (therefore) degenerate vacuum.

This is what SSB means.

  • It is a worthwhile extraneous exercise to work out the above currents in both the linear σ model (the original ambit of these observations) and the free quark QCD lagrangian, and compare, while also confirming the standard current algebra for them. Crudely, $V_\mu\sim \vec \pi \times \partial_\mu \vec \pi$, $\sim \bar q \gamma_\mu \vec \tau q$; and $A_\mu\sim \sigma \partial_\mu \vec \pi+...$, $\sim \bar q \gamma_\mu \gamma_5 \vec \tau q$.

For the above completely symmetric lagrangian, the currents are strictly conserved, and we have CVC and CAC.

However, a major triumph in the 60s and the σ model was the tapered mooting of CAC into PCAC by adding some explicit breaking (the σ term) in the lagrangian. In the above chiral model at hand, this amounts to something like $$ -2\frac{m_\pi^2 f_\pi^2}{\operatorname {Tr}M} \operatorname {Tr}~(MU+U^\dagger M), $$ where M is a matrix despoiling L and R invariance explicitly (and identified today with a diagonal u,d quark mass matrix--don't worry about it; you might as well take it to be the identity) and the dimensional parameters in front of it are chosen with the benefit of hindsight. Since it has no derivatives, it contributes nothing to the above currents.

Expanding in pions, this amounts to a constant and, importantly, a pion mass term, $m^2_\pi \vec \pi \cdot \vec \pi$.

This mass term converts the massive pions into "pseudogoldstone bosons", and, by dint of the modified equations of motion, it leads to a non vanishing divergence for the axial current, $$ \partial^\mu \vec A _\mu \sim f_\pi m_\pi^2 \vec \pi +..., \tag{#} $$ the operator PCAC fact, not hypothesis, here (only).

Taking matrix elements between the vacuum and the pion states it destroys, trivially yields the matrix relations $\langle 0|\partial \cdot A^a(0)|\pi^b\rangle \sim f_\pi m_\pi^2 \delta^{ab}$.

The imaginative leap of Gell-Mann (with a nudge by Feynman$^\dagger$) was to correctly (as most often!) abstract the essential logical structure of the picture and hypothesize that the matrix element relations also held for quark (now QCD) chiral currents, in general, as well.

Spot on. 60 years later, QCD flows in the logical channels then dug.

$\dagger$ RPF is credited with discovering (#) in the historic paper which introduced it. For the history maven, cf. here. The deeper, more general "hypothesis" version involving nucleons appears in R Norton & W Watson, Phys Rev 110 (1958) 996.

  • $\begingroup$ Excuse me, sir, if I am asking something dumb. But from expression for $V_\mu$ is seems, that this simply vanishes : we have $U^{\dagger} \partial_\mu U + U \partial_\mu U^{\dagger} = \partial_\mu (U^{\dagger} U) = 0$ $\endgroup$ Commented Apr 29, 2020 at 19:24
  • $\begingroup$ ?? The left hand side does not equal the middle and right sides. Did you check your answer? you are left-multiplying this by $\vec \tau$ before tracing! $\endgroup$ Commented Apr 29, 2020 at 19:28
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    $\begingroup$ But there is no need for tracing at this point, just use of the Leibniz rule for derivative. $\endgroup$ Commented Apr 29, 2020 at 19:32
  • $\begingroup$ O, now I see, sorry for disturbance $\endgroup$ Commented Apr 29, 2020 at 19:41
  • $\begingroup$ Weinberg's peculiar alternate path through this landscape is overviewed in this CERN talk video. $\endgroup$ Commented Sep 26, 2022 at 10:07

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