I'm really confused about how the PCAC condition should be written for the negative pion, and also confused about the two different conventions for the pion decay constant which are related by a factor $\sqrt{2}$: $F_\pi\approx 93\,\textrm{MeV}$ and $f_\pi\approx 130\,\textrm{MeV}$.

I have understood that $\pi^-(x)=\frac{1}{\sqrt{2}}\left(\pi_1(x)-i\pi_2(x)\right)$ where $\pi_{1,2,3}$ are the triplet of scalar pion field-operators which parameterize the spontaneously broken axial chiral symmetry.

$$\Sigma(x)=\exp\left(\frac{2i}{F_\pi}\pi_a(x)\tau^a\right)=\exp\left(\frac{2i}{F_\pi}\begin{pmatrix}\pi_3(x) & \pi_1(x)-i\pi_2(x) \\ \pi_1(x)+i\pi_2(x) & -\pi_3(x)\end{pmatrix}\right) \tag{1}$$

The factor of $2$ is to give a properly normalized kinetic term in the chiral Lagrangian, and $F_\pi \approx 93\,\textrm{MeV}$ is the pion decay constant (source: Schwartz, section 28.2). These pion fields create and annihilate pion states:

$$\pi_a=\int\frac{d^4k}{(2\pi)^4}\frac{1}{\sqrt{2E_k}}\left(a^{\dagger}(k,a)e^{-ikx}+a(k,a)e^{+ikx}\right) \tag{2}$$

Where $a^{\dagger}(k,a)$ creates a pion state with isospin-index $a$ and momentum $k$:

$$a^{\dagger}(k,a)|\Omega\rangle=|\pi_a(k)\rangle \tag{3}$$

These states are normalized via the usual:

$$\langle \pi_a(k)|\pi_b(k')\rangle=(2\pi)^3 2E_k \delta^3(\vec k - \vec k') \delta_{ab} \tag{4}$$

My first question is, what is the negative-pion state $|\pi^-(k)\rangle$? Is it given by the following?

$$|\pi^-(k)\rangle \overset{?}{=} \frac{1}{\sqrt{2}}\bigl( | \pi_1(k)\rangle -i|\pi_2(k)\rangle \bigr ) \tag{5}$$

My second question is regarding PCAC: $$\langle \Omega | J^5_{\mu,a}(x)|\pi_b(k)\rangle = iF_\pi k_\mu e^{-ikx}\delta_{a,b} \tag{6}$$

In the above $J^5_{\mu,a}(x)=\bar\psi (x) \tau_a\gamma_\mu\gamma_5 \psi (x)$ is the axial-vector current (associated with axial chiral rotations), and $F_\pi\approx 93\,\textrm{MeV}$ is the pion weak decay constant. In Schwartz page 571, there is a footnote saying the following:

"Another common definition for $F_\pi$ you may find in the literature is $\langle \Omega | J^5_\mu (x) |\pi^- (p)\rangle=if_\pi p^\mu e^{-ipx}$... Since $\pi^-=\frac{1}{\sqrt{2}}\left(\pi^1-i\pi^2 \right)$, this leads to $f_\pi = \sqrt{2}F_\pi=130\,\textrm{MeV}$."

First off, in the above statement, I don't know what $J^5_\mu(x)$ is supposed to mean. Does it mean the current associated with diagonal chiral rotations, $J^5_\mu(x)=\bar\psi (x)\gamma_\mu\gamma_5 \psi (x)$? I thought this current had nothing to do with the pions, only the $\eta$-meson? In the standard PCAC relation (6), this current does not appear at all.

The only sensible-looking way I can make sense of the negative-pion version of the PCAC in that footnote is if I take

$$J^5_\mu\overset{?}{=}J^5_{\mu,1}+iJ^5_{\mu,2}=\bar\psi (x)\left(\tau^1+i\tau^2 \right)\gamma_\mu\gamma_5 \psi (x) \tag{7}$$

So my second question is, what is $J^5_\mu (x)$ in the above-quoted footnote, and how does the new PCAC (involving the negative pion) relate to the original one?

The reason I'm asking this is because I'm dealing with a direct calculation of the pion parton distribution amplitudes, whose normalizations are fixed by the relevant PCAC relations. If I'm off by a factor of $\sqrt{2}$ in the PCAC, then that will affect my entire calculation by a similar factor.

  • $\begingroup$ Most likely, the 130MeV convention is used - see PDG, for instance. As for the rest of your question, I believe it was already nicely answered here. $\endgroup$
    – Lodinn
    Oct 2, 2021 at 1:53

1 Answer 1


Looks like the classic "catch my sloppiness" exercise on Schwartz. (My students got extra credit for those). I edited your question to drop P&S in favor of S, clearly your intention. Let's only deal with $F_\pi\sim 93$MeV, to avoid confusion. In your (1), you took his τs to be Pauli σs, when he clearly takes them to be the real SU(2) generators, so σ/2 s, for his (28.26)... so the last member of your (1) is flawed.

Otherwise your (5) and (7) are, indeed, correct as conjectured. You only need compare normalizations for the neutral versus charged pion, so you could be cavalier about the absolute normalizations of currents!

Indeed, Matt almost certainly means $J^5_{\mu , +}= J^5_{\mu 1,1} +i J^5_{\mu 1,2} \propto F_\pi \partial_\mu (\pi_1 + i \pi_2) + O(\pi^2)$ in his footnote.

You could convince yourself the axial current is $\propto \Sigma^\dagger \partial_\mu \Sigma - \Sigma \partial_\mu \Sigma ^\dagger\propto F_\pi \partial_\mu \pi^a \tau^a+ O(\pi^2)$ rewritten in 0, $\pm$ notation, as above, but I'll leave the "joy" for you.

The PDG thus uses $f_\pi= \sqrt {2} ~~ F_\pi $ ~ 130 MeV, as per Matt's footnote.

  • $\begingroup$ Thanks for this fantastic answer. I definitely meant Schwartz instead of P&S, and I indeed forgot the factor of $1/2$ in the $\tau$'s which are taken to be the Hermitian generators of $SU(2)$. By the way, did you mean to write $J^5_{\mu,-}=J^5_{\mu,1}+iJ^5_{\mu,2}\propto F_\pi \partial_\mu (\pi_1 - i \pi_2 )$? In order to get the right results for the negative-pion $\frac{1}{2}(\pi_1-i\pi_2)$ I need to use $\tau_1+i\tau_2$ (notice the opposite signs). $\endgroup$ Oct 4, 2021 at 21:34
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    $\begingroup$ Oh, just choose what your conventions tell you... The current annihilates a π- state. If your wrote $|\pi^-\rangle\propto \pi^- | \Omega\rangle$, then you need the current to be like π+ for charge neutrality. You seem to use the opposite convention where the field destroys the state. $\endgroup$ Oct 4, 2021 at 22:00

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