According to Peskin and Scrhroeder the pion decay constant $f_\pi$ is defined via the following matrix element $$\left\langle0|j^{\mu5a}(x)|\pi^b(q)\right\rangle=-if_\pi \delta^{ab} q^\mu e^{-iqx}$$ while $j^{\mu5a}=\bar{Q}\gamma^\mu\gamma^5\tau^a Q$ with $Q$ the up-down quark doublet $Q=(u, d)^T$. The point is, that intuitively the definition seems to imply that $f_\pi$ is a kind of amplitude for a pion to behave like a pair of quarks and vice-versa.

This intuition works, for example, in the leptonic decay of a charged pion, where the amplitude $\mathcal{M}(\pi^+\to l^+\nu_l)\propto f_\pi G_F$. The first multiplier I interpret here as the amplitude for the pion, a bound state of quarks, to decouple into two free quarks, and the Fermi constant $G_F$ then describes how those two quarks decay into a lepton pair.

My question is the following. How can I understand the appearance of the negative powers of $f_\pi$ in some other hadronic amplitudes? For example, take the $\pi^0\to2\gamma$ decay (which bothers me most, practically) with the amplitude $$\mathcal{M}(\pi^0\to2\gamma)\propto \frac{e^2}{f_\pi}$$

One can visualize this process via the Feynman diagram where the pion first splits into a quark pair, then one of the quarks emit photon and, after, annihilate with the other quark emitting the second photon. The factor $e^2$ is natural since there were two photons emitted. But I would expect the $f_\pi$ factor to appear in the numerator, just as in the case with a leptonic decay. What is the reason here that it is in fact the negative power of $f_\pi$ in the correct formula?


How to compute pion interactions in general? Since pions are (pseudo)goldstone bosons, the procedure is following: starting from lagrangian which contains quark fields $q(x)$ (and suppose that you've integrated out gluon sector), you need to extract pion degrees of freedom from them, namely $$ q(x) \equiv U\tilde{q}(x), $$ where $$ U = e^{i\gamma_{5}\frac{\pi_{a}t_{a}}{f_{\pi}}} $$ There you just need to subtitute nonzero VEV $\langle |\bar{\tilde{q}}\tilde{q} |\rangle$.

From the described picture follows that the basic object of pion effective field theory is $U$, which depends on argument $\frac{\pi}{f_{\pi}}$.

In general the relevant effective lagrangian for free pions which is needed to answer your question has the form $$ \tag 1 S = \int d^{4}x\left(\frac{f_{\pi}^{2}}{4}\text{Tr}\left[ \partial_{\mu}U\partial^{\mu}U^{+}\right] - f_{\pi}^{2}m_{\pi}^{2}\text{Tr}\left[ U^{\dagger} + U - 2\right] + ...\right) + N_{c}\Gamma_{WZ}, $$ Here $$ \Gamma_{WZ} \equiv \frac{i}{240 \pi^{2}}\int d^{5}x\epsilon^{ijklm}\text{Tr}\left[U\partial_{i}U^{-1}U\partial_{j}U^{-1}U\partial_{k}U^{-1}U\partial_{l}U^{-1}U\partial_{m}U^{-1}\right] $$ is the Wess-Zumino term, which captures all the information about anomalies of underlying theory, and dots represent higher derivative terms and mixed terms.

Note the presence of $f_{\pi}^{2}$ at the first term in $(1)$ (it is needed for canonical form of pions kinetic term) and an absense of such quantity in the Wess-Zumino term (since in fact it is just the number).

The short formal answer on your question is following. In order to describe the mentioned processes - $\pi^{+} \to l^{+} \nu_{l}, \pi^{0} \to ll^{+}, \pi^{0} \to 2\gamma$ - we need to elongate the derivatives in $(1)$. The in turns out that the first two processes are mediated by the kinetic term in chiral effective field theory action $(1)$, which has $f_{\pi}^{2}$ factor in the front, while the general contribution into process $\pi \to \gamma\gamma$ is made by the Wess-Zumino term, which hasn't any dimensional factors in front of it. The amplitudes for the first two processes are proportional to $f_{\pi}$, while for the latter it is proportional to $f_{\pi}^{-1}$.

My statement that after gauging and adding the lepton part such action contains information about processes $\pi \to \mu \bar{\nu}_{\mu}$ and $\pi \to \gamma \gamma$.

When gauging the first terms of $(1)$, we simply modify partial derivatives $\partial_{\mu}$ to $$ \partial_{\mu}U \to \partial_{\mu} - iR_{\mu}U + iUL_{\mu}, $$ where $$ L \equiv g_{2}\frac{W_{a}\tau_{a}}{2} + g_{1}W^{0}\begin{pmatrix} \frac{1}{6} & 0 \\ 0 & \frac{1}{6}\end{pmatrix}, \quad R \equiv g_{1}W^{0}\begin{pmatrix}\frac{2}{3} & 0 \\ 0 & -\frac{1}{3} \end{pmatrix} $$ (in general, $g_{i}$ are matrices of constants (CKM matrix elements)).

We then may extract $\pi^{\pm}W^{\mp}$ vertex from the kinetic term of $(1)$ (here $W^{\pm} \equiv \frac{1}{\sqrt{2}}(W_{1} \mp iW_{2})$). Since, as I've written above, pion fields as the goldstone phase are always in combination $\frac{\pi}{f_{\pi}}$ and by taking into account that there are $f_{\pi}^{2}$ in the front of it, we have that $\pi W$ term is proportional to $f_{\pi}$.

Next, gauging of the Wess-Zumino term is harder (see an answer here). But now is only one important thing - the degree of $f_{\pi}$, so we immediately may give the result: the part of gauged WZ term which contains one pion field is inversely proportional to $f_{\pi}$.

  • $\begingroup$ Again, thank you for a clear and thorough presentation. However, I would still would like my mistake to be pointed out? I'll briefly repeat the reasoning. Since $f_{\pi}$ is roughly an amplitude for a pion to behave as a pair of quarks, the diagram where pion once splits into a quark pair which then emit two photons must contribute as $f_{\pi}e^2$, not as $f_{\pi}^{-1}e^2$? $\endgroup$ – Weather Report Feb 15 '16 at 8:53
  • $\begingroup$ My guess is that this diagram might contribute, but it is heavily suppressed. Then, I have a second question: can I somehow think of a process $\pi^0\to2\gamma$ in terms of quarks? What would be the corresponding diagram? $\endgroup$ – Weather Report Feb 15 '16 at 8:55
  • $\begingroup$ @WeatherReport : it is triangle diagram for which the loop is formed by quarks. Since anomaly is nonperturbative, then such effect is scale indepe dent. This is explanation on the quark level why such amplitude is not suppressed by $f_{\pi}$. $\endgroup$ – Name YYY Feb 15 '16 at 10:06

The second equation you have written comes from anomaly matching of the mixed anomaly with electromagnetism. In the UV the anomaly comes form the triangle loop with quarks running inside. However, in the IR all quarks are confined and you got only pions well below $\Lambda_{QCD}$. Still you need to be able to reproduce the mixed anomaly with the pions. Since under the chiral transformation with parameter $\alpha$ in a photon background the action changes by $$ \propto \alpha \epsilon^{\mu\nu\rho\sigma} F_{\mu\nu}F_{\rho\sigma}\,, $$ the action that in the IR that would give that is $$ \propto \frac{\pi^0}{f_\pi} \epsilon^{\mu\nu\rho\sigma} F_{\mu\nu}F_{\rho\sigma} $$ Indeed, the $\pi^0$ is the Goldstone boson of the spontaneosuly broken chiral symmetry which acts non-linaerly on $\pi^0$ $$ \pi\rightarrow \pi^0+ f_\pi \alpha. $$ This latter transformation follows from you initial equation that implies $$ J_\mu^5 \propto f_\pi \partial_\mu\pi^0+\ldots $$ as you can check from the canonically normalized kinetic term for $\pi^0$ using the standard trick to calculate the Noether current (that is, by promoting the parameter $\alpha$ of the chiral transformation to a spacetime dependent function).

As you see, it's the Goldstone boson transformation and the anomaly matching that fix the $1/f_\pi$ behavior in the lagrangian, and hence in the amplitude for $\pi^0\rightarrow \gamma\gamma$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.