Why is there an $i$ in the definition of hadronic decay constants?

Question

The decay of (for example) a pion can be parameterized by a decay constant $f_\pi$ defined via $$ \langle 0 | \bar d \gamma_\mu \gamma^5 u |\pi^+(p) \rangle = i f_\pi p_\mu $$ $$ f_\pi \approx 131 \text{ MeV.}$$ My question is why do we include the $i$ on the RHS? In other words how do we know that the matrix element on the LHS is purely imaginary?

Answer 1

Well, you may define your conventions, and physicists are perverse enough to actually do that (watch them...), any way you want. This is the dominant convention, and the logical "chain of custody", so to speak, is, schematically, $$ \pi^+\mapsto \pi^+ + f_\pi \theta^+, \qquad \leadsto \\ J_{5}^{\mu~~+} = \frac{δ{\mathcal L}}{δ\partial_\mu \pi^+} \frac{δ\pi^+}{δ\theta^+}=f_\pi \partial^\mu \pi^+ ~~~(\sim -\bar d \gamma^\mu \gamma^5 u ),~~~\leadsto \\ \langle 0 | \bar d \gamma_\mu \gamma^5 u |\pi^+(p) \rangle = i f_\pi p_\mu ~. $$ I've been cavalier/dyslexic with signs and factors, but this is the choice for a real weak iso-raising matrix $\tau^+$. So the i pops up in the conversion from p to -i∂ . For the neutral pion (hermitean), you'd get a hermitean current, which is the standard convention.