# Π leptonic decay (current matrix element)

1. Nov 3, 2012

### Einj

I'm currently studying the pion leptonic decay and I'm getting a bit confused about some factors. Firstly, le correct lagrangian that describe the pion decay $\pi^+\rightarrow l^+ + \nu_l$ is:

$$L=\frac{4G_F}{\sqrt{2}}V_{ud}^* \bar{d}_L\gamma^\mu u_L \bar{\nu}_L \gamma_\mu l_L$$

We can't operate with the quark current and so we have to use an effective current $J^\mu_L=1/2(V^\mu-A^\mu)$ where V and A are the vector and axial curret. The vector current gives no contribution because the pion is pseudo scalar.
So we need to calculate the matrix element of the axial current. I'm a bit confused about the correct use of form factor. My professor wrote down the following matrix element:

$$\langle 0 |A_\mu|\pi^+\rangle = ip^{\pi}_\mu f_\pi$$

while in some books I found the form factor written as $F_\pi$ and realted to $f_\pi$ by some 2 or √2 factors.
Can some one tell me the exact relation between F and f?

Thanks

2. Nov 4, 2012

### Hepth

The pion decay constant:
$$\langle 0 | A_{\mu} | \pi \rangle = i p_{\mu} f_{\pi}$$
Where
$$A_{\mu} = \bar{u} \gamma_{\mu} \gamma_{5} d$$
is about 130 MeV. This is the definition.
Now, you might see $F_{\pi} = \frac{f_{\pi}}{\sqrt{2}}$ which is about 90 MeV,

The difference in notation is just different ways of writing it.
NOTE: I have noticed some DEFINE the decay constant differently in older papers where F is written as f and vice versa. I'm not completely sure but there may have been a time when the notation wasn't set in stone. (long ago)

Remember to keep track of the twos too:
If you look at the definition of my current, A, notice there are no L subscripts on the quarks. So you have:
$$\bar{d}_L \gamma^{\mu} u_L = d \frac{(1+ \gamma_5)}{2} \gamma^{\mu}\frac{(1- \gamma_5)}{2} u\\ = \bar{d} \gamma^{\mu}\frac{(1- \gamma_5)}{2} \frac{(1- \gamma_5)}{2} u\\ \\ = \bar{d} \gamma^{\mu} \frac{(1- \gamma_5)}{2} u\\ = \frac{1}{2} \bar{d} \gamma^{\mu} u - \frac{1}{2}\bar{d} \gamma^{\mu} \gamma_5 u\\ \approx \frac{1}{2} V^{\mu} - \frac{1}{2} A^{\mu}$$

So don't forget that by getting it into the form of A you pick up the extra negative 1/2.

3. Nov 4, 2012

### Einj

Yes, I kept track of it. Thanks for the help, that was exactly the answer I was looking for!