# Decay process of pions

$$\pi^+$$ and $$\pi^-$$ decay into muon(or electron) and neutrino and $$\pi^0$$ decays into photons.

So there is a weak interaction in the decay process of $$\pi^+$$ and $$\pi^-$$.

But the mean lifetime of $$\pi^0$$ is much smaller than $$\pi^+$$ and $$\pi^-$$ even though the mass of neutral pion is smaller than that of the charged pions.

It seems that these processes do not obey the uncertainty principle.

Is it related to the weak interaction?

The uncertainty principle is a kinematic envelope where the particles conjugate variables are constrained to be, not an equation.

The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time. $$\Delta x \Delta p > \frac{\hbar}{2}$$ $$\Delta t \Delta E > \frac{\hbar}{2}$$

If one measures the pion, both charged and neutral pions will be within the envelopes.

The difference in life time is because of conservation of quantum numbers and the difference in the possible interactions. Because of conservation of lepton number , the charged ones cannot go electromagnetically to just an electron/positron, the lepton number should be the same before and after the decay, and the weak interaction is called weak because it is much weaker than the electromagnetic.

Let me illustrate Anna's answer by plugging in numbers. Expecting heavier particles to have larger decay rates is an implicit acknowledgment of the phase space available when the decay products are much lighter. Otherwise, the mechanism of decay and couplings rule.

I understand you are asking how far $$m/\Gamma\geq 1/2$$ is from saturation; I am, of course, taking $$\hbar=1$$, non-dimensionalized in HEP units--anyone working in different ones in particle physics has lost their mind and was never heard from again.

Now the neutral π decays by the chiral anomaly electromagnetically, with, very crudely (you'll see why you need not do better than mere order of magnitude), $$\Gamma(\pi^0)\approx \frac{\alpha_e^2 m_\pi^3}{64\pi^3 f_\pi^2} \approx 8 eV \approx 10^{16}/s .$$ The (hadronic) pion decay constant is about $$f_\pi\approx$$ 130 MeV, close to the mass of the pion.

• Consequently, $$m/\Gamma\approx 2\cdot 10^{7}$$, freakishly far from UP saturation.

The charged pions, by celebrated contrast, decay weakly, so $$\Gamma(\pi^{\pm})\approx \frac{G^2 f_\pi^2 m_\mu^2 m_{\pi}}{8\pi} \cos ^2 \theta_C (1-m_\mu^2/m_\pi^2)^2 \approx 10^8/s \approx 6.6 ~ 10^{-8} eV,$$ where $$G\approx 1.2 \cdot 10^{-5} GeV^{-2}$$ is Fermi's constant, hence

• $$m/\Gamma\approx 2\cdot 10^{15}$$, even more extremely remote from saturation.

You may now appreciate why one all but never considers the UP in connection with particle decays in HEP...

Let me discuss my answer in two steps:

1. Why do they decay differently?

As already mentioned, there are some conserved quantities, that have to be the same before and after the decay.

1. Why do they decay at different times?

This is because the various interactions take place at different time scales. For an electromagnetic interaction, it takes around $$10^{-18}$$ s, whereas a weak interaction takes place during approximately $$10^{-10}$$ s. So an EM decay happens much faster. Now the neutral pion can decay both into two photons (EM) and into an $$e^-e^+$$ pair (weak). But since one of them happens much quicker, it almost always decay via the EM route.