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$\pi^+$ 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?

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3 Answers 3

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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.

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  • $\begingroup$ There is also the slight possibility the neutral pion might decay into a Z boson. However it is rarely likely. Decays through electromagnetism happen faster than decays through weak force. $\endgroup$ May 18, 2020 at 14:49
  • $\begingroup$ @RoghanArun not a real Z which has a mass about 80GeV,whereas the pion is ony 0.135 GeV. Thus any weak decay is also suppressed by the mass of the Z. I think also quantum numbers would enter, as the pion is a combination of quark antiquark pairs hyperphysics.phy-astr.gsu.edu/hbase/Particles/hadron.html $\endgroup$
    – anna v
    May 18, 2020 at 15:12
  • $\begingroup$ It could still be a virtual(off-shell) Z boson which could violate energy conservation(because it is a virtual particle) and then could decay into other particles. $\endgroup$ May 18, 2020 at 15:14
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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.

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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, enter image description here

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...

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  • $\begingroup$ There is no chiral anomaly for the electromagnetism. Only the weak force violates CPT symmetry. What do you mean when you say chiral anomaly electromagnetically. $\endgroup$ May 18, 2020 at 14:51
  • $\begingroup$ You are misreading this: the current violated is not the electromagnetic current. The neutral flavor global axial current is violated by electromagnetism (the 2 photons)! Review your QFT text. $\endgroup$ May 18, 2020 at 14:55
  • $\begingroup$ So you are saying electromagnetism violates neutral flavor global axial current like the weak force violated CP and T symmetry. $\endgroup$ May 18, 2020 at 15:09
  • $\begingroup$ Absolutely not. CP and T are discrete symmetries, not anomalous ones. They are violated "classically" at the EW lagrangian level. They have nothing to do with anomalies, which are violations due to quantum loop effects. $\endgroup$ May 18, 2020 at 15:27

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