Calculating lifetime of a pi meson via Heisenberg uncertainty relationship?

This is a problem from my textbook:

"A proton or neutron sometimes 'violates' conservation of energy by emitting and then reabsorbing a pi meson, which has a mass 135MeV/$c^2$. This is possible as long as the pi meson is reabsorbed within a short enough period of time $\Delta t$ consistent with the uncertainty principle. Consider $p \to p + \pi$. By what amount $\Delta E$ is energy conservation violated? (ignore any kinetic energies)"

For this, I thought it appropriate to take the "violation" of energy conservation's only constituent to be the rest energy of the pion, $E_0 = mc^2 = 135MeV$.

"For how long a time $\Delta t$ can the pi meson exist?"

Using the energy-time uncertainty relation, I do the following calculations: $$\Delta E \Delta t \approx \hbar$$ $$\Delta t \approx \frac{\hbar}{\Delta E}$$

Taking $\Delta E$ to be the rest energy of the pion calculated above, the resulting $\Delta t$ comes out to be $4.88 \times 10^{-24}s$. But checking the Wikipedia article on pi mesons, the lifetime is said to be $2.6 \times 10^{-8}s$. Is there something that I'm missing in this process? Was my choice of the rest energy for $\Delta E$ perhaps incorrect? Why am I coming out with the wrong value for the lifetime of the pion?

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Basically, DZ has given a good answer to this question. There are only a couple of points I would like to add from pure physics point of view, just for pedagodical purpose, if I may.

The time you are calculating is the time the $\pi^0$ is in 'flight' from the moment it is emitted by the proton until it is captured again. So it is ok to use the uncertainty principle to find an estimate. Since this is a strong interaction process it will be very short time, which you find. That does not mean the $\pi^0$ lives that short time. It is like you throw a fire cracker up in the air after it has been 'programmed' to explode in 1 min say. The fire cracker will fly upwards, reverse its motion and hit the ground in a fiew seconds, and then will explode - the $\pi^0$ does not do that when it is recaptured by the proton though. If the $\pi^0$ had the freedom to fly longer it would 'explode' in the programmed time of about $2.6\times10^{-8}s$. The free-$\pi^0$ life time is the time the particle actually lives in free flight so to speak. It is much longer than the other time because the mechanism that drives its decay is an electromagnetic interaction, as stated in DZ's answer. These two life times are calculated using different Feynman diagrams. I hope this clarifies the difference between the two 'lives' of the $\pi^0$.

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Thank you for the great answer! The extra details and the analogy helped a great deal. – Ataraxia Feb 24 '13 at 17:18

Without getting into too much technical detail: when Wikipedia gives the lifetime of the pion, it is talking about an on-shell pion, also known as a real pion, which is a particle-shaped wave in the pion field that satisfies the relation $E^2 - p^2 c^2 = m_\pi^2 c^4$, with the constant $m_\pi = 135\text{ MeV}$.

However, the "temporary" pions that are produced during a process like $p\to p\pi^0 \to p$ are known as off-shell or virtual pions. They are particle-shaped waves in the pion field for which $E^2 - p^2 c^2 \neq m_\pi^2 c^4$. It's one of the quirks of quantum field theory that waves like this are allowed to exist, but they're inherently unstable, in a sense; they dissipate away with a characteristic time scale related to the energy discrepancy $E^2 - p^2 c^2 - m_\pi^2 c^4$. In other words, the greater the difference between $E^2 - p^2 c^2$ and $m_\pi^2 c^4$, the quicker the disturbance disappears. For an energy discrepancy of $135\text{ MeV}$, which is quite large (in this context), you will find a very short characteristic time indeed.

Of course, you might then ask why an on-shell pion decays at all. If the energy discrepancy is zero, wouldn't the characteristic time be infinite? In fact it is; on-shell particles don't suffer from the inherent instability that I mentioned in the previous paragraph. But they can decay in a different way: for a neutral pion, it's an electromagnetic interaction between their constituent quarks. This electromagnetic process is entirely separate, and it has an independent (and much longer) time scale associated with it.

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this is true for pi0. You should add that charged pions decay by the weak interaction. – anna v Feb 9 '13 at 5:17
Since the question was about a neutral pion, I wrote my answer accordingly - but I've clarified that. – David Z Feb 9 '13 at 5:30