Particle Lifetimes from Uncertainty Principle

Classical Mechanics Level 2

Suppose an exotic particle with a very large mass of $20 \pm 0.01 \text{ TeV}$ is measured at the Large Hadron Collider. What is the correct lower bound for the lifetime of this exotic particle given no additional information, in seconds? (What does this say about the detectability of very heavy particles?)

3.3 \times 10^{-3}

3.3 \times 10^{-26}

3.3 \times 10^{12}

3.3 \times 10^{-12}

1 solution

Matt DeCross
May 10, 2016

Relevant wiki: Heisenberg Uncertainty Principle

This problem is an application of the energy-time uncertainty principle,

$\sigma_E \sigma_t \geq \frac{\hbar}{2}.$

A decaying particle is measured by a resonance in a particle accelerator. The width of this resonance is the energy uncertainty; the faster the particle decays, the wider the resonance and vice versa.

With an energy (mass) uncertainty of $.01 \text{ TeV}$ , the "time uncertainty" corresponding to the lifetime of the decaying particle is therefore:

$\sigma_t = \frac{\hbar}{2 (.01 \text{TeV})} \approx 3.3 \times 10^{-26} \text{ s}.$

One can look at this as the following: the universe may have "borrowed" energy and lent it to the decaying particle. The amount of energy that it can lend in a period of time is inversely proportional to how long the particle sticks around for. If the particle decays very quickly, the particle may have borrowed very high energy, and the uncertainty in the particle's energy is correspondingly high and vice versa.

For very heavy particles, a reasonable percent uncertainty will be proportionally larger (i.e. an error of $.01 \text{ TeV}$ is much larger than an error of $.01 \text{ GeV}$ . Thus we should expect very heavy particles to have very short lifetimes, and they will be hard to detect as a result.

Particle Lifetimes from Uncertainty Principle

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