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For example, let's say a W boson decays into an electron and a neutrino.

Is there some transition? Does the boson fade out while the decay products fade into existence? Or does this happen in the form of a non-time-differentiable "pop"?

EDIT: Okay I see we cannot measure any transition. But what about the wave function? Is there a transition or happen non-continuously?

Calmarius
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    The decaying particle's wavefunction fades out, the products' wavefunction fades in. The "pop" occurs due to wavefunction collapse of QM when the detector takes a look. – QuantumDot Jan 16 '15 at 10:00

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The decay takes place instantly on a timescale longer than those affected by Heisenberg's Uncertainty principle. In the case of reactions involving virtual particles, the Uncertainty principle says that $\Delta E\Delta t=\frac{h}{4\pi}$. This means that, if the reaction involves a virtual particle such as a W boson as you mentioned, then the particle will have an unmeasurably short lifespan due to the high mass-energy of the W boson.

By looking at the Uncertainty principle again: the larger $\Delta t$ is, less energy can be exchanged in the reaction: see this answer to a similar question for more information.

Community
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Tom Galvin
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