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In Griffith's Introduction to Quantum Mechanics, he said that " For unstable particle, that spontaneously disintegrates with a lifetime τ, the total probability of finding the particle somewhere should not be constant."

So, my question is, why disintegration of a particle leads to non constant value of probability, because according to me, if we aggregate all those disintegration, then we obtain the same particle, so why probability changes?

Problem 1.15, Page no. 22,

Second Edition,

Introduction to Quantum mechanics

Author :- David Griffith

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2 Answers2

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The text you are referring to continues like so (if we are looking at the same place): "In that case, the probability of finding the particle somewhere should not be constant, but should decrease at (say) an exponential rate: \begin{equation} P(t)=\int^{+ \infty}_{-\infty}|\Psi(x,t)|^2dx = e^{-t/\tau}." \end{equation} So what he means is that at $t=t_0=0$ (the moment of creation of the particle), the probability of finding the particle somewhere is $P(0)=1$ (it has not decayed). Once the clock starts ticking, there is a probability at time $t\neq t_0$ that the particle has already decayed, so it can not be found anywhere. That probability is governed by that exponential you see - the higher the value of $t$ is, the lower the probability of the particle existing (or equivalently finding it anywhere).

Vangi
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You're right, but then you're not talking about the probability of the particle that disintegrates but rather that of the particle that disintegrates and the particle(s) it disintegrates into. You have to make a choice of the quantum system first. In unstable particles, we talk about the original particle decaying. And this time evolution is not unitary (and hence, its probability of existence is not conserved). See also another answer that might help: https://physics.stackexchange.com/a/434912/133418

Avantgarde
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