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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

My Essential Learning
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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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  • Suppose, I have a matchstick and after lighting my matchstick , it start to disintegrate and if I collect all the ashes and smoke, then the number of atoms in the sample does not changed, it just changed its form. Similar to that, if I have a particle, if it disintegrate, then all little fragments after addition would make that that particle that had been disintegrated, according to me... or in other words, the probability would be 1 even after disintegration... I just want to know what makes the probability to be not equal to 1. Mathematically, I understood, I just want to know physical mean. – My Essential Learning Mar 31 '23 at 10:51
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    If you collect all the particles which result from the particle decay (or "disintegration") you are able to use kinematics in order to deduce the initial particle's properties e.g. mass, momentum. But you can not claim that you have found the particle itself - you have found the products of its decay - or as in your analogy, you would be collecting ash. E.g. if you have a decay of Carbon-14 into Nitrogen, $^{14}_6Ca\rightarrow^{14}_7N+e^-+\bar{\nu}$, you cannot claim that collecting the 3 products is the same particle as the carbon atom. – Vangi Mar 31 '23 at 11:13
  • All my doubts are clear. Thank you – My Essential Learning Mar 31 '23 at 11:15
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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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