To measure the lifetime of a specific particle one needs to look at very many such particles in order to calculate the average. It cannot matter when the experimentalist actually starts his stopwatch to measure the time it takes for the particles to decay. If he measures them now or in 5 minutes makes no difference, since he still needs to take an average. If he measures later there will be particles out of the picture already (those who have decayed in the last 5 min), which won't contribute and the ones his measuring now behave (statistically) the very same, of course.

I have just read the following in Introduction to Elementary Particles by Griffiths:

Now, elementary particles have no memories, so the probability of a given muon decaying in the next microsecond is independent of how long ago that muon was created. (It's quite different in biological systems: an 80-year-old man is much more likely to die in the next year than is a 20-year-old, and his body shows the signs of eight decades of wear and tear, But all muons are identical, regardless of when they were produced; from an actuarial point of view they’re all on an equal footing.)

But this is not really the view I had. I was imagining, that a particle that has existed for a while is analogous to the 80 year old man, since it will probably die (decay) soon. It just doesn't matter because we are looking at a myriad of particles, so statistically there will be about as many old ones as babies. On the other hand it is true that I cannot see if a specific particle has already lived long or not; they are all indistinguishable. Still I am imagining particles as if they had an inner age, but one just can't tell by their looks. So is the view presented in Griffiths truer than mine or are maybe both valid?

How can one argue why my view is wrong?

  • 4
    $\begingroup$ Griffiths is correct. Elementary particles has no age. No inner clock. They are all the same. $\endgroup$ – hpekristiansen Jun 28 '13 at 13:49
  • 8
    $\begingroup$ One need not argue that your view is wrong - it can be measured. You ARE wrong. The likelihood for a decay of a given particle is independent of how long you wait. $\endgroup$ – hpekristiansen Jun 28 '13 at 13:56
  • 3
    $\begingroup$ Given one particle you don't just not know the age (as always) you can't even say anything about the probability just from this one measurement. I don't see how my view contradicts any measurements. $\endgroup$ – Jack Jun 28 '13 at 13:58
  • 4
    $\begingroup$ @Hans-PeterE.Kristiansen I, and I think many here, look down on arguments from authority. Give us facts, logic, and reasoning. $\endgroup$ – Izkata Jun 28 '13 at 19:14
  • 4
    $\begingroup$ @Hans-Peter E. Kristiansen Yes, but you were referring to what I have quoted myself, because I didn't understand it. If you just say I should trust that it is true anyway, you are not adding anything helpful whatsoever. $\endgroup$ – Jack Jun 28 '13 at 20:04

It's impossible to say whether you are correct or Griffiths is correct a priori -- that is, before having any experience of how the world works. You need to do experiments, and Griffiths' version agrees with experiments better than yours.

The basic experiment involves detecting the products of decayed particles. Suppose we have some process that happens pretty quickly and produces a certain amount of some unstable particle. The particles were all produced at basically the same time, so they all have basically the same "age".

Now, if we just measure how many decays happen per second, Griffiths' version predicts that you'll see an exponential falloff in that number -- as Griffiths does a fine job of explaining. And that's what we actually see when we do experiments like this.

In your version, you would expect to see very few decays until some fixed time after the production, then a sudden rush of decays, and then the decays would mostly stop because the particles would basically all be gone. But that's just not what we see.

Again, there's no reason the laws of the universe have to work the way Griffiths says, and not work the way you say. It's just that the predictions of the two versions are testable, and only Griffiths' version agrees with experiments. That's science!

  • 2
    $\begingroup$ Ah, I see, that would be true. Okay, I understand now, thank you! $\endgroup$ – Jack Jun 28 '13 at 14:26
  • 1
    $\begingroup$ I love this answer. It gets straight to the point. But I would like to add that this also follows pretty inexorably from two very basic theoretical assumption (i) my state can be deduced from my state in the previous moment, i.e. time evolution is local and (ii) elementary particles have no internal structure. (i) is a pretty ubiquitous feature of fundamental theories and (ii) is almost the defintion of "elementary" $\endgroup$ – BebopButUnsteady Jun 28 '13 at 18:25
  • $\begingroup$ Thanks for the love. :) And I agree with you. But I'll also point out that both of those assumptions are only judged to be valid and relevant to what we call elementary particles a posteriori. For example, philosophers debated the existence of elementary particles for a long time before scientists found evidence that nature is well described by models involving them. And we always need to test whether a given "particle" actually has internal structure or not. E.g., protons do, but quarks don't seem to... $\endgroup$ – Mike Jun 28 '13 at 18:49
  • 4
    $\begingroup$ How do you know that the particles don't have pre-determined life spans, just selected from a random distribution with an exponential probability density function? $\endgroup$ – AJMansfield Jun 28 '13 at 20:44
  • $\begingroup$ @AJMansfield That's a great question (and deserves to be asked on its own). My brief answer is that we don't really know. However, we do have theories that work well (though not perfectly) without needing such a "hidden variable". In particular, Bell's Theorem from standard quantum mechanics seems to rule out the particular type of hidden variable you refer to. Of course, Bell's Theorem is not universally accepted, and there's plenty of wiggle room. So, again, we don't "know", but our models work well without such a construct. $\endgroup$ – Mike Jun 28 '13 at 21:54

Your thinking is analogous to the Gambler's Fallacy.

That is, the false belief that because HEADS hasn't come up in the last six coin tosses, it is somehow more likely to come up on the next toss.

The truth is that the event is independent of any prior events.

However, that doesn't mean that there aren't any old particles out there.

With numbers of astronomical magnitude, there's going to one coin that has come up heads for the entire 13 billion years of it's existence.

  • 1
    $\begingroup$ I don't think that this is how I am thinking. The question is more intrinsic. If it is so that a particle decay is like tossing a coin, then of course what Griffiths says follows. If it had an "inner age", then the analogy wouldn't be valid. $\endgroup$ – Jack Jun 28 '13 at 14:06
  • $\begingroup$ Where would the inner age be stored? In human it is (in short), the DNA molecules getting shorter that is related to aging. $\endgroup$ – Bernhard Jun 28 '13 at 14:21
  • 3
    $\begingroup$ It doesn't really matter how an inner age could be stored - you can rule it out regardless. It is a matter of every day experience at accelerator labs that you can create new particles which subsequently decay. The decay times are always consistent with the exponential decay law. For example, Tc99m is an important radioisotope for medical imaging which is produced in particle accelerators. Turns out the decay law works right for it. A close analogy is the radiative decay of electronically excited atoms. You can "reset the clock" on these at will and watch the decays. There is no "inner age." $\endgroup$ – Michael Brown Jun 28 '13 at 14:26
  • $\begingroup$ @Jack: I was going to say the same thing. "I was imagining, that a particle that has existed for a while is analogous to the 80 year old man, since it will probably die (decay) soon." That is a common probabilistic fallacy. A fair die has a $1/6$ chance of coming up $6$ no matter what the history of rolls is. Otherwise it would be a different kind of stochastic process (autoregressive), not uniformly random. The rest of what you said isn't related to the gambler's fallacy. $\endgroup$ – isomorphismes Jun 28 '13 at 19:15
  • 3
    $\begingroup$ The problem is rather that I really did not know/understand that the particle actually behaves in such a way. It was counter-intuitive for me that this works that way. This is why Mike's answer was really helpful, he has convinced me that the experiments really show that the lifetime of particles is like throwing dice rather than comparable with lifetimes of human beings. $\endgroup$ – Jack Jun 28 '13 at 19:22

It all comes down to indistinguishability. Even in principle we can not differentiate one identical particle from another. If particles had some sort of inner workings for measuring time, than this piece of information would have to be encoded in them somehow. This would then be the differentiating factor, which contradicts with indistinguishability.

On a deeper level, all particles are really excitations of all-permeating quantum fields. So the appropriate analogy would be imagining a sheet in the draft, which would result in a bump propagating along the sheet. There is no sense in talking about the age of the bump, because the bumps at different places are different featureless entities, i.e. particles are just manifestations of the more fundamental underlying fields. Also in a further analogy to this video, you could say that determining the age of a particle is the same as determining the age of number $3$ (and not in a way that has relation to human culture).


I'd like to add (1) something I'd like the OP to take away from this discussion and also (2) to add a "counterexample", whose contemplation will serve to strengthen the other answers, which I must say are all excellent.

  1. Firstly, it should be emphasised that exponential distribution is the $\mathbf{unique}$ distribution that is memoryless in the sense spoken about in Mike's, Chris's and Manishearth's answers. In other words, to experimentally verify that a particle doesn't "heed its age", you look for the exponential distribution. If you see it, then you are observing memorylessness, as stated in the other answers. But it's stronger than this: if you don't see an exponential distribution, you $\mathbf{know}$ there is some memory of age present. To understand this uniqueness, we encode the memorylessness condition into the basic probability law $p(A,B) = p(A) \, p(B|A)$. That is, suppose after time $\delta$ you observe that your particle has not decayed (event A). If $f(t)$ is the propability distribution of lifetimes, then the probability the partcile has lasted at least this long is $1-\int_0^\delta f(u)du$. The $a\, priori$ probability distribution function that the particle will last until time $t+\delta$ and then decay in the time interval $dt$ is $f(t+\delta) dt$. (This is events $B$ and $A$ observed together, which is the same as plain old $p(B)$ since the particle cannot last unti time $t + \delta$ without living to $\delta$ first!) Hence the conditional probability density function is $p(B|A) = \frac{f(t+\delta)\,dt}{1-\int_0^\delta f(u)du}$. But this must be the same as the unconditional probability density that the particle lasts a further time $t$ measured from any time, by assumption of memorylessness. Thus we must have $\left(1 - \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta)$, for all values of $\delta$. Letting $\delta\rightarrow 0$, we get the differential equation $f^\prime(t) = - f(0) f(t)$, whose unique solution is $f(t) = \frac{1}{\tau}\exp\left(-\frac{t}{\tau}\right)$. You can readily check that this function fulfills the general functional equation $\left(1 - \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta)$ for any $\delta > 0$ as well.

  2. There are "particles" that do remember their age, although they're not fundamental particles and almost certainly don't qualify for what the OP is thinking. But they illustrate the other answers by showing what fundamental particles would need if they were to remember their age. If we think of an excited fluorophore (instead of a quantum field in a raised state), then fluorophores generally undergo one or more changes of state in their fluorescence process. We can think of this as a psuedoparticle - a quantum superposition of free photons and raised matter states - in the same way as a polariton is thought of as a pseudoparticle. Real fluorophores are more complicated - the quantum superposition involves states other than simply the primary excited state and photon, so there is an internal state to record the "particle's" "age". I have drawn below a schematic diagram of the energy levels for something like fluorescein below. The fluorophore generally gets raised to a higher level than it will fluoresce from, and thus undergoes a series of decays between these higher states before dropping back to the ground state (or, more often, something in a band just above the ground state). So the total fluorescence lifetime is the sum of several, memoryless pdfs: the total pdf - being the pdf of the sum of exponential distributions - is the convolution of all the individual exponential distributions.

Fluorescein fluorescence

If there is one dominant higher energy state with lifetime $\tau_1$ and the main fluorescence transition has lifetime $\tau_2$, then the pdf for the overall lifetime is $\frac{1}{\tau_1\,\tau_s}\int_0^t e^{-\frac{u}{\tau_1}} e^{-\frac{t-u}{\tau_2}} du =\frac{e^{-\frac{t}{\tau_1}}-e^{-\frac{t}{\tau_2}}}{\tau_1-\tau_2}$ and I have drawn a sample function of this kind for $\tau_1 = 1$ unit and $\tau_2 = 10$ units. Mostly, raised fluorescence states look like memoryless particles in practice because the higher states are so shortlived compared with the lowest singlet state, but there are some for which the behaviour below is quite observable: i.e. there is a time throughout which an excited population is quiet, then the fluorescence comes with a rush, then goes quiet again.



If you feel that the particle should decay faster because it has already lived long and may be approaching or may have passed the average lifespan, this is the Gambler's Fallacy.

The average age of a particle can be derived from the knowledge that the particle has an x% chance of decaying in every small interval of time.

For example, if you have a die, and you stop rolling the minute you get a 6, the average lifespan of your game is around 5 rolls.

However, when calculating said average, you include the possibility that the game end in a single roll. And the possibility that it ends in two rolls (etc etc).

Now, let's say that after a single roll of the die, you get a number that is not a 6. At first, it seems obvious that the game will probably end in around 4 rolls.

However, there's a catch here -- you now have some information, and that information is that the first role was not a 6. Probabilities change as you get more information. Your initial probability calculation doesn't apply anymore as it assumed that there was a chance for the first roll to yield a 6.

The same concept applies here. Experimentally, particles follow an exponential decay distribution (The probability that a particle decays in time $t$ is $Ae^{-\lambda t}$, in other words there's a $\lambda$ probability of the particle decaying in any given second). The decay distribution supports the fact that the "probability of a particle decaying in the next moment" is constant, and thus the particle has no "age" that affects the decay.

Either way, an "age" would become a degree of freedom, which would affect the thermodynamic properties of the particle.

Also, when you model these particles mathematically, they all come out to be equivalent. I can swap two muons and I haven't changed the system at all.

  • 1
    $\begingroup$ It's not the gambler's fallacy. $\endgroup$ – Ben Jun 28 '13 at 16:47
  • $\begingroup$ @Ben Care to explain? $\endgroup$ – Manishearth Jun 28 '13 at 16:49
  • 1
    $\begingroup$ Read Jack's comment to Chris' answer. $\endgroup$ – Ben Jun 28 '13 at 18:12
  • $\begingroup$ @Ben OK, I'll edit the first sentence, but the answer is still valid as it addresses that point in the end. $\endgroup$ – Manishearth Jun 28 '13 at 18:13

Let me put it this way:

Over it's lifetime nothing inside the particles changes.

The only reason it dies is truly random - something inside it randomly tunnels out of the particle, and it no longer exist in previous form. Probability of this random process is equal for each and every nanosecond of particle's life.


You quoted the following phrase from Griffiths' book: "elementary particles have no memories, so the probability of a given muon decaying in the next microsecond is independent of how long ago that muon was created." It follows from this phrase that decay is exponential (the number of surviving particles depends on the time exponentially). This is a very good approximation, however, strictly speaking, this is not a precise law. As Khalfin noted half a century ago (please see some references in http://arxiv.org/abs/quant-ph/0408149 ), exponential decay is, strictly speaking, incompatible with quantum mechanics, so there must be deviations from exponential decay for very short and very long times. Such deviations have been found experimentally for very short times (http://george.ph.utexas.edu/papers/tunnelling.pdf ), but not for very long times (as far as I know): the deviations for very long times are difficult to observe, but there is little doubt that such deviations do exist, as quantum mechanics is in very good agreement with experiments.

As for Griffiths' phrase, I think it is warranted anyway because it is an established practice to give a simplified description in textbooks.

EDIT (07/07/2013): Looks like nonexponential decay of metastable states for long times was demonstrated experimentally: http://dro.dur.ac.uk/4234/1/4234.pdf .


protected by Qmechanic Jul 7 '13 at 6:49

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.