Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's suppose I'm in the lab and I claim that I can predict more than QM can, specifically, I can predict exactly at which moment in time a particle decays. You don't believe me (naturally) so I set up the experiment, provide a piece of paper with a time written on it, and start the clock. At the time I have written down, the particle decays.

Exactly which of the six postulates of QM would this violate? As far as I can tell, it violates none of them so long as the results from multiple identical trials of this experiment reproduce the correct particle decay time distribution.

(And yes, I'm aware of this paper http://arxiv.org/abs/1005.5173, but I would prefer a simpler explanation.)

share|improve this question
3  
"I claim I can predict how this coin will land. It will be heads." *Tosses coin. It comes up heads.* . "See? I can beat statistics. I don't need to repeat this experiment, I have made my point." What would you do if the particle had not decayed at the time you said? Blame experimental error? That's what I do when my coin demo fails... –  Floris Aug 19 at 19:21
2  
You read my post far too literally. Or you're pretending to read it too literally. –  Nick Aug 19 at 22:48
2  
I'm sorry - I thought that was how science works. –  Floris Aug 19 at 22:59
1  
There are six postulates of QM? –  Bubble Sep 27 at 22:07
2  
@Floris: Have you noticed this part of the question: "... so long as the results from multiple identical trials of this experiment reproduce the correct particle decay time distribution"? –  bright magus Dec 12 at 10:42

4 Answers 4

Exponential decay curves appear in all scientific disciplines and are a way of studying/modeling samples changing with time in a simple manner.

exponentialdecay

A quantity undergoing exponential decay. Larger decay constants make the quantity vanish much more rapidly. This plot shows decay for decay constants (λ) of 25, 5, 1, 1/5, and 1/25 for x from 0 to 5.

This quantity may be used in a number of classical seteups as long as the supposition of :

expdec

the proportionality of the change in numbers with time as above , holds.

To get it experimentally many instances have to be recorded . In the case of particle decays it is a probability curve that appears as the solution of quantum mechanical equations. In the case of pharmacology, for example, it is a probability curve for the effects of a specific medication and the exponetial comes from simple assumptions of behavior. ( no esoteric differential equations).

In both cases it is useful in describing a sample, but useless in predicting a specific instance except statistically.

Let's suppose I'm in the lab and I claim that I can predict more than QM can, specifically, I can predict exactly at which moment in time a particle decays.

You have to find an individual particle, and wait for it to decay.

You don't believe me (naturally) so I set up the experiment, provide a piece of paper with a time written on it, and start the clock. At the time I have written down, the particle decays.

Exactly which of the six postulates of QM would this violate? As far as I can tell, it violates none of them so long as the results from multiple identical trials of this experiment reproduce the correct particle decay time distribution.

Then you need a second particle, a third, a fourth ..... Getting the exponential distribution is not the problem. The violation of the concept of Quantum Mechanics is in

At the time I have written down, the particle decays.

Either you are a metaphysical ( outside the experimental setup) seer or you have an alternative theory to Quantum Mechanics as QM only predicts probability curves, the Born rule.

Basically it is against the basic tenet that associates a measurement to the solutions of the Schrodinger equation via the square of the amplitude, as a probabilistic measurement.

As Floris said, throwing dice and claiming to know every throw goes against statistical probabilities. Your knowing every throw of the decay goes against quantum mechanical probabilities.

At the moment there are no successful theories to replace quantum mechanics, and the ones that have tried just reproduce it in a more convoluted, special and far more complicated ( as Bohm's model ) way.

share|improve this answer

If in the Stern-Gerlach experiment we prepare an ensemble of particles in a superposition of spin up and spin down with respect to the z-axis and you could predict in every instance if we would obtain either spin up or spin down with respect to this axis but still reproduce the statistics corresponding to that superposition, there would at first sight be no real problem. However, according to quantum mechanics I could use the unitary operator on the state you provided before we measured the state and find out that we didn't actually start off in the superposition which leads to the contradiction in presuming we actually prepared the superposition in the first place. Then before we measured I could decide to measure along, for example, the x-axis and get results inconsistent with the preparation of a superposition.

For the case of the radioactive decay it is a bit less clear to me, but I would say that you could possibly determine the time-dependence of the metastable state.

share|improve this answer
    
"I would say that you could possibly determine the time-dependence of the metastable state" ... isn't the whole point of this question that you can't do this? Imagine a hydrogen atom prepared in an excited state, can one predict when it decays from this metastable state? The linked paper seems to state definitively no. –  CuriousKev Sep 27 at 22:38
    
That is exactly what I mean, but the question states that I were to be given the exact time of decay, in which case I would be able to use that information to do a unitary transformation to get a more defined initial state of the metastable state before it has decayed. –  Jasper Sep 27 at 22:53

There is nothing wrong about predicting the results of individual measurements in QM, for instance Bohmian mechanics gives the same results of QM(and more) but is a deterministic theory.

The problem arises when you take into account locality, however locality is not part of QM.

(rev1) As it has been already pointed out, QM talks about probabilities and expectation values, even though Born's rule say something about the probability of individual measurements the way you check probabilites is measuring a lot of times.

You can't contradict QM predicting individual measurements because QM don't say anything about individual measurements.(the only way might be predicting a value with zero probability)

Bohmian mechanics is a theory that reproduces the same results of QM but gives more because you can actually predict the value of individual measurements, however this theory is far from experimental verification.

Predicting values is related with Bell's theorem because that implies the existence of a function that allows you to predict the values $$ f(a,\lambda) $$ where $a$ are the parameters you know from experiment and $\lambda$ the parameter(s) that allows you to predict the individual measurements. Experiments violate Bell's inequalities so \lambda need to be non-local.

share|improve this answer
    
This is probably short enough to be a comment, and should include some more detail to be a complete answer. –  Sean 11 hours ago

Particle decay is not a good example here bc a lot of the stochastics come from not following the nuclear dynamics closely. A better example might be the decay of a metastable state in an atom. There we have an atom that is not in an energy eigenstate by itself but the whole system is. There are well studied toy models of an atom in a quantized electromagnetic field that allow us to predict the decay rates. But prior to measurement, the system is still in a superposition of decayed and not decayed. Measurement is what seems to be the culprit, then. But if you could predict the outcome of your measurement, then the most glaring postulate you would violate would be unitarity.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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