# Deriving the statistical interpretation from Schrodinger's equation?

1. Jan 14, 2014

### pantheid

So, there are two things in Quantum Mechanics that I understand are axioms: the first is the schrodinger equation, which cannot be derived. Okay fine, we have to start somewhere. The second axiom is that the integral from a to b of the wavefunction-mod-squared gives the probability of finding the particle between a and b. My question is: Is there any framework that can derive the statistical interpretation just by manipulating the schrodinger equation and building on other principles, or is this just treated as fundamental?

2. Jan 14, 2014

### tom.stoer

There are claims that it's possible to derive the Born-rule in the Many-Worls-Interpretation. However it is still open whether these statements are true.

We had a some discussions here in the forum, but I don't think that anybody can provide a sound and complete result.

3. Jan 14, 2014

### dextercioby

Schrödinger's equation is a theorem in a symmetry based axiomatization of QM, following the ideas of Weyl and Wigner. The symmetry based axiomatization starts off with Born's rule regarding the statistical nature of the mathematical objects describing the quantum states.

4. Jan 14, 2014

### jcsd

In Bohmian mechanics the Born rule is derived (given the odd assumption). One interesting feature though Bohmian mechanics it allows for the possibility of a system to be in a state of quantum non-equilibrium where the Born rule is not obeyed.

5. Jan 14, 2014

### pantheid

but then that's not a derivation, thats just an assumption that it works.

6. Jan 14, 2014

### atyy

Two leading approaches for deriving the Born rule are:

1) Deutsch and Wallace's decision theoretic approach
http://arxiv.org/abs/quant-ph/9906015
(Proc. R. Soc. Lond. A 8 August 1999 vol. 455 no. 1988 3129-3137)
http://arxiv.org/abs/0906.2718

2) Zurek's quantum Darwinism
http://arxiv.org/abs/0707.2832
http://arxiv.org/abs/0903.5082
(Nature Physics, vol. 5, pp. 181-188 (2009))

I believe both are best seen within the many-worlds interpretation, but it is not entirely clear which interpretation Zurek is using. As tom.stoer says, there is no consensus about their correctness.

Last edited: Jan 14, 2014
7. Jan 14, 2014

### jcsd

It depends on whether you think the underlying assumption is reasonable or not.

8. Jan 15, 2014

### tom.stoer

Even Wallace admits that it's still open whether his approach does succeed.

9. Jan 15, 2014

### bhobba

Exactly.

Personally I like Gleason's Theorem:
http://kof.physto.se/theses/helena-master.pdf

But that has an assumption - basis independence (physically this means non-contextuality).

Dextercioby also hit the nail on the head - while one can axiomatize QM in various ways most exposed to it would say the approach with the greatest elegance is to use Born's rule and symmetry to derive Schrodinger's equation.

You will find this approach in Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

He develops it from 2 axioms. The first axiom is the observable postulate (ie the eigenvalues are the possible outcomes of an observation) and the second is Born's rule.

Interestingly, via Gleason's Theorem, you can derive Born's rule from the first axiom, so QM is really just one axiom. Obviously that's a crop of the proverbial - more than one axiom is required. Its just in that approach the rest are derivable from quite reasonable further assumptions, such as the probabilities from Born rule does not depend on inertial frames.

Still it's very interesting to see just what the real key non intuitive assumption of QM is and that the rest really follow from that in a reasonable way.

Thanks
Bill

Last edited by a moderator: May 6, 2017
10. Jan 15, 2014

### Demystifier

11. Jan 15, 2014

### tom.stoer

I don't agree.

Gleason's theorem says that if a probability measure shall be introduced, then it must comply with Born's rule. But Gleason's theorem does not say that you have to introduce a probability measure at all.

12. Jan 15, 2014

### Demystifier

This indeed is an important assumption of the Gleason's theorem, but there is an even more important one: The aditivity of the expectation values for commuting observables.

As shown by Bell, hidden variable theories (such as the Bohmian one) may violate this assumption in general, and yet be compatible with all measurable predictions of QM in situations when measurements are performed.

13. Jan 15, 2014

### bhobba

I suspect you are thinking about the error in Von-Neumanns hidden variable proof which made that assumption and not Gleason's Theorem which has much weaker assumptions. The only assumption is the measure must be basis independent.

I have posted the proof - you can check it for yourself - but its well known - basis independence is innocuous mathematically, and more or less required by the fact you are dealing with a vector space - physically though it implies non-contextuality which is far from trivial.

Thanks
Bill

Last edited: Jan 15, 2014
14. Jan 15, 2014

### bhobba

Are you seriously doubting the expectation of an observation will reach a stable value?

Kolmogorov's axioms follow from Cox's axioms. Do you seriously doubt Coxes axioms can't be applied?

Of course they are assumptions but I suspect most would put them in the innocuous category.

Thanks
Bill

15. Jan 15, 2014

### atyy

Even in the context of many-worlds? (I agree there's a good argument they can, but is it really clear that the argument is completely correct and without flaw?)

16. Jan 15, 2014

### tom.stoer

Bill,

Gleason's theorem says that there is one unique probability measure on Hilbert spaces. But Gleason's theorem does not say that you must introduce a probability measure at all. You can use Hilbert spaces for many other purposes, not only QM, and in these cases you don't introduce a probability measure. The fact that you want to introduce a probability is a matter of interpretation or applicability of the Hilbert space formalism to nature.

So first you have to use two axioms like
1) QM uses Hilbert spaces
2) QM makes probabilistic predictions
Then you can use Gleason's theorem which tells you which measure to use.

Suppose you formulate classical electrodynamics using Hilbert spaces. Does Gleason's theorem force you to introduce a probability measure for electrodynamics? I would say "no".

17. Jan 15, 2014

### bhobba

The operator rule says the outcomes of an observation is an eigenvalue of the operator. It implies you get outcomes with each observation.

Are you seriously doubting, when given outcome values you can't apply probability to analyse those outcomes? If so many areas of applied mathematics go down the gurgler such as actuarial science, weather forecasting, econometrics, the list goes on and on. Its such a trivial assumption that no text in such areas, to the best of my knowledge anyway, even state it as an assumption. It is of course, but its so utterly obvious no one elevates it to that status. But for some reason in QM, there are those, when confronted with observational data, say its a real issue.

Sorry, my background in applied math tells me its so trivial it does not rate a mention as a key assumption.

Thanks
Bill

Last edited: Jan 15, 2014
18. Jan 15, 2014

### bhobba

In many worlds one interprets it as a confidence level you are in a particular world obeying Cox's axioms and derives probabilities that way.

What I am talking about here is from the formalism - not a particular interpretation. Its simply that from the first axiom, that the outcome of an observation is an eigenvalue of the observable, you get data from carrying out the same observation under that same conditions a number of times, trials or whatever terminology you want to use.

Its utterly trivial that one can do statistical analysis of such data assuming values have a certain provability of occurring.

My background is in applied math where I studied such things as mathematical statistics and stochastic models. That one can do such things is considered so trivial it is assumed without even mentioning it that values can be assigned a probability and those probabilities must add up to one.

I believe your background is biology. Do you seriously doubt we cant do things like assign a probability to the number of offspring a member of a population will have? Of course its an assumption, but its so utterly obvious its doubtful anyone would even think of questioning it. And if you did its so widely used in such areas as weather forecasting and actuarial science you would have to really push it convincing anyone it even debatable.

Thanks
Bill

19. Jan 15, 2014

### bhobba

Does EM have the axiom of QM that observables are Hermitian operators and the possible outcomes of observations are the operators eigenvalues? Its an axiom from which you get data - data implies you can assign probabilities.

Imagine someone has given you a sequence of numbers and said they can only be certain values and asked you to analyse them. Don't you think you would get a strange look if you said - can I assume a probability can be assigned to a particular numbers occurrence?

Thanks
Bill

Last edited: Jan 15, 2014
20. Jan 15, 2014

### tom.stoer

Bill, you don't get the point.

In http://arxiv.org/abs/quant-ph/0405161 I found the following text statement:

That's exactly my point.

Gleason's theorem says that if you want to construct a probability measure on a Hilbert space, then the probability measure is uniquely determined. But Gleason's theorem does not tell you why you should introduce a probability measure at all. There are applications for separable Hilbert spaces in other branches of physics, and in these other branches you do not introduced probabilities. This shows that Gleason's theorem alone is not sufficient to explain why you should do that.

Anyway, the assumption may be trivial, but it is an assumption. No mathematical theorem about a mathematical structure forces you to interpret this mathematical structure in a certain way, or to interpret it at all.

Last edited: Jan 15, 2014
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