Eigenvalue degeneracy in real physical systems

In summary: BillIn summary, according to quantum mechanics, degeneracies can be associated with symmetry or topological characteristics of the system. If a system has an odd number of electrons, for example, it will have at least a two-fold degeneracy. When all the operators in a system are represented by non-degenerate matrices, it is true that the eigenvalues are distinguishable. However, this is only true when all the observables in the system are measured. If not, the system is said to be in a superposition of different eigenstates and the collapse postulate must be taken with some grain of salt.
  • #106
ErikZorkin said:
Postulates are not formal axioms

For that matter, the article you cite in the OP is not a formal proof.
 
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  • #107
Then why are there still so many attempts at axiomatization of QM?
 
  • #108
atyy said:
For that matter, the article you cite in the OP is not a formal proof.

No, sir, it is!
 
  • #109
For this matter, such works as this are closer to what's called formal axiomatic foundation.

But I fear that this discussion went in a wrong direction a bit.
 
  • #110
ErikZorkin said:
Then why are there still so many attempts at axiomatization of QM?

There are two major lines of foundational research nowadays.

The first takes the Copenhagen interpretation (and does not attempt to solve the measurement problem), and the axioms as given for example in the articles by Paris and Busch as correct. The research is to find a more "intuitive" derivation of the axioms, eg. http://arxiv.org/abs/1011.6451 or http://arxiv.org/abs/1303.1538 or http://arxiv.org/abs/1403.4621.

The second tries to solve the measurement problem of Copenhagen, eg. Bohmian mechanics, Many-Worlds, Consistent Histories etc
 
  • #111
ErikZorkin said:
Consistency of arithmetic can't even be proven. I believe you mean not the consistency, as mathematicians define it but rather informally since it allows for accurate predicitons.

Precisely what has that got to with QM not being fully worked out yet? All physical theories are like that even when expressed in highly abstract mathematics such as Symplectic geometry. Why are you shifting context?

Thanks
Bill
 
  • #112
ErikZorkin said:
No, sir, it is!

Of course it is not. It is ordinary mathematical proof.
 
  • #113
atyy said:
Of course it is not. It is ordinary mathematical proof.

Well, since it concerns computable analysis, the proof is constructive and as such can be formalized automatically. That's called proof normalization. And there is software out there that does the job. It's the calssical proof of the spectral theorem that can't be ever formalized.
 
  • #114
ErikZorkin said:
For this matter, such works as this are closer to what's called formal axiomatic foundation.

But I fear that this discussion went in a wrong direction a bit.

Yes, the 3 links I gave in post #110 and that bhobba gave in #104 are in this spirit.

However, that people still work on deriving the axioms does not mean the axioms are not already at the same level of rigour as the article in the OP. It just means that people are looking for more "intuitive" ways to derive the axioms.
 
  • #115
bhobba said:
Why are you shifting context?

Well, if it was I, who shifted the focus, I apologize.
 
  • #116
ErikZorkin said:
Well, since it concerns computable analysis, the proof is constructive and as such can be formalized automatically. That's called proof normalization. And there is software out there that does the job. It's the calssical proof of the spectral theorem that can't be ever formalized.

Sure that's the same as I would say for the axioms of QM.
 
  • #117
atyy said:
Sure that's the same as I would say for the axioms of QM.

For God's sake of course NO!
 
  • #118
ErikZorkin said:
For God's sake of course NO!

Why not?
 
  • #119
ErikZorkin said:
For this matter, such works as this are closer to what's called formal axiomatic foundation..

I think the book I suggested does just that. Its based on the formal logic approach of Von Neumann who was hardly ignorant of such things. It starts out from logic in a formal sense.

Thanks
Bill
 
  • #120
May be way off base here, but is what you're asking equivalent to "are there physical systems for which a complete set of mutually commuting observables is known not to exist?"
 
  • #121
atyy said:
Why not?
Because those axiomatics that are used in common QM framework are not computable. But, again, there is no comprehensive formalization of QM.

ANYWAY, I'd like to stick with the spectral decomposition. So far, what I leanred from this thread is:

1. Spectral decomposition might be stated in approximate format:

For any Hermitian operator T any ε>0, there exist commuting projections P1, ... Pn with PiPj =0 and real numbers c1,...cn such that || T - ∑i=1nciPi || ≤ ε.

Whereas in doing so we take the risk of dropping off something important after projection

2. POVMs might be used instead of PVMs

POVMs are in bijection with PVMs. That's not the actual issue, as soon as we don't directly use the spectral theorem to construct POVMs. That is what I don't fully understand so far.
 
  • #122
ErikZorkin said:
Because those axiomatics that are used in common QM framework are not computable. But, again, there is no comprehensive formalization of QM.

If QM is not axiomatizable, how can you prove it is uncomputable?
 
  • #123
ErikZorkin said:
But, again, there is no comprehensive formalization of QM.

You keep saying that. Its false. I gave you the book that does just that ie develops QM from algebraic logics. The issue is QFT - not QM but that is a whole new thread.

I really do need to ask where you are getting this from because wherever it is is leading you astray.

Thanks
Bill
 
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  • #124
atyy said:
If QM is not axiomatizable, how can you prove it is uncomputable?

And even aside from that its irrelevant. Classical physics is uncomputable leading to chaos and the butterfly effect. So?

Thanks
Bill
 
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  • #125
As a simple example of what I mean, one can axiomatize number theory using Peano's axioms. It is of course true that no finite axiomatization can produce all true statements of number theory. However, defining number theory as Peano's axioms is good enough for almost all "experimental observations" - including Fermat's last theorem. We won't be able to get Paris-Harrington, but that just means we will lack knowledge of one of the "conservation laws", and we can figure out how to add it to our axioms if experiments start addressing the issue (analogous to how the second law of thermodynamics was discovered).
 
  • #126
ErikZorkin said:
POVMs are in bijection with PVMs. That's not the actual issue, as soon as we don't directly use the spectral theorem to construct POVMs.

PVM's are not the usual term used in QM - the more restrictive term resolution of the identity is usually used (its the influence of the great mathematician Von Neumann):
https://en.wikipedia.org/wiki/Borel_functional_calculus

Resolutions of the identity are simply disjoint POVM's. The reason POVM's are used is threefold. First they define general observations - resolutions of the identity only define so called Von Neumann observations. Secondly a very important theorem, Gleasons Theorem, is much easier to prove for POM's. Finally from an elegance viewpoint why impose the restriction of disjoint. Given a resolution of the identity Ei one can form the Hermitian operator O = ∑ yi Ei. This was all sorted out by Von Neuman.

Now your concern seems to be given a Hermitian operator actually determining the Ei. Of course that's a hard computational problem like many in physics but its not a big issue at the foundations of QM any more than the computability of differential equations in classical physics is - in fact its irrelevant. Its very important because like in classical physics it likely leads to things that are very interesting and of great practical importance like chaos - but foundationally its a non issue.

Thanks
Bill
 
  • #127
atyy said:
It is of course true that no finite axiomatization can produce all true statements of number theory.

That's of course true but of zero relevance to physics or applied math in general. The types of undecidable questions it can't answer is of zero relevance to physics - at least no one has found one. In computer science that's another matter - the halting problem is logically equivalent to Godel's theorem - but for physics its a non issue.

Thanks
Bill
 
  • #128
atyy said:
If QM is not axiomatizable, how can you prove it is uncomputable?

Who said it's not axiomatizable? I don't want to go deep into this discussion, but already the spectral theorem is based on classical lohic, which includes the exluded middle axiom among others. Result derived from such axiom are uncomputable in general.

bhobba said:
Now your concern seems to be given a Hermitian operator actually determining the Ei. Of course that's a hard computational problem like many in physics but its not a big issue at the foundations of QM any more than the computability of differential equations in classical physics is

Gleason's theorem admits a constructive proof. Differential equations, as soon as they fulfill Lipschitz condition, also have computable solutions.

I feel that the people here really think of finite approximations of some numbers. But computability is more general. For instance, there is an algorithm computes spectral decomposition up to arbitrary precision, but the final result will always be shifted by ε from the original operator.Regarding POVM. What's actually the logical sequence behind it? With PVM, you start off with a Hermitian measurement operator, spectral decompose it (problematic), measure eigenvalue and project the state using the projection corresponding to that eigenvalue.

With POVMs you seem to start with defining measurement operators that partition the unity somehow and then do straightforward procedures to compute probabilites and final states. Where do you actually use spectral decomposition here and who gives you the measurement operators?
 
  • #129
ErikZorkin said:
With POVMs you seem to start with defining measurement operators that partition the unity somehow and then do straightforward procedures to compute probabilites and final states. Where do you actually use spectral decomposition here and who gives you the measurement operators?

When Von Neumann first gave a rigorous account of QM it was based around resolutions of the identity. But further research showed that observations are more general than that being based on POVM's, but can be reduced to resolutions of the identity using the concept of a probe:
http://www.quantum.umb.edu/Jacobs/QMT/QMT_Chapter1.pdf

Because of that measurement theory in QM is now based on POVM's. The link I gave was a proof of QM foundations from POVM's and non contextuality - that's all there is formally to QM. Schroedingers equation etc (ie the measurement operators) comes from symmetry considerations - see Ballentine Chapter 1 to 3.

As mentioned previously operators enter into it from resolutions of the identity. If the resolution of the identity of an observation is Ei then by definition the operator associated with it is O = ∑yi Ei. yi is a real number arbitrarily assigned to outcome Ei. Its entirely arbitrary but of course if an obvious association is there then its used eg if Ei is outcome position yi then that is used.

That's all there is to QM foundations really. An even deeper justification is the following:
https://arxiv.org/pdf/quant-ph/0101012

It may seem like QM is pulled from the air so to speak. It isn't really, but that is a whole new thread.

Thanks
Bill
 
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  • #130
  • #131
bhobba said:
Because of that measurement theory on QN is now based on POVM's.

My question is, do you use spectral decomposition theorem to derive POVMs or if yes, where?

A side question: can you formulate the Stern-Gerlach experiment in terms of POVMs instead of PVMs?
 
  • #132
ErikZorkin said:
This is an informal foundation. Let's just finish discussing axiomatization of QM. That's not the topic of the thread.

Ok.

Then what is your issue precisely? That computing eigenvalues and eigenvectors is a computationally difficult problem? So?

Thanks
Bill
 
  • #133
bhobba said:
Ok.

Then what is your issue precisely? That computing eigenvalues and eigenvectors is a computationally difficult problem? So?

Thanks
Bill

It's not about "difficulty" (or what you would call complexity, if stated precisely). It's about the fact that, in general, eigenvectors/spaces and projections are UNCOMPUTABLE. There is no algorithm to compute them. If you reread the OP, you'll see why I am interested in computable versions of spectral theorem and why I am looking for more suitable explanation than Operator/Decomposition/Measurement/Projection.
 
  • #134
ErikZorkin said:
My question is, do you use spectral decomposition theorem to derive POVMs or if yes, where?

Of course. One starts with operators that come from symmetry considerations - see Chapter 3 Ballentine. The key physical assumption is those symmetry considerations lead to equations of exactly the same form as classical systems. So the natural assumption is that's how to quantise a classical system. That gives a differential equation whose solution gives the eigenvalues and eigenvectors. Sometimes its analytic - mostly it isnt. That's where one must use a computer. Also there is some very deep but not quite rigorous mathematics tied up in this:


I think you will find the above lectures very interesting and illuminating of stuff you seem interested in. It for example explains where quantisation comes from - its very very deep - but as I said not exactly rigorous..

ErikZorkin said:
A side question: can you formulate the Stern-Gerlach experiment in terms of POVMs instead of PVMs?

PVM's are not used in QM - its resolutions of the identity. A resolution of the identity is by definition also a POVM. One has outcome up |u> and outcome down |d>. |u><u| and |d><d| is the resolution of the identity. To get an operator you arbitrarily associate a number with each element of the resolution of the identity - say -1 with up and 1 with down so you have O = - |u><u| + |d><d|.

Thanks
Bill
 
  • #135
ErikZorkin said:
It's not about "difficulty" (or what you would call complexity, if stated precisely). It's about the fact that, in general, eigenvectors/spaces and projections are UNCOMPUTABLE. There is no algorithm to compute them. If you reread the OP, you'll see why I am interested in computable versions of spectral theorem and why I am looking for more suitable explanation than Operator/Decomposition/Measurement/Projection.

I have and I don't get your issue. They aren't computable but so what?

Thinking about it a bit more are you asking if it could be simulated on a computer - I would say - no. Feynman commented on this and reached the same conclusion which he always found rather amazing.

Thanks
Bill
 
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  • #136
bhobba said:
Thinking about it a bit more are you asking if it could be simulated on a computer - I would say - no. Feynman commented on this and reached the same conclusion which he always found rather amazing.

I am afraid this is a heavy misunderstanding of my question. Perhaps, it's my fault, I don't know. You seem to try to shift the focus all the time. I can't comment on philosophical aspects of whether so or so phenomenon is computable or not. There is a huge evidence for computable phyisics, however, which is the fact that anything can be measured only up to a finite precision. Since you can always (at least theoretically) increase accuracy, what you effectively do is you apply a definite procedure, or algorithm if you will, that yields effectively a computable number. Now, this number might not perfectly match with the idealistic theoretical model, but that's not a problem at all. If your theory is classical, it's speculative to say how you're supposed to compute and an uncomputable number since you always perform an algorithmic procedure. In fact, what you effectively do is you implicitly work within a constructive and computable framework that consistently describes the phenomenon whereas classical theory is simply an idealistic approximation thereof. Whether you like it or not, you can't play around with uncomputable numbers in real experiments.

But that's just a side remark which I am a bit afraid of having to do. I started with the words: let's pretend we simulate the universe. Would you argue that? Let's pretend there is a supercomputer that simulates all observed phenomena how we see it. Since we always measure something by a definite procedure and only up to finite precision, there is no fundamental limitation in making such a speculative hypothesis. Now, it turns out that in order to do so, you need to have computable procedures behind the simulation. Approximate spectral decomposition is such a procedure. And that's what actually use in reality. You can't measure a real number up to an infinite precision, roughly speaking. Effctively, there is always a bound.

Say, in Stern-Gerlach experiment, if the beam splitting is beyond Planck scale even if the beam traveled though the whole observable Universe, you can't deduce whether spin is up or down. So you land at a non-verifiable statement, scientifically speaking. You'd need a more rigorous theory.
 
  • #137
ErikZorkin said:
I started with the words: let's pretend we simulate the universe. Would you argue that?

Yes I would. But I am suspecting I am not the person to answer your query because I don't really get it. QM systems can sometimes be degenerate. I can say that 100% for sure. Degeneracy is not forced on us because it can be removed - but not in a natural way. But its relation to the other parts of your post I don't understand.

Thanks
Bill
 
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  • #138
ErikZorkin said:
Say, in Stern-Gerlach experiment, if the beam splitting is beyond Planck scale even if the beam traveled though the whole observable Universe, you can't deduce whether spin is up or down. So you land at a non-verifiable statement, scientifically speaking. You'd need a more rigorous theory.

Beam splitting is beyond Planck scale? What you mean by that beats me.

Thanks
Bill
 
  • #139
bhobba said:
Beam splitting is beyond Planck scale? What you mean by that beats me.

Thanks
Bill

Refer to my previous post.

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So, POVMs in case of Stern-Gerlach experiment would be the same as the projections obtained by spectral decomposing, Pα and Pβ?
 
  • #140
ErikZorkin said:
\So, POVMs in case of Stern-Gerlach experiment would be the same as the projections

Yes - that's basically what I said.

Thanks
Bill
 
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