Undergrad Eigenvalue degeneracy in real physical systems

[ErikZorkin]

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, ... P_n with P_iP_j =0 and real numbers c₁,...c_n such that || T - ∑_i=1ⁿc_iP_i || ≤ 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.

 

[atyy]

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?

 

[bhobba]

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

 

[bhobba]

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

 

[atyy]

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).

 

[bhobba]

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

 

[bhobba]

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

 

[ErikZorkin]

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.

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?

 

[bhobba]

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

 

[ErikZorkin]

https://arxiv.org/pdf/quant-ph/0101012

This is an informal foundation. Let's just finish discussing axiomatization of QM. That's not the topic of the thread.

 

[ErikZorkin]

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?

 

[bhobba]

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

 

[ErikZorkin]

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.

 

[bhobba]

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..

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

 

[bhobba]

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

 

[ErikZorkin]

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.

 

[bhobba]

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

 

[bhobba]

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

 

[ErikZorkin]

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

Thanks
Bill

Refer to my previous post.

So, POVMs in case of Stern-Gerlach experiment would be the same as the projections obtained by spectral decomposing, P_α and P_β?

 

[bhobba]

\So, POVMs in case of Stern-Gerlach experiment would be the same as the projections

Yes - that's basically what I said.

Thanks
Bill

 

[ErikZorkin]

Yes - that's basically what I said.

Thanks
Bill

Ok. So, the conclusion is, POVMs are not the answer.

This is the best attempt so far:

In order to do physics, we only need to know the eigenvalues to the precision of the measurement apparatus. We don't need to know the multiplicity, since we need to project onto the space of states that are close enough to the measured eigenvalue. If the numerics gives us many eigenspaces for eigenvalues close enough to the measured value, we would project onto their direct sum. If the numerics gives us fewer, degenerate eigenspaces, we would also project onto their direct sum, but we would need fewer projectors. In both cases, the numerics would provide us with a sufficiently good projector, even though we might not know whether it projects onto degenerate or non-degenerate eigenspaces.​

 

[A. Neumaier]

Ok. So, the conclusion is, POVMs are not the answer.

This is the best attempt so far:

In order to do physics, we only need to know the eigenvalues to the precision of the measurement apparatus. We don't need to know the multiplicity, since we need to project onto the space of states that are close enough to the measured eigenvalue. If the numerics gives us many eigenspaces for eigenvalues close enough to the measured value, we would project onto their direct sum. If the numerics gives us fewer, degenerate eigenspaces, we would also project onto their direct sum, but we would need fewer projectors. In both cases, the numerics would provide us with a sufficiently good projector, even though we might not know whether it projects onto degenerate or non-degenerate eigenspaces.​

This is correct as far as it goes but to be realistic one needs POVMs rather than projections. If you simulate the universe with projections it will be very far from the real universe.

 

[ErikZorkin]

This is correct as far as it goes but to be realistic one needs POVMs rather than projections. If you simulate the universe with projections it will be very far from the real universe.

Well, but, at least in case of discrete variables (and in a simulation, they will all be discrete in fact), POVMs simply coincide with PVMs, do they?

Also, and that's what I am trying to understand, how is it easier to simulate a world with POVMs if you still have to deal with uncomputable spectral decompositions? I'd understand if you could define POVMs which consistently describe measurement up to a finite precision.

 

[A. Neumaier]

Well, but, at least in case of discrete variables (and in a simulation, they will all be discrete in fact), POVMs simply coincide with PVMs, do they?

No. POVMs are far more general than their projection-valued special case.

Also, and that's what I am trying to understand, how is it easier to simulate a world with POVMs if you still have to deal with uncomputable spectral decompositions?

There are no spectral decompositions in a POVM unless you define the latter in terms of one.

Anyway, you seem not to be interested in simulations by present-day computers - if you were you would ask quite different questions.

But if the simulation is the one done by God to run the actual universe - God is not bound by Turing machines to our human concept of computability. For example, God could employ hypercomputers in which the halting problem for Turing machines is solvable, or where one can process an infinite number of statements in a finite time, etc..

 

[ErikZorkin]

There are no spectral decompositions in a POVM unless you define the latter in terms of one.

What worries me a bit is that there are often answers directly contradicting each other.

For instance, I asked:

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

bhobba: Of course.

Regarding the simulation, let's assume, as I said in the OP, that computational resources are big enough, but finite. So any realizable process would be computable.

Goind back to Stern-Gerlach and too narrow beam splitting. How would you go about that using POVMs or something else that would be consistent with the fact that the measurement can be done up to a finite precision?

 

[A. Neumaier]

My question is, do you use spectral decomposition theorem to derive POVMs or if yes, where?
bhobba: Of course.

Well, it is in your question. If you derive POVMs from the projection case in a bigger Hilbert space, you need the spectral decomposition there. But if you just take a POVM and use it to simulate, all you need is that the sume condition is satisfied.

 

[A. Neumaier]

Stern-Gerlach and too narrow beam splitting.

What does this mean? In Stern-Gerlach you only measure in which half the particle appears - and you can do the measurement only if the experiment creates the two partial beams in directions so that the spots on the screen are sufficiently far apart. Nothing fancy is needed here, and only very low precision.

 

[ErikZorkin]

Thanks. I learned something from this thread.

So do you mean that you can come up with POVMs without using the spectral theorem in the first place? I read in the materials, linked here, that they are hard to construct.

Regarding the Stern-Gerlach experiment, it's still possible to have a situation when the dots are NOT far apart enough. What do you do in this case? PVMs should tell you which eigenspace to project to, but you can't decide. Apprximation doesn't seem to make much sense since the eigenspaces are exactly the opposites.

 

[A. Neumaier]

a situation when the dots are NOT far apart enough. What do you do in this case?

In this case you get only one spot and you only measure the presence of the particle but not its spin. Thus no factorization is needed..

 

[atyy]

Thanks. I learned something from this thread.

So do you mean that you can come up with POVMs without using the spectral theorem in the first place? I read in the materials, linked here, that they are hard to construct.

Regarding the Stern-Gerlach experiment, it's still possible to have a situation when the dots are NOT far apart enough. What do you do in this case? PVMs should tell you which eigenspace to project to, but you can't decide. Apprximation doesn't seem to make much sense since the eigenspaces are exactly the opposites.

The collapse depends on what knowledge you retain of the measurement. Hence if you perform a measurement, but forget the results, the collapse is different from if you retain knowledge of the results. Terminology varies here - some people use the term "collapse" only if you remember the results. But regardless of terminology, the rule for the evolution of the state after a measurement is performed depends on the knowledge of the observer.

In this sense the "computational universe" is irrelevant, because quantum mechanics in the Copenhagen interpretation is not a theory of all of reality. It requires an observer, and concerns what probabilistic predictions the observer can make if he does certain things.