Eigenvalue degeneracy in real physical systems

[ErikZorkin]

Really, could you give a reference?

Sure.

 

[atyy]

Sure.

But one just needs to know how many distinct eigenvalues there are. Presumably this should be known from experiment.

 

[ErikZorkin]

Presumably this should be known from experiment.

No.

Take an example of Stern-Gerlach experiment and suppose that beam splitting is completely undetectable (that it's beyond Planck scale, for instance -- the process of collapse still happens but we have no idea about the final state). See the image:

 

[atyy]

No.

Take an example of Stern-Gerlach experiment and suppose that beam splitting is completely undetectable (that it's beyond Planck scale, for instance -- the process of collapse still happens but we have no idea about the final state). See the image:

Then just omit the spin variable.

 

[ErikZorkin]

Then just omit the spin variable.

Eh, what?

And by the way, how can you deduce the number of eigenvalues from the experiment? It's a mathematical property of the operator.

 

[atyy]

Eh, what?

If there is no spin, then you won't get any splitting in the Stern-Gerlach experiment. So there will be no splitting in your theory, and no observed splitting, and all will be well.

And by the way, how can you deduce the number of eigenvalues from the experiment? It's a mathematical property of the operator.

OK, I don't know. In all practical cases, we don't seem to have a problem. For example, whether one needs a degeneracy or not depends on how good the experimentalist is. For example, in non-relativistic QM, the simplest Hamiltonians have lots of degeneracy. But then experiments get better, and one sees splittings, so one adds terms or uses better computations to get theory and experiment to match, eg. Zeeman, Lamb shift etc.

 

[ErikZorkin]

If there is no spin, then you won't get any splitting in the Stern-Gerlach experiment.

There is spin and there is splitting! They are just not detectable.

OK, I don't know. In all practical cases, we don't seem to have a problem.

Well, only because physicists don't care about math much. It is a fact that spectrum is uncomputable. Rubi was right, the only thing you can measure is that an eigenvalue lies in so or so range. You can't try to measure something, that you postulated mathematically, but what it uncomputable. It's nonsense. I am simply looking for a consistent explanation that is used in physics. Classical spectral decomposition and projection postulate are simply wrong when it comes to real experiments.

 

[ErikZorkin]

so one adds terms or uses better computations

What's that? "Adjusting" operators to better match with the reality (i.e. remove degeneracy)?

 

[bhobba]

What's that? "Adjusting" operators to better match with the reality (i.e. remove degeneracy)?

The value assigned to an observational outcome is entirely arbitrary.

The link I gave before explains clearly what's going on with POVM's etc etc:
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

The value assigned to the elements of the POVM are irrelevant hence degeneracy is irrelevant and easily avoided - or not - it doesn't matter. The POVM is the important thing.

Thanks
Bill

 

[bhobba]

Well, only because physicists don't care about math much.

Why anyone would believe that has me beat. Some physicists have won Fields medals.

Thanks
Bill

 

[ErikZorkin]

Well, in the Stern-Gerlach experiment, the outcomes are quite certain, right? It's either spin up or down.

 

[bhobba]

Well, in the Stern-Gerlach experiment, the outcomes are quite certain, right? It's either spin up or down.

Spin up and spin down do not appear in the observable - you must assign it a number. That number is entirely arbitrary - it could be 1 and 0, 1 and 2, 1 and -1 or 1 an 1 - in which case you have degeneracy.

Thanks
Bill

 

[ErikZorkin]

As far as I remember, not all degenerates may be removed.

 

[bhobba]

As far as I remember, not all degenerates may be removed.

Hmmmm. Theoretically it should be possible - but likely not in a natural way.

Thanks
Bill

 

[ErikZorkin]

Theoretically it should be possible

How, for example?

 

[bhobba]

How, for example?

Read the link I gave.

An observation is a mapping to a POVM. The value assigned is entirely arbitrary.

An observation/measurement with possible outcomes i = 1, 2, 3 ... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

Thanks
Bill

 

[atyy]

What's that? "Adjusting" operators to better match with the reality (i.e. remove degeneracy)?

Yes. https://en.wikipedia.org/wiki/Fine_structure

 

[ErikZorkin]

Yes. https://en.wikipedia.org/wiki/Fine_structure

Nice example! But, again, it's not always possible to split the eigenvalues, as far as I understand.

An observation is a mapping to a POVM.

I sympathize with POVM approach as it (seemingly) avoids direct use of spectral decomposition (or?). Apparently, there has been a bit of discussion as to how it really addresses the question in the first place. For instance, POVM don't seem suitable for discrete variables.

 

[bhobba]

I sympathize with POVM approach as it (seemingly) avoids direct use of spectral decomposition (or?). Apparently, there has been a bit of discussion as to how it really addresses the question in the first place. For instance, POVM don't seem suitable for discrete variables.

I think you mean continuous variables.

Yes there is an issue - but it really needs a thread of its own.

That said continuous values don't ever actually occur.

Thanks
Bill

 

[vanhees71]

This is a phantastic example for the strange inability of mathematicians and physicists to communicate with each other. I must admit, I lost track what's the problem with "degeneracy" here. There is no such problem from a physicist's point of view, because it's all well defined in terms of observables (discribed by self-adjoint operators of Hilbert space) and states (described by a self-adjoint trace-class operator with trace 1, the statistical operator):

Any complete set of independent compatible observables $A_j$ ($j \in \{1,2,\ldots,n\}$) defines via the common (generalized) eigenvectors of the corresponding self-adjoint operators $\hat{A}_j$ (that are pairwise commuting by definition), denoted by $|a_1,\ldots,a_n \rangle$. Here the the components of the tupel $(a_1,\ldots,a_n) \in \mathbb{R}^n$ can run over discrete and/or continuous domains.

If the state of the system is represented by the statistical operator $\hat{\rho}$ the probability/probability density to simultaneously find the values $(a_1,\ldots,a_n)$ measuring the observables $A_j$ is given by (Born's rule):
$$P_{\rho}(a_1,\ldots,a_n)=\langle a_1,\ldots,a_n|\hat{\rho}|a_1,\ldots,a_n \rangle.$$
Everything else follows from the standard rules of probability theory. If I measure only von observable, e.g., $A_1$, I just have to sum/integrate over all other $a_2,\ldots,a_n$ over the corresponding spectrum, i.e.,
$$P_{\rho}(a_1)=\sum_{a_2,\ldots,a_n} P_{\rho}(a_1,\ldots,a_n).$$
This refers to measuring precisely the observable $A_1$. Of course in this case the eigenspaces of $\hat{A}_1$ alone are degenerate, because you need the other independent compatible observables to completely define the orthonormalized basis vectors (up to phase factors for each basis vector of course, but these cancel always out in physical results that are all dictated by the above given probability measure according to Born's rule). QT as a mathematical theory is consistent, as "weird" as some physicists and mathematicians seem to think it might be ;-))!

If you measure something not precisely you can envoke more complicated descriptions in terms of POVMs. Strictly speaking that you must do in any case if you measure an observable in the continuous part of the spectrum, because then you necessarily must use an apparatus with finite resolution, i.e., you can make, e.g., a position measurement of a particle, only up to a certain resolution determined by a the technology available for the measurement apparatus (which can be as fine as you like but not absolutely exact ever).

Now you can of course make this into a precise and well-defined mathematical formalism, and this is nice too, but one should not forget this not that difficult to understand physics and "metrological" constraints, which are very clear to everybody who as ever done experiments (and be it as a theoretician only in the standard lab practice mandatory to attend for every physics student wherever physics is taught).

 

[ErikZorkin]

I think you mean continuous variables.

? POVMs are problematic when coupled with continuous variables? I thought in exactly the opposite way. Nevertheless, I don't care about continuous variables. The key problem of this story is: you can't computationally define the operator's spectrum, let alone eigenstates/projections. That's it. There is certainly a way to avoid such a treatment.

 

[bhobba]

I don't care about continuous variables. The key problem of this story is: you can't computationally define the operator's spectrum, let alone eigenstates/projections.

Then there is no issue.

QM has nothing to do with the practicalities of computing eigenvalues any more than classical mechanics has anything to do with the practicalities of numerically solving a differential equation. It leas to interesting things like the butterfly - but has nothing to do with the theories validity.

Thanks
Bill

 

[ErikZorkin]

Then there is no issue.

I am pretty sure! I thought I almost found an answer (POVMs), but then I got a bit concerned about the difficuleties as pointed out by rubi. Forgive me my ignorance, but do you apply spectral theorem when dealing the POVM--based QM?

 

[bhobba]

I am pretty sure! I thought I almost found an answer (POVMs), but then I got a bit concerned about the difficuleties as pointed out by rubi. Forgive me my ignorance, but do you apply spectral theorem when dealing the POVM--based QM?

No. Please, please read the link I gave. The spectral theorem only applies to what's called resolutions of the identity which are disjoint POVM's.

Thanks
Bill

 

[atyy]

Nice example! But, again, it's not always possible to split the eigenvalues, as far as I understand.

I think as long as we have experiment as a guide, it is ok. The theorem you cite assumes finite dimensions and fails for infinite dimensions. For finite dimensions, the degeneracy cannot be greater than the dimensionality. Let the dimensionality be N. Then we guess the number of eigenvalues m to range between 1 .. N. For each guess, we compute the predictions. Although we will never know with certainty which m is correct, we just take the provisional answer to be the one that matches observations most closely.

This will be fine, because in practice, even for non-degenerate Hamiltonians, our ability to do a brute-force diagonalization is already insufficient. For typical Hamiltonians in condensed matter, we quickly run out of electrons in the universe that can do the computation, eg. http://fqxi.org/data/essay-contest-files/Swingle_fqxi2012.pdf.

 

[ErikZorkin]

QT as a mathematical theory is consistent

What's QT? Quantum mechanics or QFT? First of all, the very formal foundation of QM is not done yet. Second, you can't prove its consistency. Because your axiomatic system would necessarily include arithmetic whose consistency can't be proven in arithmetic alone.

 

[bhobba]

First of all, the very formal foundation of QM is not done yet.

Von Neumann sorted that out ages ago. Dirac's elegant formulation is now rigorous since Rigged Hilbert Spaces have been worked out.

Please, please read a good book on QM such as Ballentine.

Thanks
Bill

 

[atyy]

First of all, the very formal foundation of QM is not done yet.

Formal foundations of QM are done, eg. http://arxiv.org/abs/1110.6815 (p9), which can be generalized to continuous variables, partial example http://arxiv.org/abs/0706.3526 (Eq 3, 4).

 

[bhobba]

Because your axiomatic system would necessarily include arithmetic whose consistency can't be proven in arithmetic alone.

Godels theorem is irrelevant in this context. Its as consistent as any other physical theory ie as consistent as geometry, arithmetic etc etc.

Thanks
Bill

 

[ErikZorkin]

Von Neumann sorted that out ages ago. Dirac's elegant formulation is now rigorous since Rigged Hilbert Spaces have been worked out.

Please, please read a good book on QM such as Ballentine.

Thanks
Bill

Well, the axioms of QM are far from what's called a formal mathematical axiom. cf. axioms of ZFC

 

[ErikZorkin]

Godells theorem is irrelevant in this context. Its as consistent as any other physical theory ie as consistent as geometry, arithmetic etc etc.

Thanks
Bill

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.

 

[atyy]

@ErikZorkin, I googled around and found http://arxiv.org/abs/quant-ph/0111063 A reformulation of Hilbert's tenth problem through Quantum Mechanics, which you may find amusing.

 

[ErikZorkin]

Formal foundations of QM are done, eg. http://arxiv.org/abs/1110.6815 (p9), which can be generalized to continuous variables, partial example http://arxiv.org/abs/0706.3526 (Eq 3, 4).

Postulates are not formal axioms

 

[bhobba]

Well, the axioms of QM are far from what's called a formal mathematical axiom. cf. axioms of ZFC

If that what you want then here is the book:
https://www.amazon.com/dp/0387493859/?tag=pfamazon01-20

Be warned - it what mathematicians call non trivial - meaning its HARD. But it does what you want.

Thanks
Bill

 

[atyy]

Postulates are not formal axioms

Sure, but no one doubts they can be formulated in ZFC.