# I Eigenvalue degeneracy in real physical systems

Tags:
1. Apr 8, 2016

### ErikZorkin

I understand this question is rather marginal, but still think I might get some help here. I previously asked a question regarding the so-called computable Universe hypothesis which, roughly speaking, states that a universe, such as ours, may be (JUST IN PRINCIPLE) simulated on a large enough computer, and the question was resolved quite successfully.
This is to say that everything, that has a meaning in terms of observation, might be in principle simulated (up to a finite precision).

Now, the question. (Forgive me my mediocre knowledge) Let A be a Hermitian operator acting on an n-dimensional Hilbert space H. By the spectral theorem, we can decompose A into the sum Σi=1m λi Pi where λi-s are m mutually distinct eigenvalues of A and Pi-s are the corresponding orthogonal projections. Then, H can be rewritten as a direct sum of the corresponding subspaces. Now, if we were to simulate all (observable in real world) physical systems, we would need to know whether the eigenvalues of all Hermitian operators that correspond to the real physical systems are distinguishable. Otherwise, our "supercomputer" would be unable to determine, which eigenstate the system falls into after measurement. In particular, it is true when all the operators are represented by non-degenerate matrices.

Are there (or have there been observed) real-world physical systems known to have indistinguishable eigenvalues?

My question is motivated by the following work:

Computable Spectral Theorem

Another discussion on the topic (quite old though)

Last edited: Apr 8, 2016
2. Apr 8, 2016

3. Apr 8, 2016

### ErikZorkin

Thanks for the answer. By distinguishable I meant if it was known beforehand that some eigenvalues are equal and some are distinct.

4. Apr 8, 2016

I'm not really sure what you are asking but typically degeneracies in quantum mechanics can be associated with symmetry or topological characteristics of the system.

Take time reversal for example. In a system with an odd number of electrons you will have at least a two-fold degeneracy since T^=-1 for fermions.

5. Apr 9, 2016

### ErikZorkin

Ok, but was is degeneracy in real physical systems and what's its relation to measurement?

6. Apr 9, 2016

### A. Neumaier

For a Copenhagen-style experiment, a system described by $\psi$ before the measurement is described by $P_k\psi$ after the measurement, where $P_k$ is the projector to the eigenvalue $k$ measured. This is completely determined, and independent of the dimension of the eigenspace.

This probably makes your question moot.

7. Apr 9, 2016

### vanhees71

Let's define the projector in the previous posting a bit more specifically.

According to some flavors of the Copenhagen interpretation after a measurement (in fact it's only very special measurements calle von Neumann filter measurements which almost never are really done as measurements but as preparation procedures in an approximate way) of an observable $A$ leading to the result $a$, which is an eigenvalue of the self-adjoint operator $\hat{A}$ describing $A$ in the quantum theoretical formalism and where the eigenspace to this eigenvector is spanned by the orthonormal vectors $|a,\beta \rangle$ and the system is prepared in a pure state, described by a state vector $|\psi \rangle$ after the filter measurement the system is in a pure state described by the state vector
$$|\psi ' \rangle = \sum_{\beta} |a,\beta \rangle \langle a,\beta|\psi \rangle=\hat{P}_a |\psi \rangle.$$

8. Apr 9, 2016

### ErikZorkin

But the Pi's depend on multiplicity of eigenvalues.

9. Apr 9, 2016

### vanhees71

Yes, you project not to a specific eigenvector but just to the eigenspace. Only if you measure a complete set of compatible observables in the sense of a von Neumann filter measurement you project to the then uniquely determined state. If not, you miss information, and then you make the plausible assumption to estimate the state as the projection of the state the system was prepared in to the eigenspace with equal weights for all possibilities. In some sense you can understand it as an application of the maximum-entropy principle, i.e., you choose the state "of least prejudice".

Note, however, that the collapse postulate has to be taken with some grain of salt. It's not a necessary assumption within the minimally interpreted QT, and what really happens in a measurement process depends on the measurement device and its interaction with the measured system. As I said before, what's described here is a very special and rarely realized von Neumann filter measurement, which you can take as a state-preparation procedure.

10. Apr 9, 2016

### stevendaryl

Staff Emeritus
There is an obvious example of completely degenerate energy eigenstates, which is free particles. A free electron is infinitely degenerate, because the electron's momentum can point in a continuum of different directions, all with the same energy. Similarly, since a free electron's energy doesn't depend on it's spin, spin-up and spin-down have the same energy.

I think you were thinking along the lines of eigenstates for bound particles, and in that case, it seems that interactions tend to break any "accidental" degeneracies (ones that cannot be deduced from symmetry considerations).

11. Apr 9, 2016

### ErikZorkin

I think the computational aspect is still not addressed. But, perhaps, it's because this is not right a place to ask.

12. Apr 9, 2016

### vanhees71

Which computational aspect? A (generalized) basis of Hilbert space is determined by calculating common eigenvectors of a complete set of compatible observables. For a (non-relativistic) electron you can choose common generalized eigenvectors of the three momentum components $\hat{\vec{p}}$ and $\hat{s}_z$. The eigenvectors are $|\vec{p},\sigma_z \rangle$ with $\vec{p} \in \mathbb{R}^3$ and $\sigma_z \in \{\hbar/2,-\hbar/2 \}$. These are also eigenvectors of $\hat{H}$ with $\hat{H}=\hat{\vec{p}}^2/2m$.

13. Apr 9, 2016

### ErikZorkin

Which eigenvalue is observed determines which projection is "applied" after measurement, right? Eigenvalues, for which it is computationally undecidable, whether they are equal or distinct, it's impossible to determine which projection applies.

14. Apr 9, 2016

### Staff: Mentor

Not if there is degeneracy. Even if there is no degeneracy its very easy to create one by simply making two outcomes the same value. This is often done, for example in theoretical discussions of QM, in creating an 'indicator' operator that is one for some outcome and zero for the rest.

Thanks
Bill

15. Apr 9, 2016

### ErikZorkin

Detecting degeneracy is undecidable. Also, I am not interested in theoretical constructions, but practical.

16. Apr 9, 2016

### Staff: Mentor

There is no difference. Physically it would mean you simply change the readout on your apparatus.

Thanks
Bill

17. Apr 9, 2016

### rubi

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.

18. Apr 9, 2016

### ErikZorkin

So you can change it EXACTLY SO that degeneracy appears?

19. Apr 9, 2016

### rubi

Change what? (EDIT: Oh, I didn't realize that this was a response to bhobba.)

Here is an example:
Let $A$ be an observable given by a matrix and we have observed the value $a=5$ with a precision of $\sigma=0.5$. We don't need the projector onto the eigenspace to the eigenvalue $5$, but rather a projector $P(4.5,5.5)$ onto the space of states which is spanned by eigenstates with eigenvalues $4.5 \leq a \leq 5.5$. The numerics might give us $P(4.5,5.5) = P_{5.1}$ with multiplicity 4 or $P(4.5,5.5) = P_{4.9} + P_{5.3}+P_{5.4}$ with corresponding multiplicitties $1$, $1$ and $2$ and this decomposition might be numerically unstable and uncomputable, but since we don't care about the decomposition, but only about $P(4.5,5.5)$ itself, this uncomputability issue isn't relevant for us.

Last edited: Apr 9, 2016
20. Apr 9, 2016

### ErikZorkin

This is MUCH closer to what I was asking.

21. Apr 9, 2016

### Staff: Mentor

I think you need to see an axiomatic treatment of QM - see post 137:

Axiom 1
Associated with each Von Neumann measurement we can find a Hermitian operator O, called the observations observable such that the possible outcomes of the observation are its eigenvalues yi.

The values of those outcomes are entirely arbitrary - any operator can be made degenerate or non degenerate without changing the underlying physics.

Thanks
Bill

22. Apr 9, 2016

### vanhees71

There's nothing that determines which value is observed when measuring an observable. The state of the system determines the probabilities (and only the probabilities) with which you'll find a possible value. In terms of my notation above it's given, according to Born's rule,
$$P(a)=\mathrm{Tr}(\hat{\rho} \hat{P}_a),$$
where $\rho$ is the statistical operator, representing the system's state when the measurement is done.

23. Apr 9, 2016

### A. Neumaier

Practically you decide the spectrum by calucating it numerically. This gives you an orthonormal basis and the projection operators. Approximately of course, byt that the nature of practice.

if your spectrum is too tightly spaced it is unlikely that you perform in reality a Copenhagen measurement; hence you shouldn't simulate it as one. In this case you should look at POVMs instead.

Last edited: Apr 9, 2016
24. Apr 9, 2016

### ErikZorkin

What's the mathematical background of this?

25. Apr 9, 2016

What's POVM?