(How) can you tell that a given operator's eigenspectrum will feature degeneracy?
I am speaking about operators representing physical observables and am not interested in purely mathematical objects (if that's relevant to answering the question).
I know that a degenerate eigenvalue corresponds to an eigenspace with dim > 1, which means 'linearly independent' eigenkets, even orthogonal eigenkets, that correspond to the same eigenvalue. Thus, additional commuting operators would be necessary to form a complete set of commuting observables (CSCO), narrow down the eigenspace and to thereby be able to precisely specify any given eigenket. And as far as I'm aware the latter part of that sentence also goes the other way to define a CSCO.
Now my question is: How, if at all, can you tell if a given operator's eigenspectrum will feature degeneracy? How do I show that 'all emerging eigenspaces' have dimension one only? Or how do I show that a given combination of observables forms a CSCO?
Take the one dimensional position operator $ \hat{x} $ as an example. The respective eigenkets corresponding to the eigenvalue x' shall be denoted by $ |x'\rangle $: I 'know' or it can be shown through Plancherel's theorem that the corresponding eigenfunction $ \langle x|x'\rangle = \delta \left( x-x' \right) $ is the dirac delta distribution and I can argue to myself that there is no more paramater to tweak except applying a (phase) factor, i.e. there is no degeneracy. But what is the fundamental argument?
quantum-mechanics hilbert-space operators eigenvalue
Markus Gratis
Markus GratisMarkus Gratis
$\begingroup$ You've forgotten to define \ket and \braket commands... $\endgroup$
– Ruslan
I think you're really asking for too much here - for arbitrary classes of operators in infinite dimension, there's certainly no algorithmic way to simultaneously diagonalize them nor to prove that their simultaneous eigenspaces don't feature any degeneracy.
In the specific case of a position operator, it's quite clear that the eigenspaces are generated by the Dirac functions $ \delta(x - x_0) $ of eigenvalue $ x_0 $ - no function that's spatially spread out could be an eigenstate of the position operator by trivial considerations - and so you explicitly construct a complete eigenbasis which is nondegenerate. This is only possible because the position operator is a particularly simple operator. If you consider arbitrarily complicated operators, the problem quickly becomes impossible to solve. Sturm-Liouville problems would form a small subset of such questions, for instance.
In cases which you care about in quantum mechanics, however, it's quite typical for the system you're considering to form an irreducible representation of a certain Lie algebra (or Clifford algebra), such as $ \mathfrak{sl}_{\mathbb C}(2) $ for the Fock space. If this is the situation, then you can prove by general arguments in the theory of sufficiently nice (reductive, for instance) Lie algebras that the energy eigenstates can't feature any degeneracy, because you visibly get a subrepresentation by starting with the ground state and acting on it with creation operators.
This, however, simply pushes the burden of proof from nondegeneracy to the irreducibility of the associated representation, which depending on your concern at that moment may be easier or harder to establish. A similar argument to this one can be used to prove the nondegeneracy of the spherical harmonics, i.e. that the angular momentum operators form a maximally commuting set of observables, and thus to solve Sturm-Liouville problems, for instance. Unfortunately it's not really possible to push these methods much further than this, which is why Sturm-Liouville problems occupy such a privileged position in the theory of partial differential equations.
In a sense, however, you should keep in mind that degeneracy is a rather exceptional occurence for operators chosen "at random" - it usually only happens if the operators under consideration have a lot of structure associated with them, for instance, if they commute with certain obvious symmetry operators. (Consider linear momentum and rotational symmetries, for example.) An operator chosen "at random" would almost surely be nondegenerate, and two operators chosen at random would almost surely not commute, and thus be impossible to simultaneously diagonalize.
Ege ErdilEge Erdil
Now my question is: How, if at all, can you tell if a given operator's eigenspectrum will feature degeneracy. How do I show that 'all emerging eigenspaces' have dimension one only? Or how do I show that a given combination of observables forms a CSCO?
This is an interesting question! The answer is, basically, you can't tell--at least, not in physics. In mathematics, you can start with a given Hilbert space and since you already know the dimensionality $d$ of the Hilbert space, an operator would already be in a $d\times d$ matrix representation in some basis even if there are only, say, $n<d$ distinct eigenvalues of the given matrix. Since you're diagonalizing the operator already knowing what the true dimensionality of the Hilbert space is, you'd automatically recover multi-dimensional eigensubspaces for certain eigenvalues--just like in a linear algebra class.
Of course, you already know that, as can be inferred from your question. However, I just wanted to set the stage by saying it all out loud.
Now, in physics, you never know the true dimensionality of the Hilbert space. For example, say you start with a particle and you measure its position and you can say that if all I want to describe is the position of the particle then the Hilbert space whose basis is given by $\{|x\rangle|x\in\mathbb{R}\}$ is my Hilbert space. You can define a position operator via its specifying its matrix elements in this basis, i.e., $\hat{X}$ is such that $\langle y |\hat{X}|x\rangle=\delta(x-y)$. And the eigenspectrum of the position operator would be non-degenerate, trivially. However, if the particle is a spin $\frac{1}{2}$ particle then you will eventually discover that, in fact, there is another property of the particle that you can measure, called its spin state. It can be either up or down and you can measure this simultaneously with measuring the position of the particle. So, you say that, in fact, the Hilbert space was bigger; we just didn't know about these other degrees of freedom called spin states. Now, the Hilbert space is such that its basis is given by $\{|x,s\rangle|x\in\mathbb{R},s\in\{\frac{1}{2},-\frac{1}{2}\}\}$. So now, the position operator's eigenspectrum is degenerate with each eigenvalue corresponding to a two-dimensional eigensubspace (corresponding to the two spin states).
So, in physics, we posit what our set of commuting operators is and try to see if that describes the system that we want to describe. If it turns out that the system actually has additional independent properties then we add operators corresponding to those properties to our set of commuting operators, enlargening the Hilbert space and adding degeneracy to the spectrum of our old operators.
Dvij D.C.Dvij D.C.
$\begingroup$ Thank you! I have a followup question on terminology: Is there a specific term denoting the Hilbert space $ H_s $ that is made up of spin eigenkets and the 'fully disjoint' Hilbert space $ H_{..} $ that position AND momentum operator act in? Since the eigenkets are Fourier transforms of another and x & p are conjugate variables in Hamiltonian mechanics, and they sort of maximally do not commute, I do presume they act on the identical Hilbert space. $\endgroup$
– Markus Gratis
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