# Commuting observables and CSCO's

I've been looking at some basic quantum mechanics all day in an attempt to better my understanding of the subject. While going over the proof that commuting operators are compatible, I started getting questions relating to complete sets of commuting observables (CSCO's).

I apologize for the fact that these questions might be trivial, it's been a long day.

Suppose $A$ and $B$ are two observables that commute.

1. Do $A$ and $B$ have the same amount of eigenvalues?
2. Do $A$ and $B$ have the same amount of eigenvectors?
3. What do the previous answers mean for the cardinality of the sets (of eigenvectors of $A$, $B$ and $AB$)?

Basically, I'm wondering about dimensions and sizes of spaces and sets in the case where two observables commute.

• Id asdume so since the operators themselves are $n\times n$ matrices and multiplication for commutators wouldnt be defined if two operators had different dimensions – Triatticus Jun 4 '15 at 23:39
• For Hilberst space spanned by $N$ kets ($N$ may be infinity) you should define it's action on every ket, thus it's matrix representation has to be $N \times N$. Thus all the operators acting on the same Hilbert space have the same amount of eigenvalues/eigenvectors. If $A$ and $B$ commute - it is possible to find an orthnormal basis for the Hilbert space, in which they both are diagonal. – Alexander Jun 5 '15 at 0:55
• @Alexander If they commute then it is possible to find common eigenvectors but one operator can still have lots more eigenvectors all on it's own than the other one has all on it's own and similarly the other can have lots more eigenvalues than the one. – Timaeus Jun 5 '15 at 3:28
• @Timaeus For commuting Hermitian operators the number of linearly independent eigenvectors is the same, as well as the number of eigenvalues, if they are counted with their multiplicity. – Lagrangian Jun 5 '15 at 14:39

1. Do $A$ and $B$ have the same amount of eigenvalues?

They don't have to, $S_z$ and the identity commute. The former has two eigenvalues. The latter has one. Other examples exist too, such as the operator, $H=p^2/2m+e/r$ which has an infinite number of eigenvalues but can commute with operators with a finite number of eigenvalues such as both operators listed above. However, if you look at the dimension of each eigenspace and add up all those dimensions for each eigenvalue then you get the same number for any hermitian operator (regardless of whether they commute) because since they are hermitian that sum just gives you dimension of the whole space. And that sum is just the dimension of the whole space and it happened regardless of whether they commuted, so it is really besides the point. The answer is that there is no real relationship in general for the number of eigenvalues of operators that commute.

1. Do $A$ and $B$ have the same amount of eigenvectors?

They don't have to, $S_z$ and the identity commute. If the operators act on a spin space the former has two eigenvectors (modulo scalars) and the latter has uncountably many (a whole two dimensional eigenspace of eigenvectors with uncountably many directions). Again you could count the dimension of each eigenspace and add them up, but you'll just get the dimension of the whole space again. There will still literally be two directions that are eigen to one operator and uncountably many directions that are eigen to the other operator. And adding the dimensionality of the eigenspaces for each operator will again just give you the dimension of whole space. And again it happenes regardless of whether they commuted, so it is really besides the point. The answer is that there is no real relationship in general for the number of eigenvector directions of operators that commute.

1. What do the previous answers mean for the cardinality of the set?

Nothing, the sets could be finite or countably infinite or for eigenvectors there can be uncountably many. Even $AB$ could have uncountably many different directions of eigenvectors. If you have not just two but have a collection of mutually commuting observables where you can't add more observables that also commute with those (a CSCO) then the vectors that are common eigenvectors to all the operators will have fixed directions so now there is a fixed set of directions. And the cardinality of those directions will be the dimension of the whole space but now it is actually related to the number of eigenvector directions that are eigen to all the operators, the eigenvectors that are common to all the operators give you real directions. And you can count them, but counting them will just give you the dimensionality of the whole space.

• It may be a bit useless to speak about "uncountably many eigenvectors" of the identity since what really matters is the dimension of the eigenspaces or the number of lineary independent eigenvectors. (I could as well say S_z has uncoutably many eigenvectors, namely the 'standard' eigenvectors multiplied by arbitrary constants). – Lagrangian Jun 5 '15 at 11:37
• @Lagrangian Thank you for your comments, I expanded my answer to include mention the dimensionality of the eigenspaces and underlying space and count the directions (eigenvectors modulo scalars) instead of the eigenvectors. I didn't think discussion of cardinalities of sets eigenvectors directions was useless because that's what I read the third question as asking. – Timaeus Jun 6 '15 at 3:43

## Commuting matrices

We take two Hermitian matrices A and B that commute. Since they are Hermitian and act on the same Hilbert space they have a common (orthonormal) basis of eigenvectors. In general all Hermitian matrices have the same amount of lineary independent eigenvectors since they all have a orthonormal basis of eigenvectors.

Commuting observables don't necessarily have the same eigenvalues: the identity operator, which only has eigenvalue 1, commutes with everything. So it commutes with some hermitian operator with two different eigenvalues. However, if you count the eigenvalues with their multiplicity the numbers will match for hermitian operators since they are diagonalisable.

I don't exactly see what you mean with the thrid question.

## CSCO

First let's see what a CSCO is, it's a set of observables $\{A, B, ...\}$ with the following properties:

• they all commute with each other, meaning there exists a common basis of eigenvectors of all these observables
• this common basis of eigenvectors is unique (up multiplication with scalars)

This means we can do the following every eigenvector in this basis can be labelled with it's eigenvalues of all these observables: $|a, b, ... \rangle$. This labelling is unique: the cross-section of the eigenspace corresponding with eigenvalue a for A, the eigenspace corresponding with eigenvalue b for B and so on, is a one dimensional subspace of the Hilbert space.