Non-commutativity of operators, simultaneous eigenstates and degeneracies over finite dimensional Hilbert space

Suppose, I have two operators $\hat{A}$ and $\hat{B}$ defined over a finite dimensional Hilbert space $H$. Also assume that both the operators have their own eigenstates that span $H$(existence of eigenbasis). Now, we take their commutator: $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$. We could now have three scenarios :

1. $[\hat{A},\hat{B}]=0$
2. $[\hat{A},\hat{B}]\neq0$
3. There exist a null space(dimensionality >=1) of $[\hat{A},\hat{B}]$ which is a proper subspace of the underlying Hilbert space

I want to analyse each case separately and get some idea about the nature of eigenstates, degeneracy and simultaneous solutions.

Analysis of 1

In the first case, the commutator takes any state vector $\left|\psi\right> \in H$ to $0$. So, if we apply the commutator on $\left|\psi\right>=\left|a\right>$($\left|a\right>$ being an eigenvector of $\hat{A}$), we get: $$\hat{B}\hat{A}\left|a\right>=a\hat{B}\left|a\right>=\hat{A}(\hat{B}\left|a\right>)$$ The second equality gives: $\hat{B}\left|a\right>$ as an eigenvector of $\hat{A}$ with eigenvalue a.

Further, we can have two more subcases:

a. $\hat{B}\left|a\right>=b\left|a\right>$ $\rightarrow$ This is a kind of redundant case where $\hat{B}\left|a\right>$ is no new eigenfunction of $\hat{A}$. Instead we can now say we have $\left|a\right>$ as a simultaneous eigenfunction of $\hat{A}$ and $\hat{B}$.

b. $\hat{B}\left|a\right>=\left|c\right>$ $\rightarrow$ This means we have two linearly independent eigenvectors of $\hat{A}$(namely,$\left|a\right>$ and $\left|c\right>$) with the same eigenvalue.(degeneracy of order two).

Mathematically, we cannot gain any other insight.So I find the following statement erronous.

Two commuting operators give rise simultaneous eigenfunctions.

Since, we donot know about the existence of simultaneous solutions, we could probably say:

Also, I think its fine to say that commutability is a necessary condition for existence of simultaneous eigenfunctions but not a sufficient condition.(as my working clearly shows)

Analysis of 2

Here, there exists no state vector $\left|\psi\right>$ for which $\left([\hat{A},\hat{B}]\right)\left|\psi\right>=0$. This means that the existence of a simultaneous eigenfunction will directly contradict the given global non-commutativity(here by global I am distinguishing case 2 from case 3 of the previously outlined scenarios).

Hence, this case means absolutely no common eigenfunctions.

In addition, there an interesting situation here. We now have a new non-null operator(namely, the commutator)which will have its own eigenstates and eigenvalues. If anyone can shed light on the properties of these eigenstates and their relation with eigenstates of $\hat{A}$ and $\hat{B}$, it would be nice.

Analysis of 3

I have never come across anywhere talk about this case, but I have a hunch it might bear some relevance. Here are my ideas.

This situation exposes a new direction that I think must be understood first.

In analysis 1, we conclude about the existence of degeneracy or simultaneous eigenstates by letting the commutator act on an eigenvector of $\hat{A}$ or $\hat{B}$. We are not checking for the action of the commutator on anything but an eigenstate. Yes, action of eigenstate is known to us, but what about the information derivable from its action on other state vectors? They might also contain some information that is not related to the simultaneity/degeneracy of eigenstates of $\hat{A}$ and $\hat{B}$.Anyways, we don't bother as atleast all eigenvectors go to zero.

Here, not all eigenvectors will go to zero. Then, we have as a proper subspace of $H$ as the null space of the commutator.The classification of simultaneous solution or degeneracies will now only work for a subset of the basis of $H$. What about the other eigenvectors?(in the sense, what conclusion can one draw about the connection between the eigenbasis of $\hat{A}$ and $\hat{B}$ due to this)?

My question is evident in its analysis and how one can completely describe this situation considering all possible cases.Yet for a quick look my questions are:

1. What information can we extract about the relation between the eigenbasis of $\hat{A}$ and $\hat{B}$ from the commutation relation when we look at their(meaning the commutator) action on non-eigenvectors?(Analysis 1 and 3)
2. What is the relation between the three eigenbasis (namely, $\hat{A},\hat{B},[\hat{A},\hat{B}]$) in the 2nd case(when commutator is non-zero)?(Analysis 2)

Apart from the answer to the explicitly mentioned questions, comments on my analysis and further insights will be greatly useful.

• Nitpick: The way you've phrased it, conditions (2) and (3) can both be true: a non-zero operator such as $[\hat{A}, \hat{B}]$ can still have a non-trivial null space. I would rephrase the cases as "the null space of $[\hat{A}, \hat{B}]$ is (1) the entire space $H$, (2) trivial, or (3) a subspace of $H$." – Michael Seifert Aug 3 '16 at 15:47
• Related question on Math.SE: math.stackexchange.com/q/56307/11127 , math.stackexchange.com/q/236212/11127 and links therein. – Qmechanic Aug 3 '16 at 15:59