8

Given two Hermitian operators $A$ and $B$, such that $[A,B] = 0$, if one [or both] of these operators are degenerate, how does one define a formal way of going about simultaneously diagonalising both of them? Preferably, if possible, with some example of the method.

Previous chat discussions:

  1. I asked this question on the h-bar a while back, and while I got a good response, I think the nature of brevity of the chat meant that when I got to thinking about it more, I realised I still wasn't entirely sure of the whole thing, plus I figured it would be nice to put this question on the site as I personally couldn't find it when I needed it.

  2. Discussed in the hbar chat room here.

  3. Previous conversation in chat about the subject.

Qmechanic
  • 201,751
  • I don't understand exactly what you are asking for. Are you looking for a proof of the statement that two commuting Hermitian matrices can be simultaneously diagonalized? For a method to do it by hand with pen and paper, given two matrices? For a numerical method suitable for a computer? The three questions will have very different answers. – Federico Poloni Oct 06 '17 at 16:23
  • 1
    It seems OP is asking how to circumvent bad basis choices due to degenerate eigenvalues, like e.g. this Phys.SE post. – Qmechanic Oct 07 '17 at 07:37

4 Answers4

7

For each pair (eigenspace for $A$, eigenspace for $B$), pick a basis for their intersection. (In particular, if an intersection is 0-dimensional, it will not have/contribute with any basis vectors.) Finally concatenate all the bases.

Qmechanic
  • 201,751
6

The simplest method is to first diagonalise $A$. Then consider in turn each eigenvalue $\lambda$ and a basis of the associated eigenspace $\mathcal{E}_\lambda$: $\newcommand{\ket}[1]{|{#1}\rangle}(\ket{1},\ket{2},\cdots,\ket{n})$. You then construct the matrix $\newcommand{\bra}[1]{\langle{#1}|}M_B=(\bra{i}B\ket{j})_{1\le i,j \le n}$. There is one such matrix for each eigenvalue $\lambda$, just to be crystal clear, but I will refrain from tagging a $\lambda$ to $M_B$ to keep the notation readable. Finally you diagonalise $M_B$. This will gives you eigenvalues $\mu_1, \cdots, \mu_n$ (not necessarily distinct) and eigenvectors $u_i$, which are column vectors,

$$u_i=\begin{pmatrix} u_{i1}\\ \vdots \\u_{in} \end{pmatrix},$$

such that $M_Bu_i=\mu_iu_i$. Then you construct $\ket{i'}=\sum_j u_{ij}\ket{j}$, and now $(\ket{1'}, \cdots, \ket{n'})$ are eigenvectors for both $A$, all for the eigenvalue $\lambda$, and for $B$, for the respective eigenvalues $\mu_1,\cdots,\mu_n$.

Of course, if $\lambda$ was not degenerate in the first place, i.e. $n=1$, then there is nothing to do! If it is degenerate, most often, $n$ will be small, at least way smaller than the dimension of the eigenproblem for $A$, so the diagonalisation of $M_B$ will be comparatively easy. Then, again, don't forget you have to do this for each eigenvalue $\lambda$ of $A$. Of course, you could start by diagonalising $B$ instead: just do what looks simpler.

There are way more efficient numerical methods note, which would make a big difference for large matrices, but for your bread and butter quantum system, the method I highlighted should be tractable.

  • Sorry, I don't understand what $(u_{ij}){1 <= j <= n}$ means, what is, for instance, $u{i2}$? –  Oct 06 '17 at 12:44
  • From the previous conversation on chat with OP, liked in the question comments, I'm not sure they really understand the significance and importance of restricting this method exclusively to a basis of the eigenspace of $A$. Maybe it's clicked in the meantime, though. That said, (cont.) – Emilio Pisanty Oct 06 '17 at 13:25
  • It's a bit misleading to say "you construct the matrix $M_B$", because it implies that there is a single such matrix. Instead, there is one such matrix per eigenspace of $A$. I would strongly urge a notation change to keep this in mind. – Emilio Pisanty Oct 06 '17 at 13:26
  • I think the statement "consider in turn each eigenvalue" is clear enough but OK, I can add a caveat… –  Oct 06 '17 at 13:28
5

I suppose that the Hilbert space $H$ is finite dimensional and I indicate by $H_\lambda$ the eigenspace of $A$ with eigenvalue $\lambda$ and some dimension $d_\lambda\geq 1$.

You know that $$H_\lambda \perp H_{\lambda'}\quad \mbox{if $\lambda\neq \lambda'$}\tag{1}$$ and $$\oplus_\lambda H_\lambda= H\tag{2}\:.$$

The fundamental idea is that every $H_\lambda$ is invariant under the action of $B$, i.e., $$B(H_\lambda) \subset H_\lambda\:.$$ This is because $A(Bx)= BAx= B\lambda x= \lambda (Bx)$ if $x\in H_\lambda$.

As a consequence you can

(a) restrict $B$ to $H_\lambda$ and, noticing that $B|_{H_\lambda} : H_\lambda \to H_\lambda$ is still Hermitian as you easily prove,

(b) find an orthonormal basis of $H_\lambda$ made of eigenvectors $\{x^{(\lambda)}_n\}_{n=1,\ldots d_\lambda}$ of $B$ with corresponding eigenvalues $\mu^{(\lambda)}_n$.

Varying both

  1. $\lambda$ in the set of eigenvalues of $A$ and
  2. $n=1, \ldots , d_\lambda$,

due to (1) and (2) the set of all unit mutually orthogonal vectors $x^{(\lambda)}_n$ form an orthonormal basis of the entire $H$.

This basis is made of simultaneous eigenvectors of $A$ and $B$ because $$Ax^{(\lambda)}_n =\lambda x^{(\lambda)}_n$$ and $$Bx^{(\lambda)}_n= \mu^{(\lambda)}_n x^{(\lambda)}_n\:.$$

Obviously it may happen that $\mu^{(\lambda)}_n =\mu^{(\lambda')}_m$ for some $\lambda \neq \lambda'$.

Valter Moretti
  • 71,578
  • Sorry but what exactly does $B|{H{\lambda}}$ / 'restricting B to $H_\lambda$ mean? –  Oct 06 '17 at 15:52
  • 1
    It means restricting the domain of $B$ to the subspace $H_\lambda$ and viewing $B$ as an operator defined on a smaller space, i.e., from $H_\lambda$ to $H_\lambda$. This makes sense just because $B(H_\lambda) \subset H_\lambda$. $B|{H\lambda}$ indicates the operator $B$ when considered as an operator in the smaller space $H_\lambda$. – Valter Moretti Oct 06 '17 at 16:34
  • 1
    Note this is exactly the same as my answer. –  Oct 08 '17 at 13:28
  • 1
    An overlap exists, but I added some remarks explaining why the method works, and I did not enter into the details of diagonalisation procedure as you did instead. Maybe the original question concerned only the practical procedure so that I actually answered another question... – Valter Moretti Oct 08 '17 at 14:53
  • Yes, sure, the remark was for the OP! –  Oct 08 '17 at 14:56
3

The next simplest method (particularly useful if you only need numerical result) is to consider the operator $$ {\cal O}=\alpha A+\beta B $$ with $\alpha$ and $\beta$ chosen so that ${\cal O}$ has no repeated eigenvalues. Then the eigenvectors of ${\cal O}$ are simultaneous eigenvectors of $A$ and $B$.

ZeroTheHero
  • 45,515
  • If either or both $A$ and $B$ are allowed to be degenerate (as per OP), is it not the case that $O$ will necessarily have repeated eigenvalues? – LLlAMnYP Oct 06 '17 at 13:35
  • @LLlAMnYP Take $$ A=\left( \begin{array}{ccc} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & -1 \ \end{array} \right), ,\quad B=\left( \begin{array}{ccc} 1 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \ \end{array} \right), , $$ Then $A+\sqrt{2}B$ does not have repeated eigenvalues. – ZeroTheHero Oct 06 '17 at 13:54
  • 1
    Ok, my use of "necessarily" is too strong a word, but I still see cases where it's unavoidable (e.g. in your example have $B$ have three identical eigenvalues while $A$ remains as suggested). – LLlAMnYP Oct 06 '17 at 14:25
  • @LLlAMnYP in which case there is nothing to be done as any combination of eigenstates in the common degenerate subspace will be an eigenstate, i.e. the degenerate eigenstates of $\alpha A+\beta B$ are not uniquely defined as any linear combo of this is also an eigenstate. The same issue occurs with any method. You would need a third operator to hopefully lift the degeneracy. – ZeroTheHero Oct 06 '17 at 14:29
  • IOW, more quantum numbers are necessary to uniquely define a state. But that shouldn't really preclude us from diagonalizing both of them anyway, right? In your's and mine examples they're both diagonalized already. – LLlAMnYP Oct 06 '17 at 14:35
  • @LLlAMnYP you're correct. If you have common subspaces in A and B that are degenerate, you're stuck: you can still take combos and they are still eigenvectors. This is independent of any method you use as mathematically any combo of eigenvectors are equivalent to any other combo. FYI the trick I mention will work except in the circumstance you mention. The difficulty might be to find the factors $\alpha$ and $\beta$. Oftentimes the eigenvalues of $A$ and $B$ are rational so choosing irrational factors for $\alpha$ and $\beta$ does the trick. – ZeroTheHero Oct 06 '17 at 14:43
  • Well, in any case it's an elegant method, so +1 and degenerate matrices can be diagonalizable anyway. – LLlAMnYP Oct 06 '17 at 14:46
  • Well, that happens if you insist on $O$ having no repeated eigenvalues. In practice, I believe that for almost all choices of $\alpha$ and $\beta$, a matrix that diagonalizes $O=\alpha A + \beta B$ diagonalizes also $A$ and $B$ (no matter how many repeated eigenvalues they have). – Federico Poloni Oct 06 '17 at 16:20