# Simultaneous diagonalisation of non-hermitian operator (Bloch theorem)

1. Jul 21, 2010

### Llewlyn

A valuable math result for quantum mechanics is that if two hermitian operators (physical observables) commute, then a simultaneous basis of eigenvectors exists. Nevertheless, there are cases in which two operators commute without being both hermitians -- a really common one is when one operator is hermitian (such as an hamiltonian) and the other is unitary (say, a spatial translation operator which we'll consider a symmetry of the hamiltonian).

On QM books I didn't find any math results about what happens when an hermitian operator commutes with an unitary one, but this rarely matters. In fact, in the case above, I shall expand the unitary operator on small parameter and then consider the generator of the symmetry, which is hermitian, rather than the unitary operator itself.

But what happens when I couldn't invoke the generator? Say that a quantum system is invariant only under discrete translation (a crystal, for instance). So its hamiltonian won't commute with the momentum, but it'll surely do with the unitary spatial translation operator when the proper parameter is set to crystal period. Usually on physics books there's written "since they commute an orthonormal basis of eigenvectors exists" (see, Ashcroft and Mermin - Solid State Physcis. Bloch Theorem, first demonstration). But which math theorem am I using ? Does a basis of ort. eigenvectors exists even if one operator is hermitian and the other is unitary?

I'm sorry if the question sounds like homework, but I hope it won't.

Ll.

2. Jul 21, 2010

### DrDu

I think the theorem does not only hold for hermitian operators but generally for "normal" ones, i.e. for all operators which commute with their adjoint. This includes all unitary operators as well.

3. Jul 21, 2010

### Llewlyn

Yeah I suppose so, but I wasn't able to find a preposition which definitely ends the question. Normal operators with compact support are the ones for which the spectral theorem holds but this doesn't (trivially) mean that they are simultaneous diagonalisable.
However it's strange that QM books such as Shankar doesn't mention anything at regard.

Ll.

4. Jul 21, 2010

### jostpuur

If you are interested in finite matrices, it should be enough that the matrices are diagonalizable (and commute). Hermiticity or normality are not needed. I don't have any reference for this, but that's the way it seemed to be when I went through the proof myself.

5. Jul 22, 2010

### strangerep

Is there still a question here? (If the operators commute, then doesn't that resolve
the diagonalisability issue immediately?)

6. Jul 22, 2010

### Llewlyn

No, further hypotesis on the operators are needed. For instance the operator should be normal, which assure the existence of an ort. basis of eigenvectors (for the single operator). A rule of thumb for physicists is that if the operators are hermitian all it's fine. Unfortunately it seems to isn't enough, see Bloch theorem for istance.

Ll.

7. Jul 24, 2010

### strangerep

So... do you still have a question, or has it been answered to your satisfaction?
(If there's still an unresolved question, please restate it more clearly...)

8. Jul 25, 2010

### Llewlyn

Oh sorry, yeah I still got the question.

Why, if an hamiltonian is invariant under discrete traslations, can I diagonalize simultaneously the hamiltonian and the discrete translation operator?

From what we've said above it appears clearly that the hypotesis are weaker than requiring the hermitianity, nevertheless none weaker lemma or reference has been posted yet.

Ll.

9. Jul 26, 2010

### strangerep

OK, here's my attempt at an answer. (I'll ignore complications arising
from unbounded operators here.)

Let H be an Hermitian operator on the Hilbert space. Therefore, by the usual spectral
theorem, the eigenvectors of H (denoted $|n\rangle$ with respective eigenvalues $\lambda_n$)
form an orthonormal basis of the Hilbert space.
(Assume here that we've also normalized each eigenvector.)

Now let A be a linear operator on the Hilbert space that commutes
with H, i.e., [H,A] = 0. (No other assumptions are made about A.)

Then we have:

$$A H |n\rangle ~=~ A \lambda_n |n\rangle ~.$$

Using [H,A] = 0, this implies

$$H A |n\rangle ~=~ A \lambda_n |n\rangle ~.$$

or, written more suggestively,

$$H (A |n\rangle) ~=~ \lambda_n (A |n\rangle) ~.$$

Therefore, $(A |n\rangle)$ is an eigenvector of H with eigenvalue $\lambda_n$
and hence is proportional to $|n\rangle$. (This follows because all the
eigenvectors are mutually orthogonal.) So we have:

$$A |n\rangle ~\propto~ |n\rangle ~.$$

Hence each $|n\rangle$ is also an eigenvector of A and we denote the
constants of proportionality (ie the eigenvalues of A) as $a_n$ :

$$A |n\rangle ~=~ a_n |n\rangle ~.$$

from which it follows that the operator A can be expressed as

$$A ~=~ \sum_n a_n |n\rangle \langle n| ~~.$$

Does any of this help?

Last edited: Jul 26, 2010
10. Jul 26, 2010

### Llewlyn

However, it seems you have assumed non degeneration of the eigenvectors basis, don't you?
Degeneration makes the sim diagonalisation tricky, this is precisely why I would run into a math-statement.

Ll.

11. Jul 26, 2010

### xepma

Well, first I just emphasize that a unitary matrix is always diagonalizable + its eigenvalues lie on the unit circle -- just in case this wasn't clear.

Second, you are right that strangerep should have stated

$$A |n\rangle = \sum_{m} b_{m,n} |m\rangle$$

where the sum runs over all vectors corresponding to the degenerate subspace of H (i.e. all vectors $$|m\rangle$$ which share the same eigenvalue $$\lambda_n$$)

What this statement means is that the action of A on the degenerate subspace $$\textrm{span}\lbrace |m\rangle | H|m\rangle = \lambda_n |m\rangle\rbrace$$ is closed. As a correllary you can block diagonalize the unitary matrix. This is already enough, since the complete diagonalization of A cannot conflict with this block diagonalization.

12. Jul 26, 2010

### Llewlyn

Thanks it seems clear now.

Ll.

13. Jul 26, 2010

### Llewlyn

Thanks it seems clear now.

Ll.