Diagonalizability (in Hilbert Spaces)

1. Mar 13, 2006

Euclid

Under what circumstances is a (linear) operator $$\mathcal{H} \to \mathcal{H}$$ between a Hilbert space and itself diagonalizable? Under what circumstances does (number of distinct eigenvalues = dimension of H), i.e., there exists a basis of eigenvectors with distinct eigenvalues? Although I am interested in the answer to these questions from a mathematical point of view, I am thinking about them in the context of QM. If it makes the answer simpler, you can assume the operator is Hermitian.

2. Mar 13, 2006

Hurkyl

Staff Emeritus
(disclaimer: I don't consider myself an expert in this)

Are you thinking finite dimensions?

To be honest, I would simply say to all your questions "it is when it is." Though there are some special cases, such as all Hermitian operators being diagonalizable, and any matrix with distinct eigenvalues is diagonalizable as well.

As for having distinct eigenvalues, I can't imagine a simple criterion: I would think you'd just have to calculate it. I can imagine things like "a matrix has distinct eigenvalues iff its characteristic polynomial has no repeated factors", but that's rather trivial.

If you're working with an infinite-dimensional Hilbert space, then all of these issues become much, much stickier, and you have to appeal to things like distributions, or a spectral theorem.

If you dig through the "links" link at the top of this website, you will come to an online book "Physics for mathematicians". It treats the mathematics of elementary QM in a rather rigorous manner, including some technical details that I had never seen elsewhere. (e.g. that most interesting operators aren't even defined on all of your Hilbert space, and that there are Hermetian operators that are not self-adjoint!) I rather liked it.

3. Mar 14, 2006

Euclid

That document seems to be precisely what I have been needing. I have been reading up on the mathematical forumulation of QM out of some physics books, but I have been unhappy with the concern for details given by the authors. Thank you so much for the reference!

4. Mar 14, 2006

DavidK

In the finite dimensional case a linear operator, $$A$$, is diagonalizable if it is normal, i.e.

$$AA^{\dagger}=A^{\dagger}A.$$

5. Mar 17, 2006

mathwonk

my little 15 page linear algebra text on my webpage completely describes diagonalizability for finite dimensional spaces.

e.g. an operator is diagonalizable if and only if its minimal polynomial factors into distinct linear factors.

an operator on an inner product space (fnite dimensional) has an orthonormal basis of eigenvalues, i.e. can be diagonalized by an "orthogonal" matrix, if and only if it is symmetric. (over the reals).

in hilbert space you cannot use such simple criteria as how many distinct eigenvalues there are, since infinite sets always have proper subsets of the same cardinality. but compact hermitian operators form a nice class of operators possessing "hilbert bases" i.e. maximal orthonormal sets, of eigenvectors. These are not bases in the linear algebra sense of the word, because it takes a (convergent) infinite series of coefficients to express any vector in terms of them.

the little book by edgar lorch, spectral theory, is beautifully written by an expert, but unfortunately i have not read it. (It was even more unfortunate back in 1965 when i was i the course that I had not.)

Last edited: Mar 17, 2006
6. Mar 17, 2006

mathwonk

most interesting operators are perhaps differential or integral operators. differential operators are not defined naively unless the space consists of differentiable functions.

but limits of differentiable functions tend not to be differentiable, and hilbert space is supposed to be complete, hence closed under certain limiting operations. what to do? extend the space by adding in certain less differentiable functions to make it complete, and then extend the notion of differentiability to that larger space.

these extended notions of differentiability really depend on integrability, which is better preserved under limits, and the old friend, integration by parts, to switch the derivative of the non differentiable function onto some other test function.

operators which were "bounded" i.e. continuous on a smaller space, may become unbounded on a larger space, also posing problems of domain of definition. so all these questions are abstractions of everyday difficulties in dealing with differential and integral equations, which is where they arose.

(pretty glib for someone who does not know diddly about it, eh?)

7. Mar 18, 2006

cogito²

Normal operators can be diagonalized in a specific mathematical sense, but probably not the sense you're looking for (see the bounded and unbounded spectral theorem in its various forms). If you're looking at compact operators, it's a little cleaner and it can be done in the sense of eigenvalues/vectors (which is probably more what you're looking for). Check out this to get you started on compact operators.

8. Apr 2, 2006

Haelfix

Spectral theorems are usually valid up to the point of normal operators in Hilbert space. After that, well, it gets complicated and ugly (long proofs for very special cases of bizarre operators).

In the compact self adjoint case, the whole thing is super easy. Because in essence the step to convergence is trivial by Rellichs lemma. The actual spectrum is discrete, and of course is often found in physics.

9. Apr 3, 2006

dextercioby

That's why most physicists use the term "symmetric" for describing an operator which obeys $A\subset A^{\dagger}$. Using "hermitean" makes us think we're speaking about a bounded operator on an abstract Hilbert space, hence equal not included in its adjoint (provided the latter exists) where in fact, in physics, the only way to get to bounded operators is to use the inverse of Stone's theorem.

Here's an article on the wonderful subtleties of QM. It wouldn't attach.

Daniel.

10. Apr 3, 2006

dextercioby

Okay, here's the arxiv #

quant-ph/9907069. Enjoy

Man, i must admit that thread in the QM subforum is way over my head.

Daniel.

11. Apr 15, 2006

loopgrav

See "Quantum Mechanics in Hilbert Space" by Eduard Prugovecki. It's out of print, but you can find it at www.campusi.com.

As stated above, the results for infinite-dimensional spaces are far more difficult than for finite-dimensional spaces, and as far as I know, most of the results you are looking for don't even exist. To learn QM, you have to get over the desire for mathematical rigour. (QFT is even worse. Hence the development of axiomatic QFT.) It is generally assumed that all (hermitian) operators of interest are diagonalizable, along with the assumption that the Hilbert spaces of interest are separable. The most readable self-contained introduction I know of is Kolmogorov & Fomin "Introductory Real Analysis." It's slightly outdated in some of it's terminology, but it's a Dover paperback and well worth the price of admission.

12. Apr 19, 2006

Haelfix

Quantum mechanics can be made mathematically rigorous more or less. You need several additional structures (that often complicate and obscure intuition and don't serve much other than making it rigorous), like Rigged Hilbert spaces. Other than that, in the path integral formulation, the Wiener measure more or less summarizes vanilla finite degrees of freedom quantum mechanics.

The stuff that is sketchy is everything past that point, field theory, renormalization and so forth. In certain cases, for instance with supersymmetric conformal theories, and asymptotically free theories most things tend to go through (at least in certain formalisms) and are somewhat acceptable to most mathematicians. But if you want to get more general than that, it really hits a point where its no longer pure mathematics and more heuristic formal arguments that seem right, but are more or less unprovable. There are also of course, technical limitations that show up frequently.