# How to characterize self-adjoint operators in terms of orthogonal diagonalizability

Have a look at the following excerpt of Tosio Kato (taken from Zeidler Applied functional analysis vol. I):

The fundamental quality required of operators representing physical quantities in quantum mechanics is that they be self-adjoint which is equivalent to saying that the eigenvalue problem is completely solvable for them, that is, there exist a complete set (discrete or continuous) of eigenfunctions.

What does he mean? To me a self-adjoint operator $(A, D(A))$ on a Hilbert space $\mathcal{H}$ is a linear operator s.t. $A=A^\star$, which is equivalent to say that it is expressible in terms of a unique projection-valued measure $P_A$:

$$A=\int_{-\infty}^\infty \lambda\, dP_A(\lambda).$$

This is the best thing I can think of to match what Kato refers to. This is kind of incomplete, though. Where are those eigenfunctions Kato mentions? Also, a version of the spectral theorem holds for normal operators too. Why are they ruled out?

-

In terms of physics, there's a simple reason why you rule out normal operators: physical quantities are things that you can measure. And therefore the corresponding eigenvalues should be real. Normal operators in general admits complex eigenvalues.

If the self-adjoint operator is compact, then you know what the eigenfunctions are (the orthonormal basis you get from the spectral theorem; Kato may have meant his self-adjoint operators to be compact, but I doubt it). In the more general cases, what Kato (I assume) was thinking of is perhaps more along the line of "generalized" eigenfunctions. Two examples:

• On $L^2(\mathbb{R})$, the Laplacian is an unbounded self-adjoint operator (or rather, has a self-adjoint extension yada yada). From solving the ODE, you see that $e^{ikx}$ satisfy $\triangle e^{ikx} = -k^2 e^{ikx}$, so they look loke eigenfunctions, but of course, $e^{ikx}$ is not in $L^2(\mathbb{R})$.
• On $L^2([0,1])$, the operation $f(x) \mapsto x f(x)$ is bounded and self-adjoint. (But it is not compact.) It is easy to see by inspection that the Dirac distribution $\delta_{x_0}$ "solves" the eigenfunction equation with eigenvalue $x_0$, but of course the delta function is not an element of $L^2$.

In fact, in terms of the measure formulation, the eigenfunctions are precisely objects supported on a point $\lambda$. So if you apply the Lebesgue decomposition theorem, you see that for every $\lambda$ that is in the pure-point part of the measure $P_A$, the characteristic function of $\lambda$ is measurable, and its integral corresponds to a projection onto some subspace of your Hilbert space. Elements of those subspaces are eigenfunctions.

In any case, whenever you see sweeping statements like this made in books or articles, you should always take them with a grain of salt and treat them more like guiding principles rather than precise definitions.

-
If Kato had intended compact self-adjoint operators he wouldn't have said "complete set (discrete or continuous) of eigenfunctions", I think. – t.b. Jul 12 '11 at 15:12
+1 for the last sentence. – Mark Jul 12 '11 at 15:30
Excellent explanation as usual: thank you very much! Your last statement is something I will remember. By the way, aren't those "generalized eigenfunctions" usually presented as true eigenfunctions on physics' books? – Giuseppe Negro Jul 12 '11 at 16:11
@Theo: good point. @dissonance: depends on your definition of "physics' books" and on your probability measure used to say "usually". The "physics books" that I look at generally state very clearly that caveat. :) – Willie Wong Jul 12 '11 at 16:41