# The spectral theorem and Hamiltonians

The spectral theorem states that if $\hat H$ is a COMPACT self-adjoint operator on a Hilbert space $\mathcal H$, there is a basis for $\mathcal H$ consisting of of eigenvalues of $\hat H$. But how are we supposed to determine if a given Hamiltonian is compact? For example, in many introductions to the Born-Oppenheimer approximation, the claim is made that the electron Hamiltonian
$$\hat H_e = -\sum_{i = 1}^M \frac{\hbar^2}{2m_e} \Delta_{r_i} + V_{ee} + V_{eN} + V_{NN}$$
is self-adjoint and therefore has a basis of eigenvalues. (The three V's above are the coulombic nuclear-nuclear, nuclear-electron, and electron-electron interactions.) That $\hat H_e$ is self-adjoint is pretty obvious...but how do we know it's compact?

Physics should tell you. If you can envisage a continuous part of the spectrum say, high energy, far away from the force centers, the operator is certainly non-compact.

dextercioby
Homework Helper
A compact operator is bounded. The Hamiltonians in atomic/nuclear physics are unbounded.

strangerep
The spectral theorem states that if $\hat H$ is a COMPACT self-adjoint operator on a Hilbert space $\mathcal H$, there is a basis for $\mathcal H$ consisting of of eigenvalues of $\hat H$.
I presume you meant "...consisting of of eigenvectors of $\hat H$" ? :-)

... but how do we know it's compact?

Adding to what Bigubau said, I'll just mention that there are more general
versions of the spectral theorem that deal with unbounded operators.

I'll admit I don't know much about this topic, but wasn't the fact that the operators corresponding to observable have complete sets of eigenvectors enunciated as an axiom of quantum mechanics by Dirac? I don't think the theorems strangerep mentions are sufficiently general to cover all cases of physical interest. One of my physics professors used to say that the proof of completeness, convergence, etc., etc., in any physical case is that the experiment doesn't blow up.

The general spectral theorem: any Normal operator is unitarily equivalent to a multiplication operator-
is sufficiently general to cover the observables in QM.

Continuous spectra happens all the time in QM, the position/momentum operators for example.

strangerep
I'll admit I don't know much about this topic, but wasn't the fact that the operators corresponding to observable have complete sets of eigenvectors enunciated as an axiom of quantum mechanics by Dirac?

Initially, yes. Then others (such as von Neumann) introduced a more rigorous treatment
based on Hilbert spaces, etc, etc. Then it became clear that ordinary Hilbert spaces are
not really enough, and rigged Hilbert spaces are more desirable.

In Dirac's QFT Yeshiva lectures, he says (iirc) that "the mathematicians
got it wrong" -- referring to the initial attempts to use ordinary Hilbert space to
rigorize his ideas -- and mentions the "...more sophisticated spaces with which
mathematicians now work...", referring (I think) to Sobolev spaces and/or
rigged Hilbert space. Actually, his Yeshiva lectures advocate an essentially
algebraic approach, with as little emphasis as possible on the representation
space -- if I understand him correctly.

I don't think the theorems strangerep mentions are sufficiently general to cover
all cases of physical interest.

Which cases? The nuclear spectral theorem applicable in rigged Hilbert space
is very general.

One of my physics professors used to say that the proof of completeness, convergence, etc., etc., in any physical case is that the experiment doesn't blow up.

An amusing quip, (though a bit silly if one takes it seriously).

Actually, I can think of plenty of "experiments" that do blow up. :-)

To Strangerep,
All day I have been wrestling with this idea that rigged Hilbert spaces are "necessary" for QM. Most of the books I read, while they don't specifically call what they are doing a rigged Hilbert space use essentially the same idea by adding distributions.

The reason I bring this up is I wonder, is there an advantage to using "rigged Hilbert space" point of view as opposed to standard Hilbert space with distributions? Personally I have kind of gotten used to just using Hilbert space. I accept that the eigenfunctions of position/momentum don't exist in Hilbert space and since we can't specify position/momentum exactly, those states shouldn't exist.

From my experience: don't worry when calculating. Use any tricks that you have learned from physics books and papers. As long as you are getting reasonable answers - most probably the method used can be justified mathematically if so needed. When you are getting nonsense or infinities when physics suggests the result should be finite or different from what you are getting - then there is chance that you did something that mathematics is not allowing for. Reading about dense domains, rigged Hilbert spaces, hermitian but not self-adjoint operators, different self-adjoint extensions, distributions etc. may help you to understand what happens.

Of course knowing some of this in advance should not hurt.

strangerep
All day I have been wrestling with this idea that rigged Hilbert spaces are "necessary" for QM. Most of the books I read, while they don't specifically call what they are doing a rigged Hilbert space use essentially the same idea by adding distributions.

The reason I bring this up is I wonder, is there an advantage to using "rigged Hilbert space" point of view as opposed to standard Hilbert space with distributions?

Hilbert space with "distributions added" essentially is an example of rigged Hilbert
space (modulo some extra technical details). I pretty sure you're already working with
rigged Hilbert spaces without realizing it. (Indeed, this is true of anyone who works
with the Dirac bra-ket formalism, including delta distributions, etc.)

Check out Ballentine sect 1.4 (especially pp28-29) if you haven't already read it.

strangerep