The spectral theorem and Hamiltonians

  • #1
The spectral theorem states that if [itex]\hat H[/itex] is a COMPACT self-adjoint operator on a Hilbert space [itex]\mathcal H[/itex], there is a basis for [itex]\mathcal H[/itex] consisting of of eigenvalues of [itex]\hat H[/itex]. 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
[tex]
\hat H_e = -\sum_{i = 1}^M \frac{\hbar^2}{2m_e} \Delta_{r_i} + V_{ee} + V_{eN} + V_{NN}
[/tex]
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 [itex]\hat H_e[/itex] is self-adjoint is pretty obvious...but how do we know it's compact?
 

Answers and Replies

  • #2
1,481
4
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.
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
13,110
663
A compact operator is bounded. The Hamiltonians in atomic/nuclear physics are unbounded.
 
  • #4
strangerep
Science Advisor
3,261
1,199
The spectral theorem states that if [itex]\hat H[/itex] is a COMPACT self-adjoint operator on a Hilbert space [itex]\mathcal H[/itex], there is a basis for [itex]\mathcal H[/itex] consisting of of eigenvalues of [itex]\hat H[/itex].
I presume you meant "...consisting of of eigenvectors of [itex]\hat H[/itex]" ? :-)

... 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.
 
  • #5
139
12
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.
 
  • #6
72
0
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.
 
  • #7
strangerep
Science Advisor
3,261
1,199
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. :-)
 
  • #8
72
0
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.
 
  • #9
1,481
4
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.
 
  • #10
139
12
Hmmm...interesting. I guess I just need to learn more about functional analysis. Any book recommendations?
 
  • #12
strangerep
Science Advisor
3,261
1,199
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.
 
  • #13
strangerep
Science Advisor
3,261
1,199
Hmmm...interesting. I guess I just need to learn more about functional analysis. Any book recommendations?

If you want to learn it in reasonable depth, etc, then Kreyszig is the easiest to read (imho).
(He also gets superlative reviews on Amazon.) Sadly, he stops short of rigged Hilbert space
and the nuclear spectral theorem -- but so do almost all other FA texts that I know of, including
Lax, and even Reed & Simon. Gelfand and Vilenkin cover it, but you'll need an auxiliary
paper by Gould to correct an error therein -- search this forum for previous mentions if
you're really that keen.

Kreyszig also gives some application examples, including QM, but (imho) his treatment is
a bit labored because he doesn't use rigged Hilbert space.
 

Related Threads on The spectral theorem and Hamiltonians

Replies
5
Views
6K
Replies
5
Views
805
  • Last Post
Replies
15
Views
2K
Replies
4
Views
1K
Replies
2
Views
776
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
8K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
2K
Replies
1
Views
737
Top