The spectral theorem and Hamiltonians

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In summary: However, it will make it easier to understand why things might not be behaving as we expect.In summary, 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.
  • #1
AxiomOfChoice
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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?
 
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  • #2
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
A compact operator is bounded. The Hamiltonians in atomic/nuclear physics are unbounded.
 
  • #4
AxiomOfChoice said:
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
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
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
VKint said:
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
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
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
Hmmm...interesting. I guess I just need to learn more about functional analysis. Any book recommendations?
 
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  • #12
comote said:
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
VKint said:
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.
 

1. What is the spectral theorem?

The spectral theorem is a fundamental result in mathematics that provides a way to decompose a linear operator on a vector space into simpler parts. It states that any Hermitian operator (such as a Hamiltonian) can be represented by a diagonal matrix when expressed in an appropriate basis.

2. What is the significance of the spectral theorem in quantum mechanics?

In quantum mechanics, the spectral theorem is used to find the possible energy states of a physical system described by a Hamiltonian. It allows us to decompose the Hamiltonian into a set of eigenvalues and eigenvectors, which represent the energy levels and corresponding wavefunctions of the system. This is crucial in understanding the behavior of particles at the microscopic level.

3. How is the spectral theorem related to the uncertainty principle?

The uncertainty principle, which states that certain pairs of physical properties cannot be measured simultaneously with high precision, arises from the spectral theorem. This is because the eigenvalues of the Hamiltonian represent the possible energy levels of a system, and the corresponding eigenvectors represent the possible states of the system. Therefore, the spectral theorem shows that the more precisely we know the energy of a system, the less we know about its state, and vice versa.

4. What are the applications of the spectral theorem in other fields?

The spectral theorem has applications in various fields such as signal processing, control theory, and differential equations. It is used to analyze and understand the behavior of systems described by linear operators, which can represent a wide range of physical phenomena. It also has applications in computer science, specifically in the field of graph theory.

5. Are there any limitations of the spectral theorem?

While the spectral theorem is a powerful tool, it has some limitations. It only applies to Hermitian operators, which are symmetric matrices with complex entries. This means that it cannot be used for non-Hermitian operators, which are often encountered in quantum mechanics. Additionally, the spectral theorem only applies to finite-dimensional vector spaces and may not be suitable for infinite-dimensional spaces.

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