# Condition for completeness of eigen vectors of an operator

I am studying an article http://arxiv.org/abs/quant-ph/9907069 and having some problems understanding it.

Francois Gieres said:
If the Hilbert space operator A is self-adjoint, then its spectrum is real [6, 8][13]-[18] and the eigenvectors associated to different eigenvalues are mutually orthogonal; moreover, the eigenvectors together with the generalized eigenvectors yield a complete system of (generalized) vectors of the Hilbert space4 [19, 20, 8].

Is self adjointness of an operator a sufficient or necessary and sufficient requirement for its eigen vectors with the generalized eigenvectors (i dont know what are these) to form complete set? I want to read the proof too. The references include German texts which are not accessible for me right now.

P.S I have tried googling but couldnt find the answer.

dextercioby
Homework Helper
To my knowledge a full spectral theorem (either in the approach due to von Neumann or into the one by Gelfand-Kostyuchenko(-Maurin)) exists only for a self-adjoint operator or for a unitary one (and this due to the connection among the 2 types provided by Stone's theorem), i.e. there's no spectral theorem for weaker conditions: (closed) symmetric operators or even essentially self-adjoint ones.

AFAIK the sufficiency is the spectral theorem itself. I don't know if the condition for necessity valid for trace class operators (or a little more general for the compact ones) can be extended to the unbounded case, but, if it's true, I'd like to see a proof of that.

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DrDu
I have seen a lot of papers in the last 10 years studying the properties of non-hermitian but PT-symmetric operators. Apparently sometimes this weaker symmetry seems to be sufficient to allow for real eigenvalues. I don't know if and when they form a complete set.

To my knowledge a full spectral theorem (either in the approach due to von Neumann or into the one by Gelfand-Kostyuchenko(-Maurin))
What's the von Neumann approach? Is there an alternative to RHS to deal with the spectral theorem for unbounded operators in quantum mechanics?

dextercioby
Homework Helper
It's the method presented in any functional analysis book since 1932, the one using spectral measures. See the old book by Riesz and Nagy, for example.

It's the method presented in any functional analysis book since 1932, the one using spectral measures. See the old book by Riesz and Nagy, for example.
But I thought the Gelfand-Maurin nuclear spectral theorem was built off of the spectral theorem for unbounded operators found in functional analysis books.

dextercioby
Homework Helper
It was built off from a generalization found by von Neumann, namely the concept of <direct integral of Hilbert spaces>. Surely by writing the original article in the '50s, Gel'fand and Kostyuchenko had the advantage of having (almost) the full machinery of 'ordinary' Hilbert spaces already available + the pioneering work by Sobolev and Schwartz.

AFAIK the sufficiency is the spectral theorem itself. I don't know if the condition for necessity valid for trace class operators (or a little more general for the compact ones) can be extended to the unbounded case, but, if it's true, I'd like to see a proof of that.

Thanks for the information. Can I get the proof too, please. The spectral theorem I have seen till now is for matrices only (finite dimensional hilbert space or space with countable basis). Also I was wondering that the spectral theorem is valid for normal operators, so self-adjointness cant be a necessary condition (?)

What are generalized eigenvectors? What are bounded, unbounded and trace class operators? I have never encountered these terminologies and would love to know about them. Thanks:)

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Thanks for the information. Can I get the proof too, please. The spectral theorem I have seen till now is for matrices only (finite dimensional hilbert space or space with countable basis).
I think the way to start is to learn the spectral theorem for bounded linear operators from Kreyszig's excellent functional analysis book. But of course, first you need to know what a bounded linear operator is.

I assume you know what a normed vector space, and what a linear transformation is. Well, a linear operator is a linear transformation from a vector space to itself. And a linear operator T is bounded if there's some constant C such that ||T(x)|| ≤ C ||x|| for all x.

Attached is an excerpt from Kreyszig's book, detailing some concepts in spectral theory. (Don't worry if you don't understand everything in it; it's just so you get some familiarity with terminology.) After you've read over it, I can give you another excerpt which gives the actual statement and proof of the spectral theorem.

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• Kreyszig Spectral Excerpt.pdf
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Thanks a ton. I will report back in 2 days tops.

dextercioby
Homework Helper
Thanks for the information. Can I get the proof too, please. The spectral theorem I have seen till now is for matrices only (finite dimensional hilbert space or space with countable basis). Also I was wondering that the spectral theorem is valid for normal operators, so self-adjointness cant be a necessary condition (?)

I need to check on this. I think there's a spectral theorem for normal operators as well, and there are normal operators which are not self-adjoint.

What are generalized eigenvectors? What are bounded, unbounded and trace class operators? I have never encountered these terminologies and would love to know about them. Thanks:)

The generalized eigenvectors belong to a space of linear functionals over the Hilbert space. The other items find their definitions on wikipedia for short, but the book by Riesz & Nagy should clear it (or Kreyszig of course).

Yes, in any Hilbert Space (not necessarily Separable), any Normal Operator has a unique Spectral Decomposition.

Thanks a ton. I will report back in 2 days tops.
Are you ready for more Kreyszig? As I said, don't worry if you didn't understand everything. It was just so you could get familiarized with the concepts and terminology in sections 7.1 and 7.2.

Yes, sure. I have downloaded whole text (and it looks cool). GRE preperation is eating some time (but that is no excuse
I read the section recommended by you. I understood some part, couldnt understand much. It is perfectly fine when we work in finite dimensional spaces. It becomes abstract when I study infinite dimensional spaces.
I have read 7.1 (understood perfectly) and 7.2 (am aware of terminologies there). I just glided through 7.3 and 7.4 (consists of theorems on Banach spaces (which I understand as some kind of normed space ?)).

Yes, sure. I have downloaded whole text (and it looks cool). GRE preperation is eating some time (but that is no excuse
I read the section recommended by you. I understood some part, couldnt understand much. It is perfectly fine when we work in finite dimensional spaces. It becomes abstract when I study infinite dimensional spaces.
I have read 7.1 (understood perfectly) and 7.2 (am aware of terminologies there). I just glided through 7.3 and 7.4 (consists of theorems on Banach spaces (which I understand as some kind of normed space ?)).
OK, that's all you need from chapter 7. Now I suggest you try reading the chapter on the spectral theorem for bounded linear operators, and come ask questions if you don't understanding.

As for what a Banach space is, it's a borked vector space that is complete. In case you're not familiar with completeness and metric spaces, basically completeness means that there are no "holes" in your space. For instance, the rational numbers aren't complete, because you can take a sequence of rational numbers 3, 3.1, 3.14, ... has its terms get closer and closer together, but the sequence doesn't approach any rational number, because the rational numbers have a "hole" at pi. On the other hand, if you take a sequence of real numbers where the difference between terms goes to zero, then the sequence is guaranteed to approach some real number, so the real number system has no "holes".

And by the way, the definition of a Hilbert space is an inner product space that is complete, or to put it another way, an Banach space that is an inner product space.

Does the terminology "dense" mean the same thing?

dextercioby
Homework Helper
Yes, Q is dense everywhere in R, as any real number is the limit point of a sequence of rationals.

Thanks lugita and dexter for clearing the things.

DrDu