# Why are the inner products of the eigenfunctions of an operator with a discrete eigenvalue spectrum guaranteed to exist?

I was reading through a textbook, and the statement was made that the inner products are guaranteed to exist if the eigenvalue spectrum of the operator is discrete. I have come across no support for this claim, and the basis for this claim was not immediately apparent after quite a long time of consideration on my behalf. Furthermore, while trying to deduce the answer, I was lead to the complement of my first question: why does a continuous eigenvalue spectrum of an operator lead to non-normalizable eigenfunctions (i.e. the inner products don't exist)?

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 I think that this question is (also) suitable for a math forum. -you might get better answers. – Hans-Peter E. Kristiansen Aug 9 '12 at 22:18

For a complete explanation look at a functional analysis book.

But for discrete points of the spectrum one can explicitly write down the projection operator onto the corresponding eigenspace :

The resolvent of the operator has poles at the discrete points of the spectrum and the residue of such a pole is the projection operator onto the corresponding eigenspace , see http://en.wikipedia.org/wiki/Resolvent_formalism .

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 @Christoph also captured this in his post, and I would like to acknowledge that. All of these answers have helped me realize that the question I was asking does not have a simple answer like the author of the textbook I was reading implied; additionally, I think a full comprehension will require a few more years of study in math on my part. Thanks all. – BielsNohr Aug 12 '12 at 9:31 Comment to the answer(v1): Most mathematical books on functional analysis would just treat (bounded and unbounded) operators as living on a (standard) Hilbert space (where the inner product exists by definition). OP's title question(v1) only becomes non-trivial if one goes beyond the framework of standard Hilbert spaces, e.g. in the context of rigged Hilbert spaces. – Qmechanic♦ Aug 12 '12 at 19:58 @Qmechanics: Question (v1) is also non-trivial on standard Hilbert spaces, since not every value of the spectrum is an eigenvalue. So one has to prove that isolated points of the spectrum are indeed eigenvalues. – jjcale Aug 13 '12 at 17:48 The title question(v1) concerns just the eigenvalue spectrum (= point spectrum), not the full spectrum. – Qmechanic♦ Oct 2 '12 at 18:06

The spectrum of a bounded linear operator $A$ is by definition the set of numbers $\lambda$ where $A-\lambda$ is not invertible. In the finite dimensional case, this means $A-\lambda$ is neither injective nor surjective, and the former statement is just a fancy way of saying that there exists an eigenvector; as this eigenvector is by definition a part of the Hilbert space, it is in partiular normalizable.

However, in the infinite-dimensional case, injective and surjective are no longer equivalent and the spectrum decomposes into the point spectrum, the continuous spectrum and the residual spectrum.

The point spectrum is the part of the spectrum where the map is not injective (it might not be hard to prove that it is indeed discrete - too lazy to investigate right now). The residual spectrum is empty for normal operators (and thus in particular selfadjoint operators), which leaves the continuous spectrum where the map is injective (and thus there are no eigenvectors) but not surjective. The concept of eigenvector just doesn't make sense for the coninuous spectrum in this formalism.

The situation gets muddled (or, depending on your point of view, un-muddled) in the formalism of rigged Hilbert spaces, which is the proper framework for the treatment of unbounded operators: Rigged Hilbert spaces allow us to introduce 'eigenvectors' which aren't part of our Hilbert space. In particular, in case of the Hilbert space $L^2$, they can be non-normalizable functions (eg plane waves) or not functions at all (eg delta distributions).

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The complimentary question is the real question here. If you have a continuous set of vectors there could be a discontinuity, in which case the inner product would depend on how you approach the discontinuity. In that case we can say the inner product does not exist. This isn't a problem with a discrete set of vectors.

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I am still not entirely clear on the matter, and I believe you have simply shifted my question (I guess I am looking for something more fundamental than what you have given me). You said there "could" be a discontinuity... but what guarantees this (or is it a guarantee)? Does Hilbert space have inherent discontinuities when you go to the continuous level? (excuse my possible incorrect use of terminology math is not my forte). – BielsNohr Aug 8 '12 at 13:43

Here is a partial answer the question (v1). OP would have to provide more mathematically precise details and assumptions in order to guarantee that an eigenvector (for an operator $H$ with discrete spectrum) has finite norm, i.e. is normalizable. For instance, within the framework of rigged Hilbert spaces.

Counterexample: Domain $D:=C^{\infty}(\mathbb{R})$=infinitely often differentiable complex-valued functions on the real line $\mathbb{R}$. Norm:

$$|| f||^2 ~:=~ \int_{\mathbb{R}} \! dx~|f(x)|^2~.$$

Let the operator $H:=0$ to be the zero-operator, taking all functions $f$ to the zero-function $0$. (Let us mention that the pair $(H,D)$ can be modified to become a selfadjoint operator, but skip the details here.) The spectrum is discrete

$${\rm spec}(H)=\{0\}.$$

Any function $f\in D\backslash\{0\}$ is an eigenfunction for $H$ with eigenvalue $0$, but the norm $|| f||^2$ is not necessarily finite.

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 This is not in the spirit of the question, also there is no need for modification H is already trivially self-adjoint. It is not reasonable to call this spectrum discrete, as it is discrete with a continuous infinity of degeneracy, and the OP is clearly asking about a discrete point spectrum. – Ron Maimon Aug 10 '12 at 19:03 Answer to comments: 1. According to my reading of OP's question(v1), this answer does address the spirit of OP's question, namely whether it is possible or not to mathematically justify some physics lore OP had read. 2. The details concerning selfadjointness was skipped for brevity, since the domain of the adjoint operator was not specified. 3. An operator with discrete spectrum is allowed to have infinite-dimensional eigenspaces by definition. – Qmechanic♦ Aug 12 '12 at 14:34 For 1, it is possible to justify this "lore" (it's probably a theorem, not "lore"). For 2, there are no details! The zero operator is always self-adjoint, as are all multiples of the identity, in any domain. Refusing to say so is ignorant fear-mongering--- the (trivial and unimportant) issues of self-adjointness show up for operators with derivatives in cases where the boundary conditions don't allow integration by parts. 3. An operator with discrete spectrum in this problem is obviously not allowed this, because of this trivial example. – Ron Maimon Aug 12 '12 at 16:11 1. OP's statement only becomes a theorem if he specifies the pertinent assumptions. 2. Technically, if the domain $D(H^{\dagger})$ assigned to the adjoint operator $H^{\dagger}$ is different from the domain $D(H)$ of $H$, then $H$ is not selfadjoint (in the mathematical sense of the word). – Qmechanic♦ Aug 12 '12 at 17:37 1. You are supposed to find the reasonable assumptions (if any) that make the theorem true, in this case, that the spectrum has no continuous degeneracy. 2. I know the "mathematical sense" of self-adjoint (there is no other sense), and in quantum mechanics, the domains are always identified by the fact that it's a Hilbert space, so this complication doesn't exist, it is just an old mathematician scare tactic that doesn't work anymore, because anyone can look up self-adjoint on Wikipedia. – Ron Maimon Aug 12 '12 at 20:04