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Connection between isolated eigenvalues and normalizable eigenstates

  1. Jan 13, 2009 #1
    It seems to be true, that if some eigenvalue of a Hamilton's operator is an isolated eigenvalue (part of discrete spectrum, not of continuous spectrum), then the corresponding eigenstate is normalizable, and on the other hand, if some eigenvalue of a Hamilton's operator is not isolated, then the corresponding eigenstate is not normalizable (not a vector of a Hilbert space) (like plane waves).

    This starts to become intuitively clear once one has seen sufficiently examples obeying this pattern, but I have never encountered any general proof for this. Anyone knowing something about the proof?
     
  2. jcsd
  3. Jan 14, 2009 #2
    orthonormality is automatic for different eigenstates of a (hermitian) Hamiltonian.

    consider |m> and |n>
    <m|H|n>=En<m|n>=Em<m|n>

    so they must be orthonormal. Renormalizability is simply due to definition. We only consider renormalizable solutions to the problem.

    We wish to show that if the eigenvalues are continuous, there cannot exist renormalizable states.

    So suppose the spectrum is continuous and the states are renormalizable. The same proof applies and <m|n>=0 if Em≠En. However, if that is true, we can consider m such that the eigenvalue Em is only epsilon away from n. Since |m> and |n> are solutions to (a very similar) differential equations, they cannot possibly be orthonormal. I'm certain that you can change it into some more rigorous arguments based on analysis... but hey, we are talking about physics afterall.

    Edit: Actually, a completely rigorous proof of this fact is easy. We can rewrite Schrodinger's eq using Green's function.

    [tex]\psi_m=\int G_m(x)V(x)\psi_m(x-x')dx'[/tex]

    and just rewrite <m|n>-<n|n> using the above, you'll find that in the limit E_m->E_n, it must go to 0 contradicting <m|n> always =0.
     
    Last edited: Jan 14, 2009
  4. Jan 14, 2009 #3
    You are using the words "renormalization" and "normalization" synonymously? Is it ok? The word "renormalization" sounds like that it could be used to describe a process [tex]x\mapsto \frac{x}{\|x\|}[/tex], but I've thought that it is not supposed to be used like this, because it also means some other more complicated tasks.
     
  5. Jan 14, 2009 #4
    Did I understand correctly, that basically you are saying, that if we had an eigenvalue [itex]E[/itex] in a continuous part of a spectrum, such that eigenstates for [itex]E+\epsilon[/itex] are members of the Hilbert space (in rigor sense), then we would get a contradiction

    [tex]
    0 \neq \|\psi_E\|^2 = \int\limits_{-\infty}^{\infty} dx\;\psi_E^*(x)\psi_E(x) = \lim_{\epsilon\to 0} \underbrace{\int\limits_{-\infty}^{\infty} dx\; \psi_E^*(x)\psi_{E+\epsilon}(x)}_{=0} = 0?
    [/tex]

    There are some problems with this. If you choose a sequence [itex]E_k\to E[/itex], how do we know that the corresponding eigenstates [itex]\psi_{E_k}[/itex] approach towards [itex]\psi_E[/itex]?

    For example, consider a matrix

    [tex]
    A=\left(\begin{array}{ccccc}
    0 & 0 & 0 & 0 & \cdots \\
    0 & \frac{1}{2} & 0 & 0 & \cdots \\
    0 & 0 & \frac{1}{3} & 0 & \cdots \\
    0 & 0 & 0 & \frac{1}{4} & \cdots \\
    \vdots & \vdots & \vdots & \vdots & \ddots \\
    \end{array}\right)
    [/tex]

    acting in [itex]L^2(\mathbb{N},\mathbb{C})[/itex]. The eigenstates

    [tex]
    \langle\psi_k| = (0,\delta_{2k}, \delta_{3k}, \ldots )
    [/tex]

    corresponding to eigenvalues [itex]\frac{1}{k}[/itex] with [itex]k=2,3,\ldots[/itex] don't converge towards the eigenstate

    [tex]
    \langle\psi_0| = (1,0,0,\ldots )
    [/tex]

    corresponding to the eigenvalue 0, despite the fact that [itex]\frac{1}{k}\to 0[/itex].
     
  6. Jan 14, 2009 #5
    :uhh: :biggrin:

    It could be that the original hypothesis needs some tweaking.
     
  7. Jan 14, 2009 #6
    I missed this argument at the first glance (because I was being confused with the renormalization terminology). If functions satisfy approximately the same DE then they should be approximately the same.

    This idea aims for the proof of the fact that eigenvalues of normalizable eigenstates are isolated. What about the converse. If an eigenstate is not normalizable, does it imply that the eigenstate is not isolated?
     
    Last edited: Jan 14, 2009
  8. Jan 14, 2009 #7

    Hurkyl

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    Let T be a self-adjoint operator acting on a Hilbert space, and let [itex]\psi[/itex] be a vector. Combining the spectral theorem with the Lebesgue decomposition theorem, [itex]\psi[/itex] can be (uniquely) written as a sum of three orthogonal states [itex]\psi_1[/itex], [itex]\psi_2[/itex], and [itex]\psi_3[/itex] such that:

    [itex]\psi_1[/itex] is in the absolutely continuous part; that is, expectation values of functions of T are given by
    [tex]\langle \psi_1 \mid f(T) \mid \psi_1 \rangle = \int_{-\infty}^{+\infty} f(\lambda) g_\psi(\lambda) \, d\psi[/tex]
    for some function [itex]g_\psi[/itex]. (I assume you can go from here to define wavefunctions)

    [itex]\psi_3[/itex] is in the 'discrete' part; that is, it is a (possibly infinite) linear combination of eigenvectors* of T.

    [itex]\psi_2[/itex] is in the singular continuous part. This part is yucky. (Though wikipedia claims it's supposed to consist only of physically impossible states....)


    Some things to be aware of:
    There is no reason why any of the three parts of the spectrum should be disjoint; e.g. you can have an eigenvalue* lying in the continuous spectrum.

    The discrete spectrum doesn't necessarily consist of isolated points; I believe any countable set of points can occur.

    The singular continuous spectrum part is yucky. Heuristically, it's smeared out so that you don't have eigenvalues, but it's too concentrated to be expressed as an ordinary integral.


    *: An eigenvector is, by definition, a vector (with additional properties). e.g. a plane wave is not a vector, because it's not an element of your Hilbert space, and so it cannot be an eigenvector in the strictest sense. (Of course, it can be in a suitably generalized sense)
     
  9. Jan 15, 2009 #8
    Sorry about the confusing terminology... after going through a class on QFT I almost forgot that the word "normalization" exists. It's always renormalization... renormalization... and renormalization. I don't know too much about functional analysis, so the best argument I can give is the above based on usual calculus. The fact that
    [tex]|\psi_E-\psi_{E_k}|_\infty\rightarrow 0[/tex]
    i think can be derived from:
    [tex]|\psi_E - \psi_{E_k}|=\left|\int G_E V\psi_E dx' -\int G_{E_k}V\psi_{E_k}dx'\right|[/tex]

    V(x) might need to have some nice behavior though.
     
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