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Energy levels and Hilbert Spaces

  1. Jun 27, 2014 #1
    Hi.

    Is there a Hilbert Space for each energy level of a system? (And, in general, for every point in time?)

    I read in some book that if a equation for a problem accepts two different sets of wavefunction solutions (the case in question was the free particle and the sets of solutions in Cartesian and spherical coordinates) then the functions in one of those sets could be expressed by a linear combination of solutions of the same energy of the other set because of the completenes of the sets of eigenvectors of Hermitian operators.

    But a complete set of vectors should span the whole hilbert Space, hence my question.
     
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  3. Jun 27, 2014 #2

    vanhees71

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    There is one Hilbert space in quantum theory. The set of (generalized) eigenvectors of a self-adjoint operator (hermitian is not enough!) is complete, i.e., any Hilbert-space vector can be realized as a (generalized) linear combination of the set of orthonormal eigenvectors.

    What you are referring to is the case of degeneracy, i.e., an self-adjoint operator can have eigenvalues with more then one linearly independent eigenvector. Then you need to specify one or more other compatible observable(s) (of course all used observables must be mutually compatible, i.e., the self-adjoint operators representing them must commute) to pin down a particle's state by determining all these mutually compatible observables. If a set of such compatible observables has only one-dimensional common eigenspaces, the set is called complete. It is of course sufficient to use a minimal such complete set, i.e., it doesn't make sense to use a compatible set, where one observable is a function of the others.

    E.g., for a particle with spin 0, the momentum components are a (minimal) complete set of compatible observables. Their spectrum is [itex]\mathbb{R}^3[/itex], which is entirely continuous, and you thus have no true (normizable) eigenvectors but "generalized" ones, which can only be "normalized to a Dirac-[itex]\delta[/itex] distribution:
    [tex]\langle \vec{p}|\vec{p}' \rangle=\delta^{(3)}(\vec{p}-\vec{p}').[/tex]
    Now consider a free particle. Its Hamiltonian is
    [tex]\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2.[/tex]
    This implies that any (generalized) momentum eigenstate is also an energy eigenstate, but determining the energy of the particle only fixes [itex]\vec{p}^2[/itex] and not the three momentum components, i.e., for each energy eigenvalue [itex]E \geq 0[/itex] there are infinitely many (generalized) eigenvectors.

    Another set of compatible observables are the energy, [itex]\vec{L}^2[/itex], and [itex]L_z[/itex], where [itex]\vec{L}[/itex] is the orbital angular momentum of the particle. Now, indeed you can express any eigenvector of the Hamiltonian with a given energy eigenvalue [itex]E[/itex] as a linear combination of a complete set of other eigenvectors, i.e., you have
    [tex]|E,l,m \rangle=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \; \sin \vartheta \, A(\vartheta,\varphi) |\vec{p}(\vartheta,\varphi) \rangle,[/tex]
    with some function [itex]A(\vartheta,\varphi)[/itex]. Here [itex]\vartheta[/itex] and [itex]\varphi[/itex] parametrize the spherical shell with radius [itex]|\vec{p}|=\sqrt{2 m E}[/itex] in the usual way of spherical coordinates:
    [tex]\vec{p}=|\vec{p}| \begin{pmatrix}
    \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \sin \vartheta
    \end{pmatrix}.
    [/tex]
     
  4. Jun 27, 2014 #3
    Thanks for your answer, I think I grap what you mean, but I'm gonna need to ask a few things.

    In this case the self-adjoint operator would be the Hamiltonian, the eigenvalues the energy levels and the l.i. eigenvectors the solutions, right?
    So just with the op with degeneracy and a compatible one we would already have minimal such set?
    What do you exactly mean by "generalized" eigenvectors? Are those "generalized" vectors still in the HS?
    Tell me if I understood this part: the [itex]P^2[/itex] operator is not useful (by itself) to pin down the [itex]E_n[/itex] state because it is also degenerate (i.e. one [itex]P^2[/itex] -> many ψ(x)), so we also use [itex]L^2[/itex] and [itex]L_z[/itex]. The indices in [itex]\left|E, l, m\right\rangle[/itex] mean, respectively, that it is an eigenket of a certain E (and of [itex]\vec{p^2}=\sqrt{2mE}[/itex]) of a certain eigenvalue of [itex]L^2[/itex] and of a certain eigenvalue of [itex]L_z[/itex] Is it so?

    At the end of it all I still can't put together an answer to my question. Why can't I express [itex]\left|E, l, m\right\rangle[/itex] as a linear combination of any complete set of eigenvectors, like the ones correspondings to another state [itex]\left|E', l', m'\right\rangle[/itex] or the ones that consitute a solutions to another problem?
     
  5. Jun 27, 2014 #4

    ChrisVer

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    @Vanhees71, What's the difference between self-adjoint and hermitian operator?
     
  6. Jun 27, 2014 #5

    Matterwave

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    I can answer this one. A self-adjoint operator is a Hermitian operator with the additional condition that the domains of the operator and its adjoint are the same so that it really IS "self-adjoint". Self-adjoint is a stronger condition.
     
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