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Simple Eigenvalue Question

  1. Nov 27, 2011 #1
    So I understand the idea of eigenvalues, eigenvectors, and eigenfunctions corresponding to a given operator on some vector or function space. But I'm just wondering, why are eigenvalues so important in quantum mechanics and physics in general? What I mean is, why are scaled multiples of a given vector/function so important where as other characteristics of an operator seem to be less so?

    Like for example, the L^2 operator for angular momentum. The eigenvalues of the spherical Laplace operator are l(l+1) and the corresponding eigenvalues for L^2 are:


    But why are these the allowable values for angular momentum as opposed to any other arbitrary output of the Laplacian?

    I hope this makes sense. Thank you.
  2. jcsd
  3. Nov 27, 2011 #2


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    The straight answer is that it's an axiom.

    The handwavy answer is that from the full Schroedinger equation, when we require time-independent solutions we get Hψ=Eψ. The wavefunction ψ is continuous, and after imposing boundary conditions, we get discrete energy eigenvalues that matched the observed discrete spectra of atoms. From here we generalize that the observed values for other observables are also eigenvalues.
  4. Nov 27, 2011 #3
    ah okay...so the concept is sort of similar to the fact that operators corresponding to observables must be Hermitian since we want them to be real valued; similar in the sense that it just happens to be that specific characteristic of the operator that is useful?
  5. Nov 27, 2011 #4


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    The H in the equation in post #2 is the Hamiltonian operator.
  6. Nov 27, 2011 #5
    Right, I just meant that it seems like the use of eigenvalues is similar to the necessity that the operators be Hermitean; just a property that gives us what we want?
  7. Nov 28, 2011 #6


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    Yup. This axiom is usually called the "Born Rule". It's bizarre but it works.
    Last edited: Nov 28, 2011
  8. Nov 28, 2011 #7
    wow that is bizarre. Thanks so much atyy. There's gotta be something deeper there it seems.
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