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Discrete vs continuous eigenvalues

  1. Jul 2, 2010 #1
    What determines whether an operator has discrete or continuous eigenvalues?

    Energy and momentum sometimes have discrete eigenvalues, sometimes continuous. Position is always continuous (isnt it?) Spin is always discrete (isn't it?) Why?
  2. jcsd
  3. Jul 2, 2010 #2
    Momentum always has continuous eigenvalues (unless you use the artificial box normalization, when it always has discrete eigenvalues). The time-independent Schroedinger equation (the eigenvalue problem for the Hamiltonian of the system) gives the eigenvalues of the energy of that particular system (each system has a different Hamiltonian).

    It is a general rule that the energy spectrum is discrete for finite motions, i.e. motions in which the particle cannot be found infinitely far away. For such motions, the energy has a continuous spectrum. For motions in period structures, there is the added possibility that the energy can have any continuously varying value withing bands of finite bandwidth width separated by forbidden energy regions of finite gap.
  4. Jul 3, 2010 #3
    I think I figured it out. It depends on the range of the conjugate variable. If our space is limited to a box, we get discrete momenta. Otherwise, continuous momenta. If our system is periodic in time, we get discrete energy. Angles are necessarily finite/periodic, so angular momentum is discrete.

    I would wager that a deeper discussion would reveal that compactness is the key ingredient. That's just a guess.
  5. Jul 3, 2010 #4


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    Staff: Mentor

    I think this needs to be refined a bit. How about the simple harmonic oscillator? Discrete energy eigenvalues, but the wave function only approaches zero as x goes to infinity, never quite reaching it.
  6. Jul 3, 2010 #5
    So, what is the probability for finding the particle at distances larger than a as [itex]a \rightarrow \infty [/itex]?
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