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Hamiltonian & Degeneracy: Conditions

  1. Jun 4, 2008 #1
    Hi all,

    could someone give me a quick answer on the exact conditions for the hamiltonian to be non degenerate, i.e. to have different eigenvalues?

    thanks in advance.
     
  2. jcsd
  3. Jun 4, 2008 #2

    CompuChip

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    What conditions do you need on a matrix to get different eigenvalues? (Does it even exist?)
     
  4. Jun 4, 2008 #3

    I have found a proof of:

    does anybody know more general conditions?
     
  5. Jun 4, 2008 #4
    the identity matrix has repeated eigenvalues.

    I don't think your statement about systems with no scattering states is true. Consider a Harmonic oscillator in dimension n > 1. There is angular momentum eigenstates, this is a form of degeneracy and the Harmonic oscillator does not have any scattering states.
     
    Last edited: Jun 4, 2008
  6. Jun 4, 2008 #5
    I agree with comote. There's also Kramers theorem.
     
  7. Jun 5, 2008 #6
    ok, but when does the hamiltonian looks like the identity matrix?

    i think the statement is only true for 1-dim. systems. the proof is written for the 1-dim. case., sorry about the imprecision.

    so there can't be any magnetic fields.
     
  8. Jun 5, 2008 #7
    I was answering two questions, one about Hamiltonians and one about Matrices.

    As for spin-less, time independent Hamiltonians in one dimension, I hesitantly believe the statement although I wouldn't be too surprised if someone provided a counter-example. Would you be able to provide the proof of that statement? At least a link or a reference.
     
  9. Jun 5, 2008 #8
    i'm not quite convinced of the proof myself, but it goes like this (i took it from shankar):

    if we have

    $-\frac{d^2}{dx^2}\psi_1+V(x)\psi_1=E\psi_1$
    $-\frac{d^2}{dx^2}\psi_2+V(x)\psi_2=E\psi_2$

    then we obtain after multiplying $\psi_i$ and substracting one equation from the other:

    $-\psi_2\frac{d^2}{dx^2}\psi_1+\psi_1\frac{d^2}{dx^2}\psi_2=0$

    so that

    $\frac{d}{dx}\left(-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2\right)=0$

    thus

    $-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2=C$.

    Since our states fall off to 0 at $\pm\infty$, we deduce that $C=0$

    Now comes the step that i don't understand:

    Shankar goes on with

    $\frac{1}{\psi_1}\frac{d}{dx}\psi_1=\frac{1}{\psi_2}\frac{d}{dx}\psi_2$

    and integrates this equation to

    $\log(\psi_1)-\log(\psi_2)=c$

    and therefore

    $\psi_1=e^c\psi_2$,

    and that supposedly concludes the proof.
     
  10. Jun 5, 2008 #9
    No one wants to read that. Please take the time and put [ tex ] tags around your equations. A simple search and replace in notepad would do the trick.
     
  11. Jun 5, 2008 #10
    i'm not quite convinced of the proof myself, but it goes like this (i took it from shankar):

    if we have

    [tex]-\frac{d^2}{dx^2}\psi_1+V(x)\psi_1=E\psi_1[/tex]
    [tex]-\frac{d^2}{dx^2}\psi_2+V(x)\psi_2=E\psi_2[/tex]

    then we obtain after multiplying $\psi_i$ and substracting one equation from the other:

    [tex]-\psi_2\frac{d^2}{dx^2}\psi_1+\psi_1\frac{d^2}{dx^2}\psi_2=0[/tex]

    so that

    [tex]\frac{d}{dx}\left(-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2\right)=0[/tex]

    thus

    [tex]-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2=C[/tex].

    Since our states fall off to 0 at [tex]\pm\infty[/tex], we deduce that [tex]C=0[/tex]

    Now comes the step that i don't understand:

    Shankar goes on with

    [tex]\frac{1}{\psi_1}\frac{d}{dx}\psi_1=\frac{1}{\psi_2}\frac{d}{dx}\psi_2[/tex]

    and integrates this equation to

    [tex]\log(\psi_1)-\log(\psi_2)=c[/tex]

    and therefore

    [tex]\psi_1=e^c\psi_2[/tex],

    and that supposedly concludes the proof.
     
  12. Jun 5, 2008 #11
    Cute. It shows that if two bound state wfs have the same energy then they differ at most by an irrelevant factor.

    What part do you have trouble with?
     
  13. Jun 5, 2008 #12
    the part with the integration. e.g. what if [tex]\psi[/tex] has an infinite but discrete set of zeros, doesn't that invalidate the step of integration and transformation of measure?
     
  14. Jun 5, 2008 #13
    more evidently, the equation with the logarithms doesn't hold when the wave function is negative! so i guess shankar is wrong? (this proof can be found on page 176).

    edit: i just remembered that the logarithm can have different branches where negative values of the wavefunction (that can be chosen real) would be well-defined (complex analysis was a loong time ago). but anyway, i'm sure there's something wrong with shankar's proof. for one reason, since i couldn't find it in many other popular QM books!
     
    Last edited: Jun 5, 2008
  15. Jun 5, 2008 #14
    The proof is just fine. You don't really need to integrate in the last step. He's already demonstrated that the Wronskian of the two solutions vanishes. Thus they are linearly dependent.
     
  16. Jun 5, 2008 #15
    i'm unfamiliar with this method, but wikipedia says that only the reverse of your statement is true in general.
     
  17. Jun 5, 2008 #16
    Mmm. In the counter-example given on wikipedia, the two functions are still piecewise linearly dependent, meaning they are linearly dependent except at discrete points.

    Normally, when you patch together solutions in different regions (where potentials are defined piecewise), you have matching conditions which ensure that the wavefunction is continuous, and that the discontinuity in [tex]\psi'(x)[/tex] is given by integrating the Schrodinger equation across that point. Thus, in 1D, you specify the WF and it's derivative at 1 point (remember, the SE is a second order equation), and the matching conditions uniquely give you the wavefunction everywhere else.

    Therefore, you won't find yourself in a situation where there is an ambiguity in terms of how to patch up the solutions. It is unique.
     
  18. Jun 6, 2008 #17
    ok i think i understand your argument using basic ode theory together with the statement on wikipedia that the functions must be linearly dependent in some region, if their wronskian vanishes everywhere. thank you very much for clarifying this point!
     
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