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Learning the harmonic oscilator

  1. Dec 15, 2004 #1
    I'm currently taking graduate classes toward my phD in physics... when I was undergraduate I learn the harmonic oscillator just solving the schrodinger equation with such potential can be derive that: E=(n+1/2)hw, the wave functions (with hermite polynomial *e^-x2). that take to pages of derivations only.

    Now I'm taking QM at graduate level using the Cohen-Tannudji book, in this book the autors dedicate one chapter to the harmonic oscillator. they prove lemmas and deduce from [x,p]=ih => [a,a+]=1,...that E=(n+1/2)hw and the wave functions!

    I really like that approach!, it is so elegant to get such results without using wave mechanics (solving the schrodinger equation)but

    for proving that the eigenvectors are no degenerated they have to use the {|x>} representation for finding the ground state, and because the is only one solution it is not degenerated. why avoiding the {|x>} representation proving all those lemmas if you will have to used to prove the nondegeneracy of the eigenvector? Do you guys know how to prove that the eigenvalues are not degenarated without having to solve any equation?
    thanks in advance
  2. jcsd
  3. Dec 15, 2004 #2


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    There's a general theorem which states that in one dimension there are no degenerate bound states.

    The theorem is most easily proved by contradiction.
  4. Dec 15, 2004 #3
    where I can find that proof?
  5. Dec 15, 2004 #4

    I think the proof goes like this:

    Let 2 linearly independent solutions to the 1D Schrodinger equation be Ψ1 and Ψ2 .
    Form the Wronskian W= Ψ1'Ψ2 - Ψ2'Ψ1. Since the 2 solutions are linearly independent, we can show by diffrentiating the Wronskian with respect to x and using the fact that the solutions must satisfy the Schrodinger equation, the Wronskian must be a nonzero constant, i.e. position independent. If one of the solutions repesents a bound state, then it tends to zero at infinity( let it be Ψ1) then at infinity we are left with Ψ1'Ψ2=constant, so we see no matter what the value Ψ1' takes at infinity (ofcourse it is finite) Ψ2 is not zero at infinity and hence does not correspond to a bound state. This leaves us with only one solution that is bound and hence one dimensional bound states cannot be degenerate.
  6. Dec 15, 2004 #5
    I mean a proof without using the schrodinger equation at all, using the schrodinger equation the harmonic oscillator is reduced to two pages of math, and as I said I following the logic of cohen-tanudji when everything is proved from the commutator relation and the norm being >=0
  7. Dec 15, 2004 #6
    Well, this proof involves only one line of maths, not more. We don't solve the Schrodinger equation here, we merely use it on the way. True when you solve the harmonic oscillator problem by solving the Schrodinger equation, things get complicated but we are not solving the equation. Besides this proof applies to any 1D problem. However I don't know a proof involving only the commutator relations.
  8. Dec 15, 2004 #7


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    There is not other method to prove that the fundamental eigenstate of the Q harmonic oscillator hamiltonian is nondegenerate.U have to use the specific form of the annihilation operator.And just because u started in the coordinate representation,u'll have to use that representation as well.
    Proving that the solution of the 1-st order LODE (implied by the spectral equation [itex] \hat{a}|0> = |0> [/itex]) is unique is achived only by giving the annihilation operator a certain representation,and,because the coordinate representation was the handy one,u chose that one.
    The equation mentioned is not an abstract one.Not for the linear harmonic oscillator in QM.It is a specific model in which the Heisenberg unit algebra has a representation.U cannot prove the general nondegeneracy of the ground state of a general space of an irreductible representation of the unit Heisenberg algebra.It's just a particular vector (when generating the whole standard base by the Dirac-Fok method) and nothing more.

  9. Dec 16, 2004 #8
    thanks Daniel
  10. Dec 20, 2004 #9
    Up until that point, only certain aspects of the X and P operators had been exploited, and in order to go the last step, it became necessary to exploit two more properties. These properties serve to DEFINE the Hilbert space within which the operators a and a are said to act. Specifically, we want to invoke the properties that X itself is nondegenerate in that space, and its eigenkets {|x>} span the entire space (... and likewise for P). This itself is equivalent to saying that the Hilbert space can be represented by L2(R) relative to the {|x>} (or {|p>}) basis. Thus, by going into the {|x>} (or {|p>}) representation, we have a very appropriate setting within which to expose and exploit the said properties. Once exposed, these properties imply that the Hamiltonian itself, too, is nondegenerate and its eigenvectors form a basis of the Hilbert space in question.
  11. Dec 28, 2004 #10
    Since H doesn't commute with x, p or any combination of them it is clear that any complete set of commuting observables will have only one operator, and therefore it's eigenstates will be non-degenerate.

    In higher dimensions with potential linear in the square of the radius we would have degeneracies because L^2 and Lz would commute with H. Note that in one dimension and with coulombian potential we don't have degeneracy.
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