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Non hermiticity of Hamiltonian

  1. Mar 29, 2009 #1
    Hallo everyone, I have a question, how can I see that the hamiltonian H=p^2-x^4 is not hermitian, with p the momentum operator and x the position operator.
     
  2. jcsd
  3. Mar 29, 2009 #2

    CompuChip

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    Calculate its Hermitian conjugate: rewrite
    [tex]\int \Psi^* (\hat H \Psi) \, dx = \int (H^\dagger \Psi^*) \Psi \, dx[/tex]
    and show that [itex]H^\dagger \neq H[/itex].
     
  4. Mar 30, 2009 #3
    Thanks for the reply. the thing is I don't see why -x^4 is not hermitian. Since we know that p^2 is hermitian (from free particle problem), all we need to do is to investigate -x^4, which has to be hermitian from the integration you have shown. Maybe I miss anything?
     
  5. Mar 30, 2009 #4

    CompuChip

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    Errr, right, I missed that.
    p^2 is hermitian, and so is x. Therefore, x^4 is hermitian, and then the same holds for p^2 - x ^4.
    In fact, if H is really a Hamiltonian of a system, it should be hermitian.

    So maybe you are missing something, but then so am I :confused:
     
  6. Mar 30, 2009 #5
    Is H defined on the -infinity < x < +infinity?
     
  7. Mar 30, 2009 #6

    turin

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    Hermiticity depends on the space that the operator is acting on. You may have already seen this when (if) you dealt with angular momentum. Neither x nor p need be Hermitian, in general. What is the allowed range of x?
     
  8. Mar 30, 2009 #7
    Hey guys I think I know the solution. All Hamiltonian are subjected to boundary condition. If we allow x to be real, then the eigenfunction to this Hamiltonian will not converge when x goes to infinity (therefore we will have problem to show the Hermiticity of -x^4 in momentum representation, there partial integration will just yield infinity.) So if we extend x into complex plane (PT symmetry), it is clear that -x^4 is not Hermitian, because x is complex. See arXiv:physics/9712001, Bender.

    I hope my argument is correct. Thanks
     
  9. Mar 31, 2009 #8

    turin

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    I will suppress inessential constants, and treat everything as dimensionless, because I think that this is a mathematical issue rather than a physical issue.

    If H = -∂x2 - x4 on a < x < b,
    then I determine that <ψi|H|ψf>* = <ψf|H|ψi> + jif(b) - jif(a),
    where jif = ψf*(∂xψi) - (∂xψf)*ψi.

    So, if I simply require jif(a) = jif(b) for all i,f in my Hilbert space, then doesn't that make H Hermitian?

    Well, I would say that the functions that compose the Hilbert space on which the Hamiltonian is defined are subject to boundary conditions.

    If x→∞, then of course x4→∞. There is nothing wrong with this, and, in particular, this does not preclude the Hermiticity of x4 nor H. On the contrary, there would be something severely wrong with a finite x4 as x→∞.

    Extending x into the complex plane is not a PT operation. Usually, PxP = -x, and TxT = x*. I only put the "*" on the time-reversed x because you are suggesting an analytical continuation of the eigenfunctions of x into the complex plane. Normally, TxT = x. In particular, if x is already real-valued, then PTxPT = -x, which is also real-valued.

    I started reading that article, but it may be too far over my head. I don't think I agree with it, though. For example, PTpPT = p, but p has negative eigenvalues, so I must be missing part of their argument.
     
  10. Apr 1, 2009 #9
    The thing is we are still treating x as real, which if we use WKB to approximate the asymtotic eigenstate, you will notice that it just blows to infinity when x goes to infinity.That is what I am trying to say, if the Hamiltonian produces non L2 integrable function, then it is a not a well defined operator, hence the problem of hermiticity does not even have a place to start with.

    Your statement makes me ponder on another question: does one replaces the operator as real valued variable when going from quantum mechanics to classical mechanics? Or is complex variable also allowed? The problem is if we treat x and p as complex, then H=p^2-x^4 is not PT symmetry, as [H,PT][tex]\neq[/tex]0
     
  11. Apr 1, 2009 #10

    turin

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    By "it" you apparently mean the "wavenumber", that is, how quickly the function oscillates in x. I thought that you were talking about x4 . Anyway, the WKB method assumes the existence of the eigenfunction of H. So, if the WKB method does not give you a valid eigenfunction, then all that means is that you cannot use the WKB method to determine a valid eigenfunction. I don't think that you can use this argument to demonstrate the nonexistence of valid eigenfunctions.

    I don't know what you mean by "the Hamiltonian produces [a] ... function". The QM Hamiltonian acts on a wavefunction in order to generate time translation of that wavefunction, but QM posits the existence of the wavefunction independently of the Hamiltonian (i.e. you are free to specify an initial state). I will assume that you meant that the wavefunction on which the Hamiltonian acts must be L2(ℝn) (where I've been assuming that n = 1 for this problem). In that case, either you have been VERY frustrated with QM, or you have just recently decided to impose this condition on the "Hilbert" space of QM. See if you can figure out why I think this.
    Hints:
    - Why did I put quotation marks around "Hilbert"?
    - What is momentum space?
    - What is the norm of a position eigenstate?

    BTW, integrability is typically unrelated to the divergence of the wavenumber. For example, exp(i/x) is L2(-R,R), where R is any positive real number, even though it contains the point of explosion.

    CM variables may be complexified (e.g. ict in early formulations of relativity). You just have to decide how to interpret results in terms of observables. The correspondence between CM and QM does involve a complex phase, one way or another.

    Basically, I don't think we should be talking about "complex-valued" x and p.
     
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