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Proof of orthonormality of the hamiltonian when not real

  1. Jan 10, 2012 #1

    I know how to prove the orthonormality of the hamiltonian when it is real but am struggling to work out how to prove it when the hamiltonian is not real.

    When proving for a real hamiltonian the lefthand side equals zero as H(mn)=H(nm)complexconjugate. but if the hamiltonian is not real, i dont know how to proceed.

    Can anybody help?

  2. jcsd
  3. Jan 10, 2012 #2


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    I suspect you're using the term "orthonormality" incorrectly in this context. Did you perhaps mean "Hermitian", or maybe "self-adjoint"? (Try looking up these terms on Wikipedia if they're not familiar...)

    It's a bit hard to help when the problem is not clearly stated. (BTW, if this is homework it belongs over in the homework forums... :-)
  4. Jan 11, 2012 #3
    Thanks for the reply.

    It is not homework, but exam revision. Is that ok in this part of the forums?

    I think maybe i did not correctly explain the problem?

    We proved orthonormality for the eigenstates of the hamiltonian when assuming H is real and then the lecturer said that we would need to be able to do this when H is not real.

    To prove it we considered

    <ψm|H|ψn> = En<ψm|ψn> and <ψn|H|ψm> = En<ψn|ψm>

    and then took the complex conjugate of the second equation and subtracted it from the first eqn.

    The LHS equaled zero as Hmn=(Hnm)complex conj

    Does that make more sense?

    I am aware of how to prove that the Hamiltonian is Hermitian as this is something else we have covered, but maybe i misunderstood the lecturer?

  5. Jan 12, 2012 #4


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    Not sure. I'll leave it to the moderators to move this thread if they wish.

    A slightly more correct (and generalized) version of the theorem is that eigenstates of a self-adjoint operator A corresponding to distinct eigenvalues are orthogonal. "Hermitian" is the finite-dimensional (matrix) version of "self-adjoint". The latter applies to more general operators (e.g., derivatives) which are not ordinary matrices.

    The proof is a one-liner, once you understand the (abstract) meaning of "adjoint" and "self-adjoint", and also that eigenvalues of a self-adjoint operator are real (which also has a 1-line proof).
    (Textbooks such as Isham, or Ballentine can probably help there.)
    (a-b) \, \<\psi_a|\psi_b\>
    ~=~ a\,\<\psi_a|\psi_b\> - b\<\psi_a|\psi_b\>
    ~=~ \<A\psi_a|\psi_b\> - \<\psi_a|A\psi_b\>
    ~=~ \<\psi_a|A\psi_b\> - \<\psi_a|A\psi_b\>
    ~=~ 0 ~,
    implying [itex]\<\psi_a|\psi_b\> = 0[/itex] since a,b were assumed distinct.

    Which textbook(s) is your lecturer using?
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