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A How does a pseudo-Hermitian model differ from a Hermitian?

  1. Nov 28, 2017 #1

    SeM

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    Hi, I have not been able to learn how a pseudo-Hermitian differs from a Hermitian model. If one has a hermitian model that satisfies all the fundamental prescriptions of quantum mechanics, a non-Hermitian would not, as it yields averages with complex values. How does a pseudo-Hermitian differ from Hermitian and non-Hermitian models?

    Thanks
     
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  3. Nov 28, 2017 #2

    A. Neumaier

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    What do you mean by ''a pseudo-hermitian model"?
     
  4. Nov 28, 2017 #3

    SeM

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    A pseudo-Hermitian Hamiltonian.
     
  5. Nov 28, 2017 #4

    A. Neumaier

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    And when is a Hamiltonian pseudo-Hermitian? Do you mean non-Hermitian?
     
  6. Nov 28, 2017 #5

    dextercioby

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  7. Nov 28, 2017 #6

    SeM

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  8. Nov 28, 2017 #7

    A. Neumaier

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    The difference is quite shallow: If the inner product to be used is not the standard inner product in the Hilbert space but $$\langle \phi|\psi\rangle_{new}=\langle \phi|M|\psi\rangle_{standard},$$ with positive definite ##M## then the observables are required to be Hermitian in the new inner product, which is a condition different from Hermiticity in the standard inner product. Thus in the latter, the Hamiltonian need not be Hermitian, but it doesn't matter since it is not the product defining the physical Hilbert space.
     
  9. Nov 28, 2017 #8

    SeM

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    Maybe this is not directly related, but if the Hamiltonian is not Hermitian, but its matrix elements are in a Hilbert space L^2, and satisfy the condition:

    \begin{equation}
    \langle x,y \rangle = \int_a^bx(t)\overline{y(t)}dt,
    \end{equation}

    and the corresponding eigenvectors form an open subset M, in H, thus the Hamiltonian operates in Hilbert space, then, can one still use the hermitiian form <Phi|Phi*>=1 to normalize the non-hermitian solution? Or is the pseudo-hermitian form a method to work around the non-hermitian solution which has to be somehow normalized?
     
  10. Nov 28, 2017 #9

    A. Neumaier

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    In the pseudo-Hermitian case, the correct normalization leading to consistent probabilities must always be done using the physical inner product involving the operator ##M##.
     
  11. Nov 28, 2017 #10

    SeM

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    Can you give an example of this integral equality, i. e:

    \begin{equation}
    N\int_a^b \psi \Omega \psi* = 1
    \end{equation}
     
    Last edited: Nov 28, 2017
  12. Nov 29, 2017 #11

    A. Neumaier

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    I don't understand the connection to your question. All I said was that one uses a different inner product to define the physical Hilbert space, and one need to use it whenever one normalizes.
     
  13. Nov 29, 2017 #12

    SeM

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    Thanks.

    The normalization condition, which is normally for a Hermitian system:

    \begin{equation}
    N \int_a^b \psi\psi* = 1
    \end{equation}

    is an inner product for the Hermitian type of wavefunction. However, you wrote

    \begin{equation}
    \langle \phi \mid M \mid \phi \rangle
    \end{equation}

    is this the inner product of the "pseudo-hermitian" wavefunction?
     
  14. Nov 29, 2017 #13

    A. Neumaier

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    The inner product would typically be $$\int \phi(x)^*M(x,y)\psi(y)dxdy$$ with a positive definite kernel ##M(x,y)##. The case ##M(x,y)=\delta(x-y)## gives the standard inner product.
     
  15. Nov 29, 2017 #14

    SeM

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    Thanks Neumaier, that was clear!

    Cheers
     
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