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Fourier Transform of Hermitian Operators

  1. Oct 25, 2007 #1
    Question: Is the Fourier Transform of a Hermitian operator also Hermitian?
    In the case of the density operator it would seem that it is not the case:

    [tex]\rho(\mathbf{r}) = \sum_{i=1}^N \delta(\mathbf{r}-\mathbf{r}_i)[/tex]

    [tex]\rho_k = \sum_{i=1}^N e^{-i\mathbf{k} \cdot \mathbf{r}}[/tex]

    I have a hard time believing that the FT wouldn't be Hermitian though since an obverservable in one space should be an observable in another space.
     
  2. jcsd
  3. Oct 25, 2007 #2

    clem

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    The FT of a real function is not real, but satisfies a "reality condition"
    F*(w)=F(-w).
     
  4. Oct 31, 2007 #3
    but aren't the x and p operators Fourier Transforms of each other AND Hermitian?
     
  5. Nov 1, 2007 #4

    clem

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    The operators are not FT of each other.
     
  6. Nov 1, 2007 #5

    Hans de Vries

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    These are the rules:

    Code (Text):

    Function                   Fourier transform

    Real and Even       --->   Real and Even
    Real and Odd        --->   Imag and Odd
    Imag and Even       --->   Imag and Even
    Imag and Odd        --->   Real and Odd
    Arbitrary Real      --->   Hermitian
    Arbitrary Imag      --->   AntiHermitian
    Hermitian is (Real and Even) plus (Imag and Odd).


    The application of an Hermitian operator is:

    1) A convolution with an Hermitian function in position space.
    2) A multiplication with an arbitrary Real function in momentum space.


    The application of the position operator is:

    1) The multiplication with a Real and Odd function (= x ) in position space.
    2) The convolution with an Imag and Odd function (= -ihd/dp) in momentum space.


    The application of the momentum operator is:

    1) The convolution with an Imag and Odd function (= -ihd/dx ) in position space.
    2) The multiplication with a Real and Odd function (= p) in momentum space.


    Regards, Hans
     
    Last edited: Nov 1, 2007
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