Fourier Transform of Hermitian Operators

  • Thread starter BeauGeste
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  • #1
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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.
 

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  • #2
Meir Achuz
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The FT of a real function is not real, but satisfies a "reality condition"
F*(w)=F(-w).
 
  • #3
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but aren't the x and p operators Fourier Transforms of each other AND Hermitian?
 
  • #4
Meir Achuz
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The operators are not FT of each other.
 
  • #5
Hans de Vries
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These are the rules:

Code:
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:

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