Fourier Transform of Hermitian Operators

[BeauGeste]

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:

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

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

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.

 

[Meir Achuz]

The FT of a real function is not real, but satisfies a "reality condition"
F*(w)=F(-w).

 

[BeauGeste]

but aren't the x and p operators Fourier Transforms of each other AND Hermitian?

 

[Meir Achuz]

The operators are not FT of each other.

 

[Hans de Vries]

These are the rules:

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