# Fourier Transform of Hermitian Operators

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.

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Meir Achuz
Homework Helper
Gold Member
The FT of a real function is not real, but satisfies a "reality condition"
F*(w)=F(-w).

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

Meir Achuz
Homework Helper
Gold Member
The operators are not FT of each other.

Hans de Vries
Gold Member
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

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