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BeauGeste
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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.
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.