I have an operator $\hat{L}$ which gives
$$\hat{L} f(x) = \lambda \cdot f(x)$$
where $\lambda$ is the eigenvalue.
Now I Fourier-Transform my function $f(x)$:
$$\mathcal{F}(f)(p) = g(p)$$
Question: How do I transform my operator $\hat{L}$ such that it gives the same eigenvalues $\lambda$ when I apply it to $g(p)$?
Example:
Let's consider the special-case $f(x)=x^m$ and $\hat{L}=x\cdot \partial_x$. Therefore we have
$$\hat{L} f(x) = m \cdot f(x)$$
Now the Fourier-Transformation
$$\mathcal{F}(f)(p) = g(p) = (-i)^m \sqrt{2\pi} \delta^{(m)}(p)$$
How does my Operator $\hat{L'}$ look like, such that
$$\hat{L'} g(p) = m \cdot g(p)$$
Thanks for help!
Best Answer
The Fourier transform is a unitary operator on your space. This means that its transpose is its inverse, $\mathcal F ^* = \mathcal F^{-1}$. The typical thing to do is to replace $T$ with $\mathcal F T \mathcal F^*$. Observe that with this convention, you have $$ (\mathcal F T \mathcal F^*)\hat f=(\mathcal F T \mathcal F^*)\mathcal F f=\mathcal F Tf=\lambda\mathcal Ff=\lambda\hat f $$ provided that $f$ is an eigenfunction for $T$ with eigenvalue $\lambda$.
Edit: The Fourier transform interchanges the role of differentiation and multiplication by $x$. If I recall correctly, $-i\partial_x \mathcal F f=\mathcal F (-x f)$ and $x\mathcal F f=\mathcal F (-i\partial_x f)$, which means that in your example,
$$ \mathcal F^* x\partial_x\mathcal F f=\mathcal F^* x\mathcal F (-i x f)=\mathcal F^* \mathcal F (-\partial_x (x f))=-\partial_x (x f), $$
that is, $\mathcal F^* x\partial_x\mathcal F=-\partial_x x$.