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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!

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  • $\begingroup$ you could just set $\hat{L}' := \lambda$ $\endgroup$ – Loreno Heer Jul 5 '14 at 22:29
  • $\begingroup$ Well, but what if I dont know $\lambda$? :) I would like to know the general solution, if it exists. $\endgroup$ – NicoDean Jul 5 '14 at 22:37
  • $\begingroup$ I can't help with the general solution but something like $$\hat{L}' = x^{-m} \hat{L} F^{-1}$$ works for the special-case (I think) $\endgroup$ – Loreno Heer Jul 5 '14 at 22:57
  • $\begingroup$ As i wrote above, I dont know the eigenvalue $\lambda$ or $m$, so this is not particularly useful for me unfortunatly. $\endgroup$ – NicoDean Jul 5 '14 at 23:06
  • $\begingroup$ I was thinking it might be able to generalize it to something like: $\hat{L}'(f) = f^{-1} \cdot \hat{L} F^{-1} (f)$ $\endgroup$ – Loreno Heer Jul 5 '14 at 23:08
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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, $\mathcal F^* x\partial_x\mathcal F=-\partial_x x$.

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  • $\begingroup$ Wonderful answere, thanks! One question for clarification: Can you please show how I would apply this to my special-case example above? Its not entirely clear to how to apply it to an operator yet - thank you. Btw, it would be useful to change your name such that people can remember you. (Saying this as it has been said to me some time ago :) ). $\endgroup$ – NicoDean Jul 5 '14 at 23:21
  • $\begingroup$ I am still puzzled on how to apply $(\mathcal F T \mathcal F^*)$ to a given operator $T$. Could you please provide an example for the operator given in my Example above? Thanks! $\endgroup$ – NicoDean Jul 6 '14 at 18:47
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    $\begingroup$ I've edited my answer to include you example. $\endgroup$ – Jonas Dahlbæk Jul 6 '14 at 19:29
  • $\begingroup$ This sounds correct. Still I am confused about how to do the last step: As I wrote above, I have $\mathcal F f(x) = g(p) = (-i)^m \sqrt{2\pi} \delta^{(m)}(p)$. Now how do I see that $(-\partial_p \cdot p) g(p) = m \cdot g(p)$? Thanks alot for your help! $\endgroup$ – NicoDean Jul 6 '14 at 20:40
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    $\begingroup$ The problem here is that $x^m$ is not an $L_2$ function, so it is not immediate what its Fourier transform should be. You have calculated the Fourier transform in the sense of distributions, but what you end up with is not a function, but a proper distribution. It is not immediately clear how the 'Fourier transform' of your operator should act on this distribution. $\endgroup$ – Jonas Dahlbæk Jul 6 '14 at 20:52
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For clarity I will use a slightly different notation. Starting from $L_xf(x)=\lambda f(x)$ is there an operator $L_p$ such that $L_p \hat{f}(p)=\lambda \hat{f}(p)$ ($\hat{f}(p)$ is the Fourier transform of $f(x)$) ? The particular example is $f(x)=x^m$ and $L_x = x\partial_x$. In this case $\hat{f}(p)=(-i)^m\sqrt{2\pi}\delta^{(m)}(p)$. First a word of warning: Care has to be exercised when manipulating Fourier transform, in particular when reversing the order of operations. An excellent textbook which explains such things is Robert Strichartz: A Guide to Distribution Theory and Fourier Transforms. From the observation that $${\cal F}[xf(x)](p) = -i\partial_p\hat{f}(p)$$ and $${\cal F}[\partial_x f(x)](p) = -ip\hat{f}(p),$$ we are led to propose that for this particular example $$L_p=-\partial_p p,$$ hence $$L_p\hat{f}(p)=-\hat{f}(p)-p\hat{f}^{(1)}(p).$$ Let us now investigate if $L_p \hat{f}(p)=\lambda \hat{f}(p)$ holds. We start with the right-hand side. Since the Fourier transform of $x^m$ is given as the multiple derivative of a Dirac delta function, it is clear that we shall have to work with distributions. I will use the short-hand notation $$<f,\varphi>=\int_{-\infty}^{\infty} f(p)\varphi(p)dp,$$ where $\varphi$ is a suitable test function, notably allowing for integration by parts with vanishing boundary terms. This leads to the formula $$<\delta^{(m)},\varphi>=(-1)^m<\delta,\varphi^{(m)}>=\varphi^{(m)}(0).$$ We can therefore evaluate the right-hand term as follows: $$\lambda<\hat{f},\varphi>=m(-i)^m\sqrt{2\pi}<\delta^{(m)},\varphi>=mi^m\sqrt{2\pi}<\delta,\varphi^{(m)}>.$$ For the first term on the left-hand side we get $$<-\hat{f},\varphi>=-(-i)^m\sqrt{2\pi}<\delta^{(m)},\varphi>=-i^m\sqrt{2\pi}<\delta,\varphi^{(m)}>.$$ For the second term on the left-hand side we get $$<-p\hat{f}^{(1)},\varphi>=-(-i)^m\sqrt{2\pi}<\delta^{(m+1)},p\varphi>=i^m\sqrt{2\pi}<\delta,(p\varphi)^{(m+1)}>.$$ In order to calculate the derivative on the right we use the general Leibniz rule $$(fg)^n=\sum_{k=0}^n \binom{n}{k}f^{(n-k)}g^{(k)}.$$ In our case the expansion will truncate after two terms: $$(\varphi p)^{(m+1)}=\binom{m+1}{0}\varphi^{(m+1)}p+\binom{m+1}{1}\varphi^{(m)}=\varphi^{(m+1)}p+(m+1)\varphi^{(m)}.$$ For the second left-hand term we therefore get $$<-p\hat{f}^{(1)},\varphi>=i^m\sqrt{2\pi}\left(<\delta,\varphi^{(m+1)}p>+(m+1)<\delta,\varphi^{(m)}>\right).$$ However, $<\delta,\varphi^{(m+1)}p>=0$, so after combining terms we obtain $$<-\partial_p p \hat{f},\varphi>=mi^m\sqrt{2\pi}<\delta,\varphi^{(m)}>,$$ which is in fact identical to the right-hand term, so the relation $L_p \hat{f}(p)=\lambda \hat{f}(p)$ holds.

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