Exponential of the product between $x$ the derivative operator of $x$ acting in a $f(x)$

Question

The question

I'm stuck here trying to figure out how to compute and prove, the following operator action in a function:

$\exp(\varepsilon x \partial_x) f(x) = f(x \exp(\varepsilon) )$

where $\varepsilon$ is a constant.

I saw this result and I failed in my attempt to reproduced it. What I did was to expand $\exp(\varepsilon x \partial_x)$ in Taylor's series as:

$\begin{align*} \exp(\varepsilon x \partial_x)f(x)& = \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon x \partial_x)^m f(x)\\ &= \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon x)^m\frac{\partial^m}{\partial x^m}f(x) \\ \end{align*}$

I took this procedure because I already know how to compute $e^{\partial_x}f(x)$. Let me show you what I did in this case:

The translation operator

The Taylor series of a function f is

\begin{equation} f(x)=\sum_{n=0}^\infty\frac{(\partial_x^nf)(a)}{n!}(x-a)^n \end{equation}

Expanding about $x+b$ and letting $a=x$:

\begin{equation} f(x+b)=\sum_{n=0}^\infty\frac{(\partial_x^nf)(x)}{n!}b^n=\sum_{n=0}^\infty\frac{((b\partial_x)^nf)(x)}{n!} \end{equation}

By definition:

\begin{equation} e^{b\partial_x}=\sum_{n=0}^\infty\frac{(b\partial_x)^n}{n!} \end{equation}

Hence

\begin{equation} f(x+b)=(e^{b\partial_x}f)(x) \end{equation}

Returning to my question

I tried to generalize or make something similar for the previous case I discussed but I didn't get anywhere. Anyone can give me a tip or recommend a book or paper?

Thanks!!

Answer 1

You are assuming that multiplication by $x$ and the derivative commute, and that's not the case. For clarity, let me write $M_x$ for the operator of multiplication by $x$. If $f(x)=x^k$, then $$ [M_x\partial_x f](x)=kx^k,\ \ [(M_x\partial_x)^2f](x)=k^2x^k,\ \ \cdots\ ,\ \ [(M_x\partial_x)^mf](x)=k^mx^k. $$

Then \begin{align*} [\exp(\varepsilon x \partial_x)f](x)& = \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon x \partial_x)^m f(x)\\ &= \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon )^mk^mx^k \\ &=x^k\,\exp(\varepsilon k)=x^k (\exp(\varepsilon)^k\\ &=f(x\exp(\varepsilon)). \end{align*}

Thus we get by linearity that, for any polynomial $p$, $$\tag1 [\exp(\varepsilon x \partial_x)p](x)=p(x\exp(\varepsilon)). $$ Analytic functions are differentiable term by term, so the differential operator gets inside the series. This allows us to extend $(1)$ to $f$ analytic.

Answer 2

Before Martin Argerami came up with a solution I friend of mine thought in something but he's not so sure if it's right... I like to think that it is a physicist solution. I'll explain why, but first, let me show you what he thought.

If we assume that $f(x)$ has a Fourier Transform, we can write:

\begin{align*} e^{\varepsilon x \partial_x}f(x) &= \int d^3k \, f(k)e^{\varepsilon x \partial_x}e^{ikx}\\ &= \int d^3k\, f(k) \sum_{n,m} \frac{(\varepsilon x \partial_x)}{m!}\frac{ikx}{n!} \end{align*}

Notice that

\begin{align*} (x \partial_x)^m x^n = n^m x^n \, \, , \end{align*}

where commutation relation is respect! And then we can resum the series in $m$ and $n$ to obtain:

\begin{align*} e^{\varepsilon x \partial_x}f(x) &= \int d^3k\, f(k) \sum_{n} e^{(\varepsilon)^n}\frac{(ikx)^n}{n!}\\ &= \int d^3k\, f(k) \sum_{n} \frac{(e^{\varepsilon}ikx)^n}{n!} \\ &= \int d^3k \, f(k) e^{ike^\varepsilon x} = f(e^\epsilon x) \end{align*}

It's a physicist solution because we're assuming that every operation is well defined.