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}


\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?



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.

  • $\begingroup$ When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $\partial_x$. Very thank you for the solution, Martin. $\endgroup$ – Alexssandre de Oliveira J Feb 1 at 19:30

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.