A translation operator

The Taylor series of a function $f$ is


where $\partial_x$ is the derivative operator. Expanding about $x+b$:


Letting $a=x$:


By definition




Hence $e^{\partial_x}=T$ where $T$ is the translation operator and $(Tf)(x)=f(x+1)$.

A scaling operator

We can also find a closed form for a scaling operator $S$ where $(Sf)(x)=f(ax)$.

$$f(xa)=f(e^{\log{xa}})=f(e^{\log x+\log a})=f(e^{y+\log a})$$

where $y=\log x$. Letting $g(z)=f(e^z)$:

$$f(xa)=g(y+\log a)$$

By our first theorem, $g(y+b)=(e^{b\partial_y}g)(y)$. Letting $b=\log a$:

$$f(xa)=(e^{(\log a)\partial_y}g)(y)=(a^{\partial_y}g)(y)=(a^{\partial_{\log x}}f)(e^y)$$


$$\frac{\partial}{\partial \log x}=\frac{\partial x}{\partial \log x}\frac{\partial}{\partial x}=x\frac{\partial}{\partial x}$$


$$f(xa)=(a^{x\partial_x}f)(e^{\log x})=(a^{x\partial_x}f)(x)$$

Therefore $S=a^{x\partial_x}$ defines our scaling operator.

A general operator

Suppose we want an operator $G$ such that $(Gf)(x)=f(g(x))$. Consider:


where $y=h(x)$. Letting $j=f\circ h^{-1}$ yields:


Hence solving the functional equation


for $h(x)$ allows us to define our general operator $G=e^{\partial_{h(x)}}$.

For example, letting $g(x)=xa$, the function $h(x)=\frac{\log x}{\log a}$ is a solution:

$$h^{-1}(y)=a^y$$ $$h^{-1}(h(x)+1)=a^{h(x)+1}=a^{\frac{\log x}{\log a}+1}=a^{\frac{\log x}{\log a}}a=e^{\log x}a=xa$$

Hence the corresponding operator takes the form

$$e^{\partial_{h(x)}}=e^{\partial_{\frac{\log x}{\log a}}}=e^{\log a\partial_{\log x}}=a^{x\partial_x}$$

This is the scaling operator we derived before.

The case $e^{\partial_{h(x)}}=e^{x^n\partial_x}$ or equivalently $h(x)=\frac{x^{1-n}}{1-n}$ corresponds to the basis for the Witt algebra.

The question

My question is as follows: Can a similar procedure be used to find $e^{{\partial_x}^2}$, or $e^{{\partial_{h(x)}}^n}$ in general? Note that the commutator of $x$ and $\partial_x$ is nonzero and given by:


Furthermore, given the product rule $Dab=(Da)b+aDb$, it seems to be the case that

$$(x\partial_x)^2=x\partial_x+x^2\partial_x^2$$ $$(x\partial_x)^3=x\partial_x+3x^2\partial_x^2+x^3\partial_x^3$$ $$(x\partial_x)^4=x\partial_x+7x^2\partial_x^2+6x^3\partial_x^3+x^4\partial_x^4$$

and in general

$$(x\partial_x)^n=\sum_{k=1}^n \genfrac{\lbrace}{\rbrace}{0pt}{}{n}{k} x^k\partial_x^k$$

where $\genfrac{\lbrace}{\rbrace}{0pt}{}{a}{b}$ are the Stirling numbers of the second kind. I have a strong suspicion the answer might have to do with properties of the Fourier and Laplace transforms, as seen here and here.


It is known that $e^{\partial_x^2}$ is formally the Weierstrass transform (there is a short derivation in the link). What is not so well known is that the (unitary, angular frequency) Fourier transform is $e^{\frac{\pi i}{4}(\partial_x^2-x^2+1)}$. This comes from the fact that the eigenfunctions of the number operator in the quantum harmonic oscillator:

$$\hat{N} \psi_n(x) = n \psi_n(x)$$

are also eigenfunctions of the Fourier transform with eigenvalue $(-i)^n$, so formally the Fourier transform is $(-i)^{\hat{N}} = (-i)^{\frac12 (x^2-\partial_x^2-1)}$. You can also get the expression for the ordinary frequency Fourier transform with a change of variable.

In general you can show that the linear canonical transformations, of which the Fourier, Weierstrass and Laplace transforms are special cases, are of the form $e^{P_2(x,\partial_x)}$, where $P_2(x,y)$ is a complex polynomial in two variables of order $\leq 2$.

I don't know much about exponentials of higher order derivatives, but I can say that they will in general correspond to integral operators (the kernel will be the action of the operator on a delta function). As with the Fourier transform, you could formally set up a quantum mechanical problem where the Hamiltonian is

$$\hat{H} = P(x,\partial_x)$$

and solve the Schrödinger equation

$$\hat{H} \psi(x,t) = \frac{\partial\psi}{\partial t}$$

such that the unitary evolution operator is $e^{tH} = e^{tP(x,\partial_x)}$ (t is a parameter that you can set to 1). With the initial condition $\psi(x,0)=f(x)$ you can hopefully solve for $\psi(x,1)$ and work out what the operator does.

Edit: in the special case $P(x,\partial_x) = \partial_x^n$, the kernel turns out to be (proportional to) the Fourier transform of $e^{x^n}$, which is a generalized hypergeometric function as per this paper.

  • $\begingroup$ If I let $P=x\partial_x$, then $H=x\partial_x$, which yields $\psi(x,t)=C(t+\ln x)$. Using $\psi(x,0)=f(x)$, I get $C(\ln x)=f(x)$. I'm not really sure how this helps me to see what $e^{x\partial_x}$ does. Is there a good source that you could recommend which discusses linear canonical transformations in the form of $e^{P_2(x,\partial_x)}$? $\endgroup$ – user85503 Feb 4 '17 at 18:24
  • $\begingroup$ Ultimately, what I'm really interested in is $P=x\partial_x^2$. $\endgroup$ – user85503 Feb 4 '17 at 18:55
  • $\begingroup$ @user85503 Setting $t=1$ yields $C(\ln x +1)) = C(\ln (ex)) = f(ex)$. In general, the operator $e^{ax \partial_x}$ scales the argument of $f$ by $e^a$. $\endgroup$ – pregunton Feb 4 '17 at 19:33
  • $\begingroup$ @user85503 I didn't find the reference I was looking for, but you can check this, especially section 9.3, which shows examples of the correspondence between $e^{P_2(x,\partial_x)}$ and linear canonical transformations. As for the case $P=x \partial_x^2$, I don't see an easy way of solving the PDE, but you can try asking here as a separate question and see if somebody more knowledgeable than me can help you. $\endgroup$ – pregunton Feb 4 '17 at 23:50

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.