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A translation operator

The Taylor series of a function $f$ is

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

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

$$f(x+b)=\sum_{n=0}^\infty\frac{(\partial_x^nf)(a)}{n!}(x+b-a)^n$$

Letting $a=x$:

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

By definition

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

Hence

$$f(x+b)=(e^{b\partial_x}f)(x)$$

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 $b=\log a$:

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

Since

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

Then

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

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:

$$(e^{\partial_{h(x)}}f)(x)=(e^{\partial_y}f)(x)=(e^{\partial_y}f)(h^{-1}(h(x)))=(e^{\partial_y}f)(h^{-1}(y))$$

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

$$(e^{\partial_{h(x)}}f)(x)=(e^{\partial_y}j)(y)=j(y+1)=f(h^{-1}(y+1))=f(h^{-1}(h(x)+1))$$

Hence solving the functional equation

$$h^{-1}(h(x)+1)=g(x)$$

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:

$$[\partial_x,x]=\partial_xx-x\partial_x=1$$

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.

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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.

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  • $\begingroup$ 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

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