How to evaluate the action of a fractional differential momentum operator?

Question

I need to understand how a fractional operator works if before being applied on a test function it acts on another (well known)function: $$(-i\hbar\frac{\partial}{\partial x}\cdot\delta(x))^\epsilon\psi(x)$$ Of course i'm interested in non integer values of $\epsilon$. Actually my problem is to understand if there exist a method to apply the operator between brakets on the $\psi$ function. I tried in the following way, starting from the first integer derivatives: $$\frac{\partial}{\partial x}(\delta(x)\psi(x))=\frac{ \partial\delta}{\partial x}\psi(x)+\delta\frac{\partial \psi}{\partial x}$$ $$\frac{\partial^2}{\partial x^2}(\delta(x)\psi(x))=\frac{\partial}{\partial x}(\frac{ \partial\delta}{\partial x}\psi(x)+\delta\frac{\partial \psi}{\partial x})=\frac{\partial^2 \delta}{\partial x^2}\psi(x)+\frac{\partial \delta}{\partial x}\frac{\partial \psi}{\partial x}+\frac{\partial \delta}{\partial x}\frac{\partial \psi}{\partial x}+\delta \frac{\partial^2 \psi}{\partial x^2}$$ Is there any way to continue this evaluation also for a non integer exponent $\epsilon$? Or even a recursive formula that allows you to understand how to move inside the brakets the test function $\psi$?

Note that $\delta$ is not the Dirac delta function

P.S. I neglected the constants $i\hbar$ in my trial because the problem relies in the action of the operator and adding a constant would give a $(-i\hbar)^\epsilon$ factor.

Answer 1

δ(x) is very confusing, so, as per comment, I'll use f(x) instead. Moreover, since there is only one independent variable, I'll simplify notation to $\partial= \frac{\partial}{\partial x}$. You then started evaluating $$ (f'+ f\partial)^2 = (f'^2+f'' f) +3 f'f \partial + f^2 \partial^2\\ \equiv g(2;1)+ g(2;2) \partial + f^2 \partial^2 ,\\ (f'+ f\partial)^3 =g(3;1)+g(3;2)\partial +g(3;3)\partial^2 + f^3\partial^3, \\ (f'+ f\partial)^n=g(n,1)+...+ g(n;n)\partial^{n-1}+f^n \partial^n, $$ where $$ g(2;1)= \tfrac{1}{2} (f^2 )'', \qquad g(2;2)=\tfrac{3}{2} (f^2)' ;\\ g(3;1)= \tfrac{1}{2} (f(f^2)'')' ,\qquad g(3;2)= \tfrac{1}{2} f(f^2)''+\tfrac{3}{2}(f(f^2)')' ,\\ g(3;3)=\tfrac{5}{2} f(f^2)'+ f'f^2;\\ ... $$ and the simple dimensional structure of the coefficient functions $g(n,k)$ is evident.

But without further context it's hard to see what you are up to, and where you intend to go...

E.g., if f is a simple function such as $f=x$, define $y=\ln x$, hence $x\partial=\partial_y$, so you simply have a commutative binomial, $$ (1+\partial_y)^n= 1+n\partial_y+ \frac{n(n-1)}{2!}\partial_y^2+...+ n\partial_y^{n-1}+\partial_y^n, $$ which devolves to Newton's infinite series for the fractional exponent binomial, $$ (1+\partial_y)^\epsilon= 1+\epsilon\partial_y+ \frac{\epsilon(\epsilon-1)}{2!}\partial_y^2+ \frac{\epsilon(\epsilon-1)(\epsilon-2)}{3!}\partial_y^3+... . $$ Consequently, $$ (1+\partial_y)^\epsilon \psi(e^y) = \psi(x)+\epsilon x\psi'(x) \\ + \frac{\epsilon(\epsilon-1)}{2} \bigl(x\psi'(x)+x^2\psi''(x)\bigr )\\ +\frac{\epsilon(\epsilon-1)(\epsilon-2)}{3!} \bigl (x\psi'(x)+3x^2\psi''(x)+ x^3\psi'''(x)\bigr)+... $$ Note how nicely this coincides with the above expression for f=x when $\epsilon=3$ which truncates the series!