I have a differential operator
$$D_x = \frac{1}{x} \left[ (x^2 – a^2) \frac{d}{dx} \left( \frac{1}{x} \frac{d}{dx} \right) + 2 \frac{d }{dx} \right], $$
and I would like to compute
$$ e^{i \lambda D_x} ~j_0(x) = \sum_{n=0}^\infty \frac{(i\lambda)^n}{n!} [D_x]^n ~j_0(x), $$
where $~j_0(x) = \sin(x)/x~$ is the spherical Bessel function.
Is there some nice way I could do this?
I tried using the Zassenhaus formula, using the splitting
$$ D_x = A_x + B_x, ~~{\rm where}~~
A_x=\frac{(x^2 – a^2)}{x}\frac{d}{dx} \left( \frac{1}{x} \frac{d}{dx} \right), ~~B_x = \frac{2}{x} \frac{d }{dx},$$
but it does not seem like a promising path, given that higher commutators $[A_x,B_x],~[A_x,[A_x,B_x]],\ldots$ get more and more complicated.
I also tried just brute force derivation that would give me a
series representation as
$$\sum_{n=0}^\infty \frac{(i\lambda)^n}{n!} [D_x]^n ~j_0(x)
= \sum_{n=0}^\infty \frac{(i\lambda)^n}{n!} \left[ f_n(x) \sin(x) + g_n(x) \cos(x) \right],$$
but I couldn't find close formulas for $f_n(x)$ and $g_n(x)$ functions.
What else could I do?
Best Answer
The answer can be represented as an integral which appears at the bottom of this answer. The first step is to transform the operator $D_x$ to one $D_w$ via the transformations $x=\sqrt{u}$ and then $w=u-a^2.$ We find that $$D_x=(x^2-a^2)\frac{1}{x}\frac{d}{dx}\frac{1}{x}\frac{d}{dx} + \frac{2}{x} \frac{d}{dx} \to 4D_w,\, D_w=w\,\frac{d^2}{dw^2}+\frac{d}{dw} .$$ The $D_w$ operator is related to the Laguerre polynomials which allows us to make significant progress. The OP's problem then becomes $$\exp{(i\lambda D_x)}\, j_0(x) \to \exp{(c D_w)}\, j_0(\sqrt{w+a^2}) \,\,\,, (c=4i\lambda). $$ Expand the sinc function, using the abbreviation $y=a^2$, $$ j_0(\sqrt{w+y}) = \sum_{k=0}^\infty (-1)^k\frac{(w+y)^k}{(2k+1)!}= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \sum_{m=0}^{k} \binom{k}{m} y^{k-m}\,w^m .$$ We now need to see how $\exp(c\,D_w)$ behaves on monomials $w^m.$ Expand the exponential and note how powers of $D_w$ behave:
$$ \big(D_w\big)^0 w^m = w^m $$ $$ \big(D_w\big)^1 w^m = \big(w\,\frac{d^2}{dw^2}+\frac{d}{dw}\big)w^m=m^2\, w^{m-1} $$ $$ \big(D_w\big)^2 w^m = \big(w\,\frac{d^2}{dw^2}+\frac{d}{dw}\big)\, m^2w^{m-1}=(m\,(m-1))^2\, w^{m-2} $$ $$ \big(D_w\big)^j w^m = \Big( j!\binom{m}{j} \Big)^2 \, w^{m-j} $$ Thus it is easy to see that $$ \exp(c\,D_w) w^j = \sum_{j=0}^m c^j j! \binom{m}{j}^2 w^{m-j} = m! \sum_{j=0}^m c^{m-j} \binom{m}{j} \frac{w^{j}}{j!} $$ where in the last step a series rearrangement has been performed. Use the definition of the Laguerre polynomials and we have $$ \exp(c\,D_w) w^j = m!c^m \, L_m(-w/c) .$$ Thus we have a representation of a solution, $$ \exp(c\,D_w) j_0(\sqrt{w+y}) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \sum_{m=0}^{k} \binom{k}{m} y^{k-m} c^m \, m!\,L_m(z) \, \, , (z=-w/c).$$ This representation can be transformed to a integral. Use the integral that can be found on Wolfram's website $$ m!L_m(z) = e^z \int_0^\infty e^{-t} t^m J_0(2\sqrt{t\,z}) dt. $$ Pull the integral through the finite sum, then the infinite sum (justifiable by analytic continuation in a region that the integral at the end makes sense). $$\exp(c\,D_w) j_0(\sqrt{w+y}) =e^z \int_0^\infty dt\,e^{-t} J_0(2\sqrt{t\,z}) \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \sum_{m=0}^{k} \binom{k}{m} y^{k-m} (ct)^m$$ $$=e^z \int_0^\infty dt\,e^{-t} J_0(2\sqrt{t\,z}) \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \big(y + ct\big)^k =e^z \int_0^\infty dt\,e^{-t} J_0(2\sqrt{t\,z}) \frac{\sin{(\sqrt{y+ct})}}{\sqrt{y+ct}} $$ In terms of the original variables this becomes $$\exp{(i\lambda D_x)}\, j_0(x) = e^{\big((x^2-a^2)/(-4i\lambda)\big)} \int_0^\infty dt\,e^{-t} J_0\Big(2\sqrt{t\,\frac{(x^2-a^2)}{-4i\lambda}}\Big) \frac{\sin{(\sqrt{a^2+4i\lambda t})}}{\sqrt{a^2+4i\lambda t}}.$$ Not knowing the domains of the parameters, I can't say if the integral is finite; it is probable that it is considering the nice functions in the integrand.