Symbolic expression for BlattWeisskopfSquared for arbitrary L

[mmikhasenko]

Problem description

Polynomials in the BlattWeisskopfSquared implementation are complicated.

Proposed solution

They can be replaced by symbolic expressions

(the screenshot is from a review)

What should the interface look like?

No response

Additional context

No response

[redeboer]

Thanks for the reference, that's great! Indeed would be super easy to implement this with SymPy it seems. And will also make the lamdification for parametrized $l$ way nicer than this casewise implementation

[redeboer]

First some clarity on the notation of

$$F_l^2(z) = \frac{1}{z\left|h^{(1)}_l\left(q/q_R\right)\right|^2}.$$

With the definition $z(s) = \left[q(s)/q_R\right]^2$, does that mean we get

$$F_l^2(z) = \frac{1}{z\left|h^{(1)}_l(z)\right|^2}\,?$$

(If I strictly replace though, it would be something like $q/q_R=\pm\sqrt{z}$.)

If so, does this mean the 'Hankel-normalized' Blatt–Weisskopf factor, like in 10.1002/andp.19955070504, p. 415 (in AmpForm denoted $B^2_l$) becomes

$$B_l^2(z) = \frac{z^l\left|h^{(1)}_l(1)\right|^2}{z\left|h^{(1)}_l(z)\right|^2}\,?$$

It's a bit handwaving here — trying to get the same order of $z$ in the nominator and getting that factor in front from the Hankel function.

[redeboer]

Unfortunately it seems that SymPy's hankel1 has no algebraic implementation for specific $l$ or $z$ values for $h_l^{(1)}(z)$. (I assume $H$ is just the same as $h$). The expression class lambdifies to scipy.special.hankel1.

But if Equation (96) is sufficient, we could implement a specialized Hankel1 expression class that evaluates to that series.

[redeboer]

if Equation (96) is sufficient, we could implement a specialized Hankel1 expression class that evaluates to that series.
#417 (comment)

With a custom Hankel1 class and some hacking to avoid letting the sympy.Sum evaluate if $l$ is symbolic, I get:

If I strictly replace though, it would be something like $q/q_R=\pm\sqrt{z}$.

#417 (comment)

To get the same polynomials as in the current implementation, this indeed required using $h_l^{(1)}\left(\sqrt{z}\right)$.