# Angular momentum partial components of a $k$-dependent pairing potential

I am going over this review on pairing in unconventional superconductors :http://arxiv.org/abs/1305.4609v3

which on page 21 states that for a "regular" function $U(\theta)$, partial components $U_l$ of angular momentum $l$ scale as $\exp(-l)$ for large $l$. I tried to prove this statement but am not satisfied with my awnser, and would greatly appreciate some insight. Below is what I did so far.

I assume here that "regular" means infinitely derivable. The partial component $U_l$ is defined as such :

$U_l = \int_{0}^{\pi} U(\theta) P_l(\cos \theta) \sin \theta d\theta$

such that

$U(\theta) = \sum_{l=0}^{\infty} U_l P_l(\cos \theta),$

where $P_l(\cos \theta)$ is the $l$-th order Legendre polynomial. Let us make use of Rodrigue's formula :

$P_l(\cos \theta) = \frac{1}{2^l l!} \frac{d^l}{dx^l} [(x^2-1)^l] |_{x=\cos \theta}$.

The highest-order term of this polynomial is $\frac{1}{2^l l!} \frac{(2l)!}{l!} \cos^l \theta$. So in the development of $U(\theta)$, the contribution to the term of order $\cos^l \theta$ coming from the $l$-th order Legendre polynomial is $U_l \frac{1}{2^l l!} \frac{(2l)!}{l!} \cos^l \theta$. There are also other contributions to this order coming from the higher-order Legendre polynomials, but they will be proportional to some $U_k$ with $k>l$. As we want to prove that $U_l$ is exponentially small as $l$ gets large, we can neglect these contributions for now.

Let us now try to find an equivalent of $U_l \frac{1}{2^l l!} \frac{(2l)!}{l!}$ as $l$ goes to infinity. We can make use of Stirling's formula :

$l! \sim (\frac{l}{e})^l \sqrt{2 \pi l}$

which gives us

$U_l \frac{1}{2^l l!} \frac{(2l)!}{l!} \sim U_l 2^l \frac{1}{\sqrt{l \pi}}$.

If we want $U(\theta)$ to be a regular function, we need its high-order components in $\cos^l \theta$ to get smaller and smaller as $l$ gets large, as $\cos^l \theta$ behaves in a singular manner when $l$ goes to infinity. Thus, we need to have

$U_l \sim a^{-l}$, with $a>2$, as $l$ goes to infinity.

Why I am not happy with this awnser :

• I neglected higher-order components in the contribution to $\cos^l \theta$
• Maybe the $U_l$ could behave in a complicated oscillating way to make the $l$-th order term converge, without being exponentially small.

Does anyone have an alternate way of proving the fact that $U_l$ has to be exponentially small as $l$ gets large, or a way to complete the above proof ? Thanks for your help.

Just focusing on the $\cos^l\theta$ term is probably not going to get you anywhere, since $\cos^l\theta$, being a completely analytical function, is by no means singular (and following your argument you get exponential growth of $U_l$, not decay). It is a fairly well-known fact that for analytical functions (this is what "regular" means, roughly speaking the function is equal to its Taylor expansion. It is a stronger condition than being infinitely differentiable. And here, we require analyticity in a finite region on the complex plane), the expansion coefficient with Legendre polynomial (also for Chebyshev, etc) decays exponentially. The proof is not that trivial, as you can see from the requirement of complex analyticity it uses contour integral. You can find the proof in many textbooks, for example
• I meant $U_l \sim a^{-l}$ at the end of my post, not $a^l$, I just edited it. Thanks a lot for your reference, I'll look into it. – Dimitri Dec 11 '15 at 17:44