I'm following this paper, and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \begin{equation} \langle f_s|f_{s'}\rangle = \int_{0}^{1} \frac{2 \, du}{(1-u)^2} f_s(u)^*\, f_{s'}(u) \,.\tag{67} \end{equation} The singular contribution of the integral arises at the endpoint $u=1$ of the integral, and in the limit $u \to 1$, the function $f_s(u)$ takes on the form \begin{equation} f_s(u) \approx a_s (1-u)^{1/2 + is} + a_{s}^* (1-u)^{1/2-is}\tag{70} \end{equation} $f_{s'}(u)$ is also given by this expression but with $s \to s'$, where $s,s' \in \mathbb{R}_0^+$ and where $a_s,a_{s'} \in \mathbb{C}$ are complex coefficients in terms of $s, s'$, respectively. Using these approximations, they immediately present in equation (70) the normalization condition \begin{equation} \langle f_s | f_{s'} \rangle = 4 \pi |a_s|^2 \delta(s-s')\tag{70} \end{equation} which is what I'm trying to obtain. My attempt at solving this involved splitting the integral into two, as follows \begin{equation} \int_{0}^{1} \frac{2 \, du}{(1-u)^2} f_s(u)^*\, f_{s'}(u) = \int_{0}^{1-e^{-1/\varepsilon}} \frac{2 \, du}{(1-u)^2} f_s(u)^*\, f_{s'}(u) + \int_{1-e^{-1/\varepsilon}}^{1} \frac{2 \, du}{(1-u)^2} f_s(u)^*\, f_{s'}(u) \end{equation} and solving for the latter, which contains the divergent part. I used the $u \to 1$ approximation presented above and simplified the expression to a sum of $4$ terms. Each term is an integral of the form (singular endpoint - problem arises here) \begin{equation} \mathcal{I} = \int_{1-e^{-1/\varepsilon}}^{1} (1-u)^{i(s+s')-1} \, du = -\frac{i e^{-\frac{i(s+s')}{\varepsilon}}}{s+s'}. \end{equation} After integrating all terms and using using dirac delta identities involving trigonometric functions, I got \begin{equation} \mathcal{I} = -4 \pi |a_s|^2 \delta(s-s'). \end{equation} I believe the problem comes from the way I solved the $\mathcal{I}$ integral, which leads to a sign discrepancy between what I obtained and the paper.
1 Answer
In these sort of problems it is easiest to get the normalization coefficients from the completeness relation.
For example consider the Bessel function solutions to the eigenproblem on $[0,\infty]$ $$ y^2\left(- \frac{d^2\chi}{dy^2}+ k^2\right)K_{i\nu}(|k|y) = {(\nu^2+1/4})K_{i\nu}(|k|y)=0, $$ The eigenvalue is $\lambda= ({\textstyle\frac 14 +\nu^2})$. We know that the Bessels are are a complete set of eigenfuctions, and want to find their normalization and hence their orthogonality relation.
We know that for small $x$, we have $$ {\rm K}_{i\nu}(x)\sim \frac{i\pi}{2\sinh \pi \nu}\left\{ \frac{(x/2)^{i\nu}}{\Gamma(1+i\nu)}- \frac{(x/2)^{-i\nu}}{\Gamma(1-i\nu)}\right\}\nonumber\\ =i \sqrt\frac{\pi}{\nu \sinh \pi \nu} \left\{ e^{i\alpha(\nu)} (x/2)^{i\nu} - e^{-i\alpha(\nu)} (x/2)^{-i\nu}\right\}\nonumber $$ for some real $\alpha(\nu)$. We have used, at the last step, $$ \Gamma(1+i\nu)\Gamma(1-i\nu)= \frac{\pi \nu}{\sinh \pi \nu}. $$
Knowing the $K_\nu$ are solutions to a Hermitian eigenvalue problem, we can use the
the small-$x$ formulae, together with
$$
\int_{-\infty}^\infty |x|^{i\nu} |x'|^{-i\nu}d\nu = 2\pi \delta(\ln |x|-\ln |x'|)=2\pi x\delta(x-x'), \quad x,x'>0,
$$
to show that the completeness relation when evaluated for small $x$ is
$$
\frac{1}{\pi^2} \int_0^\infty 2\nu\sinh \nu\pi \,{\rm K}_{i\nu}(x){\rm K}_{i\nu}(x')\,d\nu= x \delta(x-x'),
$$
But the normalization coefficients are independent of $x$ so this result is valid for all $x$.
Hence the orthogonality property can be read off as $$ \frac{1}{\pi^2}\int_0^\infty \frac{dx}{x} {\rm K}_{i\mu}(|k|x){\rm K}_{i\nu}(|k|x)= \frac{\delta(\mu-\nu)}{2\nu \sinh \nu \pi}. $$
Some of the functions in the Kitaev article are discussed in J. D. Fay Fourier coefficients of the resolvent for a Fuchsian group, Journal f"ur die reine und angewandte Mathematik (1977) vol.\ 0293_0294, pp143-203 (see eq 17) and in Kazuto OSHIMA A Complete Set for the Maass Laplacians on the Pseudosphere Progress of Theoretical Physics, Vol. 81, No.2, February 1989
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$\begingroup$ Wow that's a really clever way of going about it. Thank you very much for the reply. I just want to be sure of something, in the orthonality condition you presented, the dirac delta $\delta(\mu-\nu)$ is read off the assumption that these form a complete set of eigenfunctions and so you already expect that the integral will have the form of some norm. constant times the dirac delta, with the constant being deduced from the completeness, right? $\endgroup$MultipleSearchingUnity– MultipleSearchingUnity2024-10-28 05:42:23 +00:00Commented Oct 28, 2024 at 5:42
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1$\begingroup$ That's right. I think I learned this trick from Audrey Terras' delightful book Harmonic Analysis on Symmetric Spaces―Euclidean Space, the Sphere, and the Poincaré Upper Half-Plane. I have corrected a typo in my ODE and added some references. I also am reading the Kitaev paper and am stuck on his complex embedding section! $\endgroup$mike stone– mike stone2024-10-28 12:20:50 +00:00Commented Oct 28, 2024 at 12:20
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$\begingroup$ I just have one more question, in the small x limit of the Bessel function, you get a sum of four terms two of which will yield the dirac delta $\delta(x-x')$ and the other two which will vanish, correct? Consider for example the term (ignoring constants) $\int_0^\infty e^{2i\alpha(\nu)} \left( \frac{x}{2}\right)^{i\nu} \left( \frac{x'}{2}\right)^{i\nu} d\nu$, this won't immediately give the dirac delta due to the exponential factor that contains a function of $\nu$. How did you deal with this one? $\endgroup$MultipleSearchingUnity– MultipleSearchingUnity2024-10-28 16:21:37 +00:00Commented Oct 28, 2024 at 16:21
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1$\begingroup$ I thiught that those terms in my expression give $\delta(x+x')$'s, but I'll have to think about the phase factor. $\endgroup$mike stone– mike stone2024-10-28 17:26:05 +00:00Commented Oct 28, 2024 at 17:26