Quantum Mechanics derivation of Wallis' Formula?

Recently there was a proof of the Wallis Product using quantum mechanics on the arXiv. However, there are many proofs of the result, Wikipedia has 4.

Fine Print the first proof has on Wikipedia, the Euler product as an input, $$\boxed{\sin \pi x = \pi x \prod_{n \in \mathbb{N}} \left(1 - \frac{x^2}{n^2}\right)}$$ and this follows from Weierstrass Factorization, which is a significant result. One proof by Oscar Ciaurri circumvents this. Another by Lars Holst.

Quantum Mechanics This derivation starts with the Hydrogen atom Schrödinger equation:

$$H \psi = \left( - \frac{\hbar}{2m}\nabla^2 - \frac{e^2}{r} \right) \psi = E \psi$$

Since this equation is radially symmetric there is an $SO(3)$ action on the solutions to this equation preserving the energies $E$. So these eigenspaces can be indexed by the representations of $SO(3)$, $E_{\ell, m}$ and $\psi_{\ell , m}$. Can we guess the values of $E$ ?

A variational approach involves minimizing as best we can, using the point-to-line distance formula: $$E_{\ell, 0} \leq \frac{\langle \psi | H | \psi \rangle }{\langle \psi| \psi \rangle}$$ This only works for the ground state energy.

They have a guesstimate of $\langle H_\ell \rangle_{min} = \frac{1}{\ell + \frac{1}{2}}\left[ \frac{\ell!}{(\ell + \frac{1}{2})!}\right]$ while the exact answer is $E_{0, \ell} = \frac{1}{(\ell + 1)^2}$

Bohr correspondence principle How do we know for $\ell \gg 1$ these two values are the same? $$\lim_{\ell \to \infty} \frac{\langle H_\ell \rangle_{min}}{E_{0, \ell}} = 1$$ This ultimately leads to the Wallis Product Formula using identities of the Gamma function

Is quantum mechanics necessary here? I think we are using Feynman-Hellman or Rayleigh-Ritz but I think there is something even more basic facts about Hermitian matrices as in these notes.

• Euler's product formula for $\sin^2$ does not need the general apparatus of Weierstrass or Hadamard products: in brief, first use Liouville's theorem to prove the partial fraction expansion of $1/\sin^2 x$, second regroup and observe that this is the derivative of $\cot x$, third, observe that this is the logarithmic derivative of $\sin x$. E.g., as math.umn.edu/~garrett/m/complex/08a_product_sine.pdf – paul garrett Mar 31 '16 at 16:19
• @paulgarrett Every time I see Liouville Theorem I think any bounded entire function is constant "gee that's nice" -- now I see it in your notes and I am like wow, we get $\zeta(2) = \frac{\pi^2}{6}$ for free! – john mangual Mar 31 '16 at 16:26
• Yes, Liouville's theorem is useful, not static. – paul garrett Mar 31 '16 at 16:27
• Here is a nice commentary on the use of the variational principle for the spectrum of the hydrogen atom to find Wallis formula. I am a bit uncertain on what the OP is seeking; an alternative derivation of the Wallis formula? (there are many); an alternative proof using the variational principle for a Hermitian eigenvalue problem? – Carlo Beenakker Mar 31 '16 at 17:38
• I think the observation is no big deal and has been blown out of proportion, particularly in the piece mentioned by @CarloBeenakker. The authors have two different ways of calculating some quantity, and the fact that these two ways must agree boils down to Wallis formula. It's a curiosity, and many similar ones exist. – Marcel Mar 31 '16 at 20:06

Actually, to prove that $$\lim_{n\to\infty}\frac{n^2}{n+\frac{1}{2}}\left[\frac{\Gamma(n)}{\Gamma(n+\frac{1}{2})}\right]^2=1,$$ there is no need in the Bohr's correspondence principle. Stirling's Series can be used to get $$\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)}\sim z^{\alpha-\beta},$$ as $z\to\infty$ (see, for example, https://projecteuclid.org/euclid.pjm/1102613160 - The asymptotic expansion of a ratio of gamma functions, by A. Erdelyi and F. G. Tricomi). Then $$\lim_{n\to\infty}\frac{n^2}{n+\frac{1}{2}}\left[\frac{\Gamma(n)}{\Gamma(n+\frac{1}{2})}\right]^2=\lim_{n\to\infty}\frac{n^2}{(n+\frac{1}{2})n}=1.$$