# $\pi$ and quantum mechanics [closed]

I read paper of Friedmann and Hagen Quantum Mechanical Derivation of the Wallis Formula for $\pi$

I am not a physicist but I know how to solve Schrödinger's differential equation for the hydrogen atom but in this article I don't understand why $$\psi_{\alpha l m}=r^le^{-\alpha r^2}Y^m_l(\theta,\phi)$$ and how to prove that $$\langle H \rangle_{\alpha l} \equiv \frac{\langle \psi_{\alpha l m}|H(r)| \psi_{\alpha l m}\rangle} {\langle \psi_{\alpha l m} |\psi_{\alpha l m}\rangle } =\frac{\hbar^2}{2m}\bigg(l+\frac{3}{2}\bigg)2\alpha-e^2 \frac{\Gamma(l+1)}{\Gamma(l+\frac{3}{2})}\sqrt{2 \alpha}$$ sorry for my English

• – Qmechanic Dec 6 '15 at 20:26
• This is a variational approach. The wave function that is given is a test wave function used to estimate the ground-state energy with $\alpha$ some parameter that is tuned to give the smallest possible energy for that family of wave functions. You can check for yourself that $\langle H \rangle_{\alpha l} > -e^2/2a_0$ – Praan Dec 7 '15 at 22:12
• Variational method in quantum mechanics. – Praan Dec 7 '15 at 22:22
• I try to calculate $\langle H \rangle_{\alpha l}$, but Mathematica says that integral does not converge on ${[- \infty ,\infty]}$.. $\langle \psi_{\alpha l m}|H(r)| \psi_{\alpha l m}\rangle=\int_{-\infty}^{\infty}\big[\frac{-\hbar^2}{2m} \psi(r)^* \frac{d^2}{dx^2}\psi(r) +V(r)|\psi(r)|^2\big]dr,$ and $\langle \psi_{\alpha l m}| \psi_{\alpha l m}\rangle=\int_{-\infty}^{\infty}|\psi(r)|^2dr$. Am I right? – vito Dec 8 '15 at 15:12

Usually you use an seperation Ansatz for the wave function which is $$\Psi = R(r)\Phi(\phi)\Theta(\theta)$$ and then you use this ansatz to solve the differential equation and it just pops up. Do you know how to do seperation ansatz?
• I use "separation of variable" $\psi(r,\theta,\varphi)=R(r)F(\theta,\varphi)$, then solve radial equation $R(r)=r^l L^{2l+1}_{n-l-1}(2br)e^{-br}$ and the angular equation $F(\theta,\varphi)=Y^m_l(\theta,\varphi)$. set of solutions are given by $\psi_{n,\theta,\varphi}(r,\theta,\varphi)=Nr^l L^{2l+1}_{n-l-1}(2br)e^{-br}Y^m_l(\theta,\varphi)$. but what is $\psi_{\alpha l m}=r^le^{-\alpha r^2}Y^m_l(\theta,\phi)$? – vito Dec 7 '15 at 9:36