/sci/ - Science & Math

>>14823503
My IPrange is banned (not my fault, prob someone with similar IP) so i cant make new posts for a while, so I gotta hope someone can help me here. It's gonna take two posts so bear with me.

My question is about the hydrogen atom and it's angular momentum, involving its numbers [math]n,l,m[/math], in which [math]\Psi_{nlm} = R_n Y_l e^{im\phi}[/math]


1. Using the ladder operators [math]L_{\pm} = L_x + i L_y[/math] on [math]L^2[/math] and [math]L_z[/math], we find that applying [math]L^2[/math] onto the spherical Legendre harmonics, [math]Y_l[/math], gives us

> [math]L^2 Y_l = \hbar^2 l(l+1)Y_l[/math].


2. With crazy analytic techniques, solving the atom's Schrodinger Equation (Coulomb potential for a proton and electron) gives us the ~Rydberg Formula

[math]E_n = -\dfrac{\hbar^2k^2}{2m} = -\dfrac{\hbar^2}{2ma_0^2}\dfrac{1}{n^2} [/math]. (Where [math]k = \frac{1}{a_0n}[/math] is the wavenumber and [math]a_0 = \frac{\hbar^2}{mkee}[/math] is the Bohr Radius).

Using some classical ideas here, Virial Theorem for the potential [math]r^{-2}[/math] tells us that

[math]T = -E_n = \frac{1}{2}mv^2[/math],

and the centripetal force is

[math]F = -\dfrac{mv^2}{r} = -\dfrac{kee}{r^2}[/math].

Solving for [math]v^2[/math] from [math]E_n[/math], plugging it into [math]F[/math] and solving for [math]r[/math] ends up to be [math]r = a_0n^2[/math], or rather

[math]\langle r \rangle = \langle\Psi|r|\Psi\rangle = a_0n^2[/math].