Answer to Question #125263 in Quantum Mechanics for momo

Question

Discuss some applications of Legendre polynomial in physics. Derive in detail
Spherical harmonics Laguerre polynomials.

Answer 1

Let's consider the Hamiltonian of the hydrogen atom (all constants are 1):

$\widehat H = \frac{\widehat p^2}{2m} + \frac{1}{\widehat r}$

In coordinate representation of this operators looks like:

$\widehat p = -i\nabla\newline \frac{1}{\widehat r} = \frac{1}{\sqrt(x^2_1 + x^2_2+x^2_3)}$

Schrodinger equation in this case will be:

$\widehat H |\Psi> = E|\Psi>\newline \left(-\frac{\Delta}{2m} + \frac{1}{\sqrt(x^2_1 + x^2_2+x^2_3)}\right)\Psi(\vec x) = E\Psi(\vec x)$

We will search the solution of the one, by assuming such decomposition:

$\Psi(\vec x) = R(r)Y(\theta,\phi)$

The Laplasian could be represented in such way:

$\Delta = \frac{1}{r}\frac{\partial}{\partial r}\left( \frac{1}{r}\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\Delta_\theta,_\phi$

Taking into account this facts we can write the new equation:

$-\frac{1}{R}\frac{1}{r}\frac{\partial}{\partial r}\left( \frac{1}{r}\frac{\partial R}{\partial r}\right) -\frac{1}{Y} \frac{1}{r^2}\Delta_\theta,_\phi(Y) + \frac{1}{r} = E\newline \left(-\frac{r}{R}\frac{\partial}{\partial r}\left( \frac{1}{r}\frac{\partial R}{\partial r}\right)+ r - Er^2\right) -\frac{1}{Y} \Delta_\theta,_\phi(Y) = 0$

The solution of the problem:

$\Delta_\theta,_\phi(Y_l,_m) = l(l+1)Y_l,_m$

So, there is Legendre polynoms: $Y_l,_m=P_l,_m(cos(\theta))$ . This solution depends from $\phi$ also.

It's spherical harmoncs for sphere for $r=1$ . Now, the previous equation have taken the view:

$\left(-\frac{r}{R}\frac{\partial}{\partial r}\left( \frac{1}{r}\frac{\partial R}{\partial r}\right)+ r - Er^2 - l(l+1)\right) = 0$

This is equation for Lauguerre polynomials.