Answer to Question #125263 in Quantum Mechanics for momo

Question #125263
Discuss some applications of Legendre polynomial in physics. Derive in detail
Spherical harmonics Laguerre polynomials.
1
Expert's answer
2020-07-07T10:03:05-0400

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.



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