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):

H^=p^22m+1r^\widehat H = \frac{\widehat p^2}{2m} + \frac{1}{\widehat r}

In coordinate representation of this operators looks like:

p^=i1r^=1(x12+x22+x32)\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:

H^Ψ>=EΨ>(Δ2m+1(x12+x22+x32))Ψ(x)=EΨ(x)\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:

Ψ(x)=R(r)Y(θ,ϕ)\Psi(\vec x) = R(r)Y(\theta,\phi)

The Laplasian could be represented in such way:

Δ=1rr(1rr)+1r2Δθ,ϕ\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:

1R1rr(1rRr)1Y1r2Δθ,ϕ(Y)+1r=E(rRr(1rRr)+rEr2)1YΔθ,ϕ(Y)=0-\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:

Δθ,ϕ(Yl,m)=l(l+1)Yl,m\Delta_\theta,_\phi(Y_l,_m) = l(l+1)Y_l,_m

So, there is Legendre polynoms: Yl,m=Pl,m(cos(θ))Y_l,_m=P_l,_m(cos(\theta)) . This solution depends from ϕ\phi also.

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

(rRr(1rRr)+rEr2l(l+1))=0\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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