Isotropic Harmonic Oscillator

The radial part of the Schrödinger equation for a particle of mass M in an isotropic harmonic oscillator potential $V(r)={\frac {1}{2}}Mw^{2}r^{2}$ is given by:

$-{\frac {\hbar ^{2}}{2M}}{\frac {\partial ^{2}u_{nl}(r)}{\partial r^{2}}}+\left({\frac {\hbar ^{2}}{2M}}{\frac {l(l+1)}{r^{2}}}+{\frac {1}{2}}Mw^{2}r^{2}\right)u_{nl}(r)=Eu_{nl}(r)$

We look at the solutions $u_{nl}\!$ in the asymptotic limits of $r\!$.

As $r\rightarrow 0\!$, the equation reduces to

$-{\frac {\hbar ^{2}}{2M}}{\frac {\partial ^{2}u_{l}(r)}{\partial r^{2}}}+{\frac {\hbar ^{2}}{2M}}{\frac {l(l+1)}{r^{2}}}u_{l}(r)=Eu_{l}(r)$

whose nondivergent solution is given by $u_{l}(r)\simeq r^{^{l+1}}$.

On the otherhand, as $r\rightarrow \infty$, the equation becomes

$-{\frac {\hbar ^{2}}{2M}}{\frac {\partial ^{2}u(r)}{\partial r^{2}}}+{\frac {1}{2}}Mw^{2}r^{2}u(r)=Eu(r)$

whose solution is given by $u(r)\simeq e^{-}{\frac {Mwr^{2}}{2\hbar }}$.

Combining the asymptotic limit solutions we choose the general solution to the equation as

$u_{l}(r)=f_{l}(r)r^{l+1}e^{-}{\frac {Mwr^{2}}{2\hbar }}$

Substituting this expression into the original equation,

${\frac {\partial ^{2}f_{l}(r)}{\partial r^{2}}}+2\left({\frac {l+1}{r}}-{\frac {Mw}{\hbar }}r\right){\frac {\partial f_{l}(r)}{\partial r}}+\left[{\frac {2ME}{\hbar ^{2}}}-(2l+3){\frac {Mw}{\hbar }}\right]f_{l}(r)=0$

Now we try the power series solution

$f_{l}(r)=\sum _{n=0}^{\infty }a_{n}r^{n}=a_{0}+a_{1}r+a_{2}r^{2}+a_{3}r^{3}+...+a_{n}r^{n}+...$

Substituting this solution into the reduced form of the equation,

$\sum _{n=0}^{\infty }\left[n(n-1)a_{n}r^{n-2}+2\left({\frac {l+1}{r}}-{\frac {Mw}{\hbar }}\right)na_{n}r^{n-1}+\left[{\frac {2ME}{\hbar ^{2}}}-(2l+3){\frac {Mw}{\hbar }}\right]a_{n}r^{n}\right]=0$

which reduces to the equation

$\sum _{n=0}^{\infty }\left[n(n+2l+1)a_{n}r^{n-2}+\left(-{\frac {2Mw}{\hbar }}n+{\frac {2ME}{\hbar ^{2}}}-(2l+3){\frac {Mw}{\hbar }}\right)a_{n}r^{n}\right]=0$

For this equation to hold, the coefficients of each of the powers of r must vanish seperately.

So,when $n=0\!$ the coefficient of $r^{-2}\!$ is zero, $0.(2l+1)a_{0}=0$ implying that $a_{0}\!$ need not be zero.

Equating the coefficient of $r^{-1}\!$to be zero, $1.(2l+2)a_{1}=0\!$ implying that $a_{1}\!$ must be zero.

Equating the coefficient of $r^{-n}\!$ to be zero, we get the recursion relation which is:

$\sum _{n=0}^{\infty }(n+2)(n+2l+3)a_{n+2}=\left[-{\frac {2ME}{\hbar ^{2}}}+(2n+2l+3){\frac {Mw}{\hbar }}\right]a_{n}$

The function $f_{l}(r)\!$ contains only even powers in n and is given by:

$f_{l}(r)=\sum _{n=0}^{\infty }a_{2n}r^{2n}=\sum _{n^{'}=0,2,4}^{\infty }a_{n^{'}}r^{n^{'}}$

Now as $n\rightarrow \infty \!$ , $f_{l}(r)\!$ diverges so that for finite solution, the series should stop after $r^{n^{'}+2}\!$ leading to the quantization condition:

${\frac {2M}{\hbar ^{2}}}E_{n^{'}l}-{\frac {Mw}{\hbar }}(2n^{'}+2l+3)=0$

$E_{n^{'}l}=\left(n^{'}+l+{\frac {3}{2}}\right)\hbar w,n^{'}=0,1,2,3,...$

As a result, the energy of the isotropic harmonic oscillator is given by:

$E_{n}=\left(n+{\frac {3}{2}}\right)\hbar w,n=0,1,2,3,...$ with $n=n^{'}+l\!$

The degeneracy corresponding to the nth level is:

$g={\frac {1}{2}}(n+1)(n+2)$

The total wavefunction of the isotropic Harmonic Oscillator is given by:

$\psi _{nlm}(r,\theta ,\phi )=r^{l+1}f_{l}(r)Y_{lm}(\theta ,\phi )e^{-}{\frac {Mw}{2\hbar }}r^{2}=R_{nl}(r)Y_{lm}(\theta ,\phi )$