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| We now solve the isotropic harmonic oscillator using the formalism that we have just developed. While it is possible to solve it in Cartesian coordinates, we gain additional insight by solving it in spherical coordinates, and it is easier to determine the degeneracy of each energy level. | | We now solve the isotropic harmonic oscillator using the formalism that we have just developed. While it is possible to solve it in Cartesian coordinates, we gain additional insight by solving it in spherical coordinates, and it is easier to determine the degeneracy of each energy level. |
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| The radial part of the [[Schrödinger Equation|Schrödinger equation]] for a particle of mass <math>M\!</math> in an isotropic [[Harmonic oscillator spectrum and eigenstates|harmonic oscillator]] potential <math>V(r)=\frac{1}{2}M\omega^{2}r^2</math> is given by: | | The radial part of the [[Schrödinger Equation|Schrödinger equation]] for a particle of mass <math>M\!</math> in an isotropic [[Harmonic Oscillator Spectrum and Eigenstates|harmonic oscillator]] potential <math>V(r)=\frac{1}{2}M\omega^{2}r^2</math> is given by: |
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| :<math>-\frac{\hbar^2}{2M}\frac{\partial^2u_{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)</math> | | :<math>-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\left(\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2} + \frac{1}{2}M\omega^{2}r^2\right)u_{nl}=Eu_{nl}.</math> |
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| We look at the solutions <math>u_{nl}\!</math> in the asymptotic limits of <math> r\!</math>.
| | Let us begin by looking at the solutions <math>u_{nl}\!</math> in the limits of small and large <math>r.\!</math> |
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| As <math>r\rightarrow 0\!</math>, the equation reduces to | | As <math>r\rightarrow 0\!</math>, the equation reduces to |
| :<math>-\frac{\hbar^2}{2M}\frac{\partial^2u_{l}(r)}{\partial r^2}+\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2}u_{l}(r)=Eu_{l}(r)</math>
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| whose nondivergent solution is given by <math>u_l(r)\simeq r^{^{l+1}}</math>.
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| On the otherhand, as <math> r\rightarrow \infty</math>, the equation becomes
| | <math>-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2}u_{nl}=Eu_{nl}.</math> |
| :<math>-\frac{\hbar^2}{2M}\frac{\partial^2u(r)}{\partial r^2}+\frac{1}{2}Mw^{2}r^2u(r)=Eu(r)</math>
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| whose solution is given by <math>u(r)\simeq e^-{\frac{Mwr^2}{2\hbar}}</math>.
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| Combining the asymptotic limit solutions we choose the general solution to the equation as
| | The only solution of this equation that does not diverge as <math>r\rightarrow 0</math> is <math>u_{nl}(r)\simeq r^{l+1}.</math> |
| :<math>u_l(r)=f_l(r)r^{l+1}e^-\frac{Mwr^2}{2\hbar}</math>
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| Substituting this expression into the original equation, | | In the limit as <math>r\rightarrow \infty,</math> on the other hand, the equation becomes |
| :<math>\frac{\partial^2f_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</math>
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| | <math>-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\frac{1}{2}M\omega^{2}r^2u_{nl}=Eu_{nl}</math> |
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| | whose solution is given by <math>u_{nl}(r)\simeq e^{-M\omega r^2/2\hbar}.</math> |
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| | We may now assume that the general solution to the equation is given by |
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| | <math>u_{nl}(r)=r^{l+1}e^{-M\omega r^2/2\hbar}f_{nl}(r).</math> |
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| | Substituting this expression into the original equation, we obtain |
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| | <math>\frac{\partial^2f_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</math> |
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| Now we try the power series solution | | Now we try the power series solution |
Revision as of 00:14, 1 September 2013
We now solve the isotropic harmonic oscillator using the formalism that we have just developed. While it is possible to solve it in Cartesian coordinates, we gain additional insight by solving it in spherical coordinates, and it is easier to determine the degeneracy of each energy level.
The radial part of the Schrödinger equation for a particle of mass
in an isotropic harmonic oscillator potential
is given by:

Let us begin by looking at the solutions
in the limits of small and large
As
, the equation reduces to
The only solution of this equation that does not diverge as
is
In the limit as
on the other hand, the equation becomes
whose solution is given by
We may now assume that the general solution to the equation is given by
Substituting this expression into the original equation, we obtain
Now we try the power series solution

Substituting this solution into the reduced form of the equation,
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24873401d1209b1aeaf8f4d2c1fec73b8621a98d)
which reduces to the equation
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8d9a61f2e496e08807dace77831fd98ab01783a)
For this equation to hold, the coefficients of each of the powers of r must vanish seperately.
So,when
the coefficient of
is zero,
implying that
need not be zero.
Equating the coefficient of
to be zero,
implying that
must be zero.
Equating the coefficient of
to be zero, we get the recursion relation which is:
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/120c9f46e40979f46f2ab40ce065219547dc9a66)
The function
contains only even powers in n and is given by:

Now as
,
diverges so that for finite solution, the series should stop after
leading to the quantization condition:


As a result, the energy of the isotropic harmonic oscillator is given by:
with 
The degeneracy corresponding to the nth level is:

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