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| | {{Quantum Mechanics A}} |
| The radial part of the [[Schrödinger equation]] for a particle of mass M in an isotropic [[Harmonic oscillator spectrum and eigenstates|harmonic oscillator]] potential <math>V(r)=\frac{1}{2}Mw^{2}r^2</math> is given by: | | The radial part of the [[Schrödinger equation]] for a particle of mass M in an isotropic [[Harmonic oscillator spectrum and eigenstates|harmonic oscillator]] potential <math>V(r)=\frac{1}{2}Mw^{2}r^2</math> is given by: |
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Revision as of 16:42, 31 August 2011
The radial part of the Schrödinger equation for a particle of mass M in an isotropic harmonic oscillator potential
is given by:

We look at the solutions
in the asymptotic limits of
.
As
, the equation reduces to

whose nondivergent solution is given by
.
On the otherhand, as
, the equation becomes

whose solution is given by
.
Combining the asymptotic limit solutions we choose the general solution to the equation as

Substituting this expression into the original equation,
![{\displaystyle {\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9af28f69669ff04ae10eb8cafb49b31b3f5531e)
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:
