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| {{Quantum Mechanics A}} | | {{Quantum Mechanics A}} |
| A free particle is a specific case when <math>V_0=0\!</math> of the motion in a uniform potential <math>V(r)=V_0\!</math>. So it's more useful to consider a particle moving in a uniform potential. The [[Schrödinger equation]] for the radial part of the wave function is: | | A free particle is a specific case when <math>V_0=0\!</math> of the motion in a uniform potential <math>V(r)=V_0,</math> so it is more useful to consider a particle moving in a uniform potential. The [[Schrödinger Equation|Schrödinger equation]] for the radial part of the wave function is |
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| :<math>\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}+V_0\right)u_l(r)=Eu_l(r)</math>
| | <math>\left(-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}+V_0\right)u_l=Eu_l.</math> |
| let <math>k^2=\frac{2m}{\hbar^2}|E-V|</math>. Rearranging the equation gives
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| :<math>\left(-\frac{\partial^2}{\partial r^2}+\frac{l(l+1)}{r^2}-k^2\right)u_l(r)=0</math>
| | Let <math>k^2=\frac{2m}{\hbar^2}|E-V|.</math> Rearranging the equation gives us |
| Letting <math>\rho=kr\!</math> gives the terms that <math>\frac{1}{r^{2}}=\frac{k^{2}}{\rho ^{2}}</math> and <math>\frac{\partial ^{2}}{\partial r^{2}}=k^{2}\frac{\partial ^{2}}{\partial \rho ^{2}}</math>. Then the equation becomes:
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| :<math>\left(-\frac{\partial^2}{\partial\rho^2}+\frac{l(l+1)}{\rho^2}\right)u_l(\rho)=u_l(\rho)=d_ld_l^+u_l(\rho)</math>
| | <math>\left(-\frac{d^2}{dr^2}+\frac{l(l+1)}{r^2}-k^2\right)u_l=0.</math> |
| where <math>d_l\!</math> and <math>d_l^{\dagger}\!</math> become the raising and lowering operators: | | |
| | If we now let <math>\rho=kr,\!</math> then the equation reduces to the dimensionless form, |
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| | <math>\left(-\frac{d^2}{d\rho^2}+\frac{l(l+1)}{\rho^2}\right)u_l(\rho)=u_l(\rho)=d_ld_l^+u_l(\rho),</math> |
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| | where <math>d_l\!</math> and <math>d_l^{\dagger}\!</math> are the raising and lowering operators, |
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| | <math>d_l=\frac{\partial}{\partial\rho}+\frac{l+1}{\rho}</math> |
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| | and |
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| | <math>d_l^\dagger=-\frac{\partial}{\partial\rho}+\frac{l+1}{\rho}.</math> |
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| :<math>d_l=\frac{\partial}{\partial\rho}+\frac{l+1}{\rho}, </math>
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| :<math>d_l^\dagger=-\frac{\partial}{\partial\rho}+\frac{l+1}{\rho}</math>
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| Being <math>d_l^{\dagger}d_l=d_{l+1}d_{l+1}^{\dagger}</math>, it can be shown that | | Being <math>d_l^{\dagger}d_l=d_{l+1}d_{l+1}^{\dagger}</math>, it can be shown that |
| :<math>d_l^\dagger u_l(\rho)=c_l u_{l+1}(\rho)</math> | | :<math>d_l^\dagger u_l(\rho)=c_l u_{l+1}(\rho)</math> |
Revision as of 23:22, 31 August 2013
A free particle is a specific case when
of the motion in a uniform potential
so it is more useful to consider a particle moving in a uniform potential. The Schrödinger equation for the radial part of the wave function is
Let
Rearranging the equation gives us
If we now let
then the equation reduces to the dimensionless form,
where
and
are the raising and lowering operators,
and
Being
, it can be shown that

For
,
, gives the solution as:

The raising operator can be applied to the ground state in order to find high orders of
;

By this way, we can get the general expression:
,
where
is spherical Bessel function and
is spherical Neumann function.
Explicit Forms of the Spherical Bessel and Neumann Functions






The spherical Hankel functions of the first and second kind can be written in terms of the spherical Bessel and spherical Neumann functions, and are defined by:

and

The asymptotic form of the spherical Bessel and Neumann functions (as z
large) are given by:

and

The first few zeros of the spherical Bessel function:


The derivatives of the spherical Bessel and Neumann functions are defined by:

and
