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| {{Quantum Mechanics A}} | | {{Quantum Mechanics A}} |
| [[Phy5645/AngularMomentumProblem|Worked Problem]] about angular momentum.
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| We will now find the orbital angular momentum eigenfunctions <math>|l,m\rangle</math> in terms of position. Recall from the previous section that | | We will now find the orbital angular momentum eigenfunctions <math>|l,m\rangle</math> in terms of position. Recall from the previous section that |
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| We may now separate out the <math>\phi</math> dependence from the <math>r\!</math> and <math>\theta\!</math> dependences; i.e., | | We may now separate out the <math>\phi</math> dependence from the <math>r\!</math> and <math>\theta\!</math> dependences; i.e., |
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| <math>\psi_{l,m}(r,\theta,\phi)=g(r,\theta)\Phi(\phi).\!</math> | | <math>\psi_{l,m}(r,\theta,\phi)=g_l(r,\theta)\Phi(\phi).\!</math> |
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| Solving for the <math>\phi\!</math> dependence, we obtain | | Solving for the <math>\phi\!</math> dependence, we obtain |
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| <math>\psi_{l,m}(r,\theta,\phi)=g(r,\theta)e^{im\phi}.\!</math> | | <math>\psi_{l,m}(r,\theta,\phi)=g_l(r,\theta)e^{im\phi}.\!</math> |
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| We may now determine the <math>\theta\!</math> dependence by using the fact that | | We may now determine the <math>\theta\!</math> dependence by using the fact that |
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| <math>\hat{\mathbf{L}}^2|l,m\rangle=l(l+1)\hbar^2|l,m\rangle.</math> | | <math>\hat{L}_+|l,l\rangle=0.</math> |
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| Writing this in the position basis as before, we obtain
| | In the position basis, the raising and lowering operators are given by |
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| :<math>L^2=-\hbar^2\left(\frac{1}{\sin \theta^2}\frac{d^2}{d\phi^2}+\frac{1}{\sin \theta}\frac{d}{d \theta}\left(\sin\theta\frac{d}{d\theta}\right)\right)</math>
| | <math>\hat{L}_\pm=\frac{\hbar}{i}e^{\pm i\phi}\left(\pm i\frac{\partial}{\partial\theta}-\cot \theta \frac{\partial}{\partial\phi}\right).</math> |
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| Where the eigenvalues of <math>L^2\!</math> are:
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| :<math>L^2|l,m\rangle= \hbar^2 l(l+1)|l,m\rangle</math>
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| We proceed by using the property of <math>L_+\!</math> and <math>L_-\!</math>, defined by
| | <math>\left(\frac{\partial}{\partial \theta}-l\cot\theta\right)g_l(r,\theta)=0.</math> |
| :<math>L_\pm=\frac{\hbar}{i}e^{\pm i\phi}\left(\pm i\frac{d}{d\theta}-\cot \theta \frac{d}{d\phi}\right)</math>
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| to find the following equation
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| :<math>\langle r,\theta,\phi|L_+|l,l\rangle=-i\hbar e^{i\phi}\left(i\frac{\partial}{\partial\theta}-\cot \theta \frac{\partial}{\partial\phi}\right)\langle r,\theta,\phi|l,l\rangle=0</math>
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| Using the above equations, we get
| | Solving the above equation, we find that the full wave function is |
| :<math>\left(\frac{\partial}{\partial \theta}-l\cot\theta\right)\langle r,\theta,\phi|l,l\rangle=0</math>
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| And the solution is
| | <math>\psi_{l,l}(r,\theta,\phi)=f_l(r)e^{il\phi}(\sin\theta)^l,\!</math> |
| :<math>\langle r,\theta,\phi|l,l\rangle=f(r)e^{il\phi}(\sin\theta)^l</math>
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| where <math>f(r)\!</math> is an arbitrary function of <math>r\!</math>. We can find the angular part of the solution by using <math>L_-\!</math>. It turns out to be | | where <math>f_l(r)\!</math> is an arbitrary function of <math>r.\!</math> Note that this function may (and, as we will see in the next chapter, does) depend on <math>l.\!</math> We may now find the wave functions <math>\psi_{l,m}(r,\theta,\phi)\!</math> by repeated application of <math>\hat{L}_-.</math> It turns out to be |
| :<math>P_l^m(x)=\frac{(-1)^m}{2^l l!}(1-x^2)^{m/2}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l</math>
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| And we know that <math>Y_l^m(\theta, \phi)\!</math> are the spherical harmonics defined by
| | <math>\psi_{l,m}(r,\theta,\phi)=f_l(r)e^{im\phi}P_l^m(\cos\theta),</math> |
| :<math>Y_l^m(\theta, \phi)=(-1)^l \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(\cos \theta)e^{im\phi}</math>
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| where the function with cosine argument is the associated Legendre polynomials defined by:
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| :<math>P_l^m(x)=(-1)^m (1-x^2)^{m/2}\frac{d^m}{dx^m}P_l(x)</math>
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| :<math>P_l(x)=\frac{1}{2^l l!}\frac{d^l}{dx^l}(x^2-1)^l</math>
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| And so we then can write:
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| :<math>P_l^m(x)=\frac{(-1)^m}{2^l l!}(1-x^2)^{m/2}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l</math>
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| Central forces are derived from a potential that depends only on the distance r of the moving particle from a fixed point, usually the coordinate origin. Since such forces produce no torque, the orbital angular momentum is conserved.
| | <math>P_l^m(x)=\frac{(-1)^m}{2^l l!}(1-x^2)^{m/2}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l</math> |
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| We can rewrite the angular momentum as
| | is an associated Legendre function. One may also write this as |
| :<math>\mathbf L=\mathbf r\times \frac{\hbar}{i}\nabla</math>
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| As has been shown, angular momentum acts as the generator of rotation.
| | <math>\psi_{l,m}(r,\theta,\phi)=f_l(r)Y_l^m(\theta,\phi),\!</math> |
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| | where |
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| | <math>Y_l^m(\theta, \phi)=(-1)^l \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(\cos \theta)e^{im\phi}</math> |
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| | are the spherical harmonics. |
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| | In the next chapter, we will be considering particles in central potentials, which are potentials that depend only on the distance <math>r\!</math> of the moving particle from a fixed point, usually the coordinate origin. Since the resulting forces produce no torque, the orbital angular momentum is conserved. In quantum mechanical terms, this means that the angular momentum operator commutes with the Hamiltonian. Therefore, the results developed throughout this chapter will be very useful in discussing such potentials. |
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| | ==Problems== |
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| | [[Phy5645/AngularMomentumProblem|Worked Problem]] about angular momentum. |
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| [[Phy5645/AngularMomentumExercise|An exercise]] with angular momentum. | | [[Phy5645/AngularMomentumExercise|An exercise]] with angular momentum. |
We will now find the orbital angular momentum eigenfunctions
in terms of position. Recall from the previous section that
If we act on the left with a position eigenvector
then this becomes
or, introducing
We may now separate out the
dependence from the
and
dependences; i.e.,
Solving for the
dependence, we obtain
We may now determine the
dependence by using the fact that
In the position basis, the raising and lowering operators are given by
We thus obtain
Solving the above equation, we find that the full wave function is
where
is an arbitrary function of
Note that this function may (and, as we will see in the next chapter, does) depend on
We may now find the wave functions
by repeated application of
It turns out to be
where
is an associated Legendre function. One may also write this as
where
are the spherical harmonics.
In the next chapter, we will be considering particles in central potentials, which are potentials that depend only on the distance
of the moving particle from a fixed point, usually the coordinate origin. Since the resulting forces produce no torque, the orbital angular momentum is conserved. In quantum mechanical terms, this means that the angular momentum operator commutes with the Hamiltonian. Therefore, the results developed throughout this chapter will be very useful in discussing such potentials.
Problems
Worked Problem about angular momentum.
An exercise with angular momentum.