Orbital Angular Momentum Eigenfunctions

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Revision as of 22:34, 19 June 2011 by StephanieReynolds (talk | contribs) (New page: [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/AngularMomentumProblem Worked Problem] about angular momentum. Now we construct our eigenfunctions of the orbital angular momentum expl...)
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Worked Problem about angular momentum.

Now we construct our eigenfunctions of the orbital angular momentum explicitly. The eigenvalue equation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_z|l,m\rangle=m\hbar|l,m\rangle}

in terms of wave functions, becomes:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle r,\theta,\phi|L_z|l,m\rangle=-i\hbar\frac{\partial}{\partial \phi}\langle r,\theta,\phi|l,m\rangle=m\hbar \langle r,\theta,\phi|l,m\rangle}

Solving for the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\!} dependence, we find

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle r,\theta,\phi|l,m\rangle=e^{im\phi}\langle r,\theta,0|l,m\rangle}

We construct the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta\!} dependence using the differential operator representation of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^2\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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)}

Where the eigenvalues of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^2\!} are:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^2|l,m\rangle= \hbar^2 l(l+1)|l,m\rangle}

We proceed by using the property of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_+\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_-\!} , defined by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_\pm=\frac{\hbar}{i}e^{\pm i\phi}\left(\pm i\frac{d}{d\theta}-\cot \theta \frac{d}{d\phi}\right)}

to find the following equation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}

Using the above equations, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{\partial}{\partial \theta}-l\cot\theta\right)\langle r,\theta,\phi|l,l\rangle=0}

And the solution is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle r,\theta,\phi|l,l\rangle=f(r)e^{il\phi}(\sin\theta)^l}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(r)\!} is an arbitrary function of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\!} . We can find the angular part of the solution by using Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_-\!} . It turns out to be

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}

And we know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y_l^m(\theta, \phi)\!} are the spherical harmonics defined by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}}

where the function with cosine argument is the associated Legendre polynomials defined by:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_l^m(x)=(-1)^m (1-x^2)^{m/2}\frac{d^m}{dx^m}P_l(x)}

with

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_l(x)=\frac{1}{2^l l!}\frac{d^l}{dx^l}(x^2-1)^l}

And so we then can write:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}

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.

We can rewrite the angular momentum as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf L=\mathbf r\times \frac{\hbar}{i}\nabla}

As has been shown, angular momentum acts as the generator of rotation.

An exercise with angular momentum.