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11.6 Spherical harmonics

The key issue about three-dimensional motion in a spherical potential is angular momentum. This is true classically as well as in quantum theories. The angular momentum in classical mechanics is defined as the vector (outer) product of r and p,

L = r x p. (11.26)

This has an easy quantum analog that can be written as

$\displaystyle \hat{\vec{L}}$ = $\displaystyle \hat{\vec{r}}$ x $\displaystyle \hat{\vec{p}}$. (11.27)

After exapnsion we find

$\displaystyle \hat{\vec{L}}$ = - i$\displaystyle \hbar$$\displaystyle \left(\vphantom{ y \frac{\partial}{\partial z}-z \frac{\partial}{...
...\partial z},x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}}\right.$y$\displaystyle {\frac{\partial}{\partial z}}$ - z$\displaystyle {\frac{\partial}{\partial y}}$, z$\displaystyle {\frac{\partial}{\partial x}}$ - x$\displaystyle {\frac{\partial}{\partial z}}$, x$\displaystyle {\frac{\partial}{\partial y}}$ - y$\displaystyle {\frac{\partial}{\partial x}}$ $\displaystyle \left.\vphantom{ y \frac{\partial}{\partial z}-z \frac{\partial}{...
...\partial z},x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}}\right)$ (11.28)

This operator has some very interesting properties:

[$\displaystyle \hat{\vec{L}}$,$\displaystyle \hat{\vec{r}}$] = 0. (11.29)

Thus

[$\displaystyle \hat{\vec{L}}$,$\displaystyle \hat{H}$] = 0! (11.30)

And even more surprising,

[$\displaystyle \hat{L}_{x}^{}$,$\displaystyle \hat{L}_{y}^{}$] = i$\displaystyle \hbar$$\displaystyle \hat{L}_{z}^{}$. (11.31)

Thus the different components of L are not compatible (i.e., can't be determined at the same time). Since L commutes with H we can diagonalise one of the components of L at the same time as H. Actually, we diagonalsie $ \hat{\vec{L}}^{2}_{}$, $ \hat{L}_{z}^{}$ and H at the same time!

The solutions to the equation

$\displaystyle \hat{\vec{L}}^{2}_{}$YLM($\displaystyle \theta$,$\displaystyle \phi$) = $\displaystyle \hbar^{2}_{}$L(L + 1)YLM($\displaystyle \theta$,$\displaystyle \phi$) (11.32)

are called the spherical harmonics.

Question: check that $ \hat{\vec{L}}^{2}_{}$ is independent of r!

The label M corresponds to the operator $ \hat{L}_{z}^{}$,

$\displaystyle \hat{L}_{z}^{}$YLM($\displaystyle \theta$,$\displaystyle \phi$) = $\displaystyle \hbar$MYLM($\displaystyle \theta$,$\displaystyle \phi$). (11.33)


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Next: 11.7 General solutions Up: 11. 3D Schrödinger equation Previous: 11.5 Now where does

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk