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 ×p.
| (11.26) |
This has an easy quantum analog that can be written as
|
\hat{L} =\hat{ r}×\hat{p}.
| (11.27) |
After exapnsion we find
|
\hat{L} = −iℏ\left (y {∂\over
∂z} − z {∂\over
∂y},z {∂\over
∂x} − x {∂\over
∂z},x {∂\over
∂y} − y {∂\over
∂x}\right )
| (11.28) |
This operator has some very interesting properties:
|
[\hat{L},\hat{r}] = 0.
| (11.29) |
Thus
|
[\hat{L},\hat{H}] = 0!
| (11.30) |
And even more surprising,
|
[\hat{{L}}_{x},\hat{{L}}_{y}] = iℏ\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{{L}}^{2}, \hat{{L}}_{z} and H at the same time!
The solutions to the equation
|
\hat{{L}}^{2}{Y }_{
LM}(θ,ϕ) = {ℏ}^{2}L(L + 1){Y }_{
LM}(θ,ϕ)
| (11.32) |
are called the spherical harmonics.
Question: check that \hat{{L}}^{2} is independent of r!
The label M corresponds to the operator \hat{{L}}_{z},
|
\hat{{L}}_{z}{Y }_{LM}(θ,ϕ) = ℏM{Y }_{LM}(θ,ϕ).
| (11.33) |