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Eigenstates of Angular Momentum
Let us find the simultaneous eigenstates of the angular momentum
operators and . Since both of these operators can
be represented as purely angular differential operators, it stands to
reason that their eigenstates only depend on the angular coordinates
and . Thus,
we can write
Here, the
are the eigenstates in question, whereas the dimensionless
quantities and parameterize the eigenvalues of and ,
which are and
, respectively. Of course,
we expect the to be both mutually orthogonal and properly normalized
(see Sect. 4.9), so that
|
(558) |
where
is an element of solid angle,
and the integral is over all solid angle.
Now,
where use has been made of Eq. (543). We, thus, conclude that
when the operator operates on an eigenstate of corresponding to the eigenvalue it converts it to an
eigenstate corresponding to the eigenvalue .
Hence, is known as the raising operator (for ). It is
also easily demonstrated that
|
(560) |
In other words, when operates on an eigenstate of corresponding to the eigenvalue it converts it to an
eigenstate corresponding to the eigenvalue .
Hence, is known as the lowering operator (for ).
Writing
we obtain
|
(563) |
where use has been made of Eq. (541). Likewise,
|
(564) |
where use has been made of Eq. (540). It
follows that
These equations are satisfied when
|
(567) |
Hence, we can write
Next: Eigenvalues of
Up: Orbital Angular Momentum
Previous: Representation of Angular Momentum
Richard Fitzpatrick
2010-07-20