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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   (556)   (557)

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,     (559)

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   (561)   (562)

we obtain (563)

where use has been made of Eq. (541). Likewise, (564)

where use has been made of Eq. (540). It follows that   (565)   (566)

These equations are satisfied when (567)

Hence, we can write   (568)   (569)   Next: Eigenvalues of Up: Orbital Angular Momentum Previous: Representation of Angular Momentum
Richard Fitzpatrick 2010-07-20