Chapter 9 The Two-Body Problem

1.2. Determination of the Angular Momentum Eigenvalues
As we noticed before and commute, and both are self-adjoint operators and thus observables. We will use the factorization technique to find the eigenvalues of and . Since the same method can be used to obtain the eigenvalues of the spin operator, we use the symbol for a general operator satisfying the fundamental commutation relation ,

Now let us introduce two non-Hermitian operators by

These are the raising and lowering operators as we will see later.
From (9.52) and (9.53) it follows that


and

In addition from (9.53) we find

We denote the eigenvalues of and by and respectively, i.e.

and

From the condition , it follows that

Now we operate on (9.59) with ;

where we have substituted for on the left-hand side of (9.61) from the commutator (9.56). We can rewrite Eq. (9.61) as

This relation shows that is an eigenfunction of with the eigenvalue . Since commutes with , the ket is also an eigenstate of . As (9.60) shows has to be larger or equal to . Thus there must be a maximum eigenvalue of otherwise the inequality (9.60) will be violated. If is the eigenstate corresponding to the largest eigenvalue of , then we have , or ,

Noting that is an eigenstate of both and , from (9.63) it follows that

Let us now investigate that action of the lowering operator on (9.59) when is equal to

This relation is found exactly as we obtained (9.62) but with rather than . If we repeat this operation times we find

Now by choosing large enough we can violate (9.60) unless for some integer

We can operate on the last relation with and use (9.57) by writing it as

to get

Noting that is an eigenstate of and therefore we have the following relation between the eigenvalues;

where is the largest eigenvalue of . By substituting for from (9.64) in we find

where is zero or is an integer. Therefore values are restricted to

From Eqs. (9.64) and (9.71) we find

Thus we have

and

Matrix Elements of and From the fact that commutes with

we find

But as Eq. (9.62) shows while the eigenvalue of has not changed, by the action of the eigenvlaue of has increased by . Let us assume that is normalized, but is not. To normalize this ket we determine its norm;

Since is normalized therefore

is the normalized eigenvector. From this and the corresponding relation for we find the matrix elements of and

and

In this representation where both and are diagonal, we have a discrete sequences of values for quantum numbers and . It is convenient to write these matrices as an infinite set of finite matrices of the form

where each represented by a given value of and is a matrix.
Suppressing the superscript for all , we note that for , and are represented by null matrices of unit rank.
For we obtain

This case will be discussed in detail later.
For the next one, , the matrices are given by


Similar matrices can be found for .
We are thus led to the conclusion that the eigenvalues of are determined by quantum numbers , as is given in Eq. (9.74). Now to each value of the quantum number , there corresponds eigenfunctions , Eq. (9.75). The fact that the eigenvalues of are not dependent on (in other words the degeneracy of the eigenvalues) is associated with the invariance of the motion in a central potential under rotations.