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# Eigenvalues of Orbital Angular Momentum

Suppose that the simultaneous eigenkets of and are completely specified by two (dimensionless) quantum numbers, and . These kets are denoted . The quantum number is defined by

 (4.25)

Thus, is the eigenvalue of divided by . It is possible to write such an equation because has the dimensions of angular momentum. Note that is a real number, because is an Hermitian operator.

We can write

 (4.26)

without loss of generality, where is some real dimensionless function of and . Later on, we will show that . Now,

 (4.27)

assuming that the have unit norms. However,

 (4.28)

 (4.29)

where is a general ket, and an Hermitian operator. The proof follows from the observation that

 (4.30)

where , plus the fact that for a general ket . [See Equation (1.22).] It follows from Equations (4.27)-(4.29) that

 (4.31)

Consider the effect of the ladder operator on the eigenket . It is easily demonstrated that

 (4.32)

where use has been made of Equation (4.26), plus the fact that and commute. It follows that the ket is an eigenstate of corresponding to the same eigenvalue as the ket . Thus, the ladder operator does not affect the magnitude of the angular momentum of any state that it acts upon. However,

 (4.33)

where use has been made of Equation (4.22). The previous equation implies that is proportional to . We can write

 (4.34)

where is a (dimensionless) number. It is clear that if the operator acts on a simultaneous eigenstate of and then the eigenvalue of remains unchanged, but the eigenvalue of is increased by . For this reason, is called a raising operator.

Using similar arguments to those just given, it is possible to demonstrate that

 (4.35)

where is a (dimensionless) number. Hence, is called a lowering operator.

The ladder operators, and , respectively step the value of up and down by unity each time they operate on one of the simultaneous eigenkets of and . It would appear, at first sight, that any value of can be obtained by applying these operators a sufficient number of times. However, according to Equation (4.31), there is a definite upper bound to the values that can take. This bound is determined by the eigenvalue of . [See Equation (4.26).] It follows that there is a maximum and a minimum possible value that can take. Suppose that we attempt to raise the value of above its maximum value, . Because there is no state with , we must have

 (4.36)

This implies that

 (4.37)

However,

 (4.38)

so Equation (4.37) yields

 (4.39)

The previous equation can be rearranged to give

 (4.40)

Comparison of this equation with Equation (4.26) yields the result

 (4.41)

But, when operates successively on it generates , , et cetera. Because the lowering operator does not change the eigenvalue of , all of these states must correspond to the same value of ; namely, . Thus,

 (4.42)

At this stage, we can give the unknown quantum number the value , without loss of generality. We can also write the previous equation in the form

 (4.43)

It is easily seen that

 (4.44)

Thus,

 (4.45)

However, we also know that

 (4.46)

where use has been made of Equations (4.34) and (4.35). It follows that

 (4.47)

Consider the following:

 (4.48)

where use has been made of the fact that and are Hermitian operators. The previous equation reduces to

 (4.49)

with the aid of Equations (4.34) and (4.35).

Equations (4.47) and (4.49) can be combined to give

 (4.50)

The solution of the previous equation is

 (4.51)

Note that is undetermined to an arbitrary phase-factor [i.e., we can replace , given previously, by , where is real, and we still satisfy Equation (4.50)]. We have made the arbitrary, but convenient, choice that is real and positive. This is equivalent to choosing the relative phases of the eigenkets . According to Equation (4.49),

 (4.52)

We have already seen that the inequality (4.31) implies that there is a maximum and a minimum possible value of . The maximum value of is denoted . What is the minimum value? Suppose that we try to lower the value of below its minimum value . Because there is no state with , we must have

 (4.53)

According to Equation (4.35), this implies that

 (4.54)

It can be seen from Equation (4.52) that . We conclude that the quantum number can take a ladder'' of discrete values, each rung differing from its immediate neighbors by unity. The top rung is , and the bottom rung is . There are only two possible choices for . Either it is an integer (e.g., , which allows to take the integer values ), or it is a half-integer (e.g., , which allows to take the half-integer values ). In fact, we shall prove, in the next section, that an orbital angular momentum can only take integer values of .

In summary, just using the fundamental commutation relations (4.8)-(4.10), plus the fact that , , and are Hermitian operators, we have shown that the eigenvalues of can be written , where is an integer, or a half-integer. Without loss of generality, we can assume that is non-negative. We have also demonstrated that the eigenvalues of can only take the values , where lies in the range . Finally, if denotes a properly normalized simultaneous eigenket of and , belonging to the eigenvalues and , respectively, then we have shown that

 (4.55) (4.56)

where are the so-called ladder operators.

Next: Rotation Operators Up: Orbital Angular Momentum Previous: Orbital Angular Momentum
Richard Fitzpatrick 2016-01-22