We found the eigenvalues of $\hat{L}^{2}$ and $\hat{L}_{z}$ by expressing these orbital angular-momentum operators as differential operators and solving the resulting differential equations. We now show that these eigenvalues can be found using only the operator commutation relations. The work in this section applies to any operators that satisfy the angular-momentum commutation relations. In particular, it applies to spin angular momentum (Chapter 10) as well as orbital angular momentum.
We used the letter $L$ for orbital angular momentum. Here we will use the letter $M$ to indicate that we are dealing with any kind of angular momentum. We have three linear
operators $\hat{M}{x}, \hat{M}{y}$, and $\hat{M}_{z}$, and all we know about them is that they obey the commutation relations [similar to (5.46) and (5.48)]
\(
\begin{equation}
\left[\hat{M}{x}, \hat{M}{y}\right]=i \hbar \hat{M}{z}, \quad\left[\hat{M}{y}, \hat{M}{z}\right]=i \hbar \hat{M}{x}, \quad\left[\hat{M}{z}, \hat{M}{x}\right]=i \hbar \hat{M}_{y} \tag{5.107}
\end{equation}
\)
We define the operator $\hat{M}^{2}$ as
\(
\begin{equation}
\hat{M}^{2}=\hat{M}{x}^{2}+\hat{M}{y}^{2}+\hat{M}_{z}^{2} \tag{5.108}
\end{equation}
\)
Our problem is to find the eigenvalues of $\hat{M}^{2}$ and $\hat{M}_{z}$.
We begin by evaluating the commutators of $\hat{M}^{2}$ with its components, using Eqs. (5.107) and (5.108). The work is identical with that used to derive Eqs. (5.49) and (5.50), and we have
\(
\begin{equation}
\left[\hat{M}^{2}, \hat{M}{x}\right]=\left[\hat{M}^{2}, \hat{M}{y}\right]=\left[\hat{M}^{2}, \hat{M}_{z}\right]=0 \tag{5.109}
\end{equation}
\)
Hence we can have simultaneous eigenfunctions of $\hat{M}^{2}$ and $\hat{M}{z}$.
Next we define two new operators, the raising operator $\hat{M}{+}$and the lowering operator $\hat{M}_{-}$:
\(
\begin{align}
& \hat{M}{+} \equiv \hat{M}{x}+i \hat{M}{y} \tag{5.110}\
& \hat{M}{-} \equiv \hat{M}{x}-i \hat{M}{y} \tag{5.111}
\end{align}
\)
These are examples of ladder operators. The reason for the terminology will become clear shortly. We have
\(
\begin{gather}
\hat{M}{+} \hat{M}{-}=\left(\hat{M}{x}+i \hat{M}{y}\right)\left(\hat{M}{x}-i \hat{M}{y}\right)=\hat{M}{x}\left(\hat{M}{x}-i \hat{M}{y}\right)+i \hat{M}{y}\left(\hat{M}{x}-i \hat{M}{y}\right) \
=\hat{M}{x}^{2}-i \hat{M}{x} \hat{M}{y}+i \hat{M}{y} \hat{M}{x}+\hat{M}{y}^{2}=\hat{M}^{2}-\hat{M}{z}^{2}+i\left[\hat{M}{y}, \hat{M}{x}\right] \
\hat{M}{+} \hat{M}{-}=\hat{M}^{2}-\hat{M}{z}^{2}+\hbar \hat{M}_{z} \tag{5.112}
\end{gather}
\)
Similarly, we find
\(
\begin{equation}
\hat{M}{-} \hat{M}{+}=\hat{M}^{2}-\hat{M}{z}^{2}-\hbar \hat{M}{z} \tag{5.113}
\end{equation}
\)
For the commutators of these operators with $\hat{M}_{z}$, we have
\(
\begin{gather}
{\left[\hat{M}{+}, \hat{M}{z}\right]=\left[\hat{M}{x}+i \hat{M}{y}, \hat{M}{z}\right]=\left[\hat{M}{x}, \hat{M}{z}\right]+i\left[\hat{M}{y}, \hat{M}{z}\right]=-i \hbar \hat{M}{y}-\hbar \hat{M}{x}} \
{\left[\hat{M}{+}, \hat{M}{z}\right]=-\hbar \hat{M}{+}} \
\hat{M}{+} \hat{M}{z}=\hat{M}{z} \hat{M}{+}-\hbar \hat{M}_{+} \tag{5.114}
\end{gather}
\)
where (5.107) was used. Similarly, we find
\(
\begin{equation}
\hat{M}{-} \hat{M}{z}=\hat{M}{z} \hat{M}{-}+\hbar \hat{M}_{-} \tag{5.115}
\end{equation}
\)
Using $Y$ for the common eigenfunctions of $\hat{M}^{2}$ and $\hat{M}_{z}$, we have
\(
\begin{align}
& \hat{M}^{2} Y=c Y \tag{5.116}\
& \hat{M}_{z} Y=b Y \tag{5.117}
\end{align}
\)
where $c$ and $b$ are the eigenvalues. Operating on Eq. (5.117) with $\hat{M}_{+}$, we get
\(
\hat{M}{+} \hat{M}{z} Y=\hat{M}_{+} b Y
\)
Using Eq. (5.114) and the fact that $\hat{M}_{+}$is linear, we have
\(
\begin{gather}
\left(\hat{M}{z} \hat{M}{+}-\hbar \hat{M}{+}\right) Y=b \hat{M}{+} Y \
\hat{M}{z}\left(\hat{M}{+} Y\right)=(b+\hbar)\left(\hat{M}_{+} Y\right) \tag{5.118}
\end{gather}
\)
This last equation says that the function $\hat{M}{+} Y$ is an eigenfunction of $\hat{M}{z}$ with eigenvalue $b+\hbar$. In other words, operating on the eigenfunction $Y$ with the raising operator $\hat{M}{+}$ converts $Y$ into another eigenfunction of $\hat{M}{z}$ with eigenvalue $\hbar$ higher than the eigenvalue of $Y$. If we now apply the raising operator to (5.118) and use (5.114) again, we find similarly
\(
\hat{M}{z}\left(\hat{M}{+}^{2} Y\right)=(b+2 \hbar)\left(\hat{M}_{+}^{2} Y\right)
\)
Repeated application of the raising operator gives
\(
\begin{equation}
\hat{M}{z}\left(\hat{M}{+}^{k} Y\right)=(b+k \hbar)\left(\hat{M}_{+}^{k} Y\right), \quad k=0,1,2, \ldots \tag{5.119}
\end{equation}
\)
If we operate on (5.117) with the lowering operator and apply (5.115), we find in the same manner
\(
\begin{gather}
\hat{M}{z}\left(\hat{M}{-} Y\right)=(b-\hbar)\left(\hat{M}{-} Y\right) \tag{5.120}\
\hat{M}{z}\left(\hat{M}{-}^{k} Y\right)=(b-k \hbar)\left(\hat{M}{-}^{k} Y\right) \tag{5.121}
\end{gather}
\)
Thus by using the raising and lowering operators on the eigenfunction with the eigenvalue $b$, we generate a ladder of eigenvalues, the difference from step to step being $\hbar$ :
\(
\ldots \quad b-2 \hbar, \quad b-\hbar, \quad b, \quad b+\hbar, \quad b+2 \hbar, \quad \ldots
\)
The functions $\hat{M}{ \pm}^{k} Y$ are eigenfunctions of $\hat{M}{z}$ with eigenvalues $b \pm k \hbar$ [Eqs. (5.119) and (5.121)]. We now show that these functions are also eigenfunctions of $\hat{M}^{2}$, all with the same eigenvalue $c$ :
\(
\begin{gather}
\hat{M}{z} \hat{M}{ \pm}^{k} Y=(b \pm k \hbar) \hat{M}{ \pm}^{k} Y \tag{5.122}\
\hat{M}^{2} \hat{M}{ \pm}^{k} Y=c \hat{M}_{ \pm}^{k} Y, \quad k=0,1,2, \ldots \tag{5.123}
\end{gather}
\)
To prove (5.123), we first show that $\hat{M}^{2}$ commutes with $\hat{M}{+}$and $\hat{M}{-}$:
\(
\left[\hat{M}^{2}, \hat{M}{ \pm}\right]=\left[\hat{M}^{2}, \hat{M}{x} \pm i \hat{M}{y}\right]=\left[\hat{M}^{2}, \hat{M}{x}\right] \pm i\left[\hat{M}^{2}, \hat{M}_{y}\right]=0 \pm 0=0
\)
We also have
\(
\left[\hat{M}^{2}, \hat{M}{ \pm}^{2}\right]=\left[\hat{M}^{2}, \hat{M}{ \pm}\right] \hat{M}{ \pm}+\hat{M}{ \pm}\left[\hat{M}^{2}, \hat{M}_{ \pm}\right]=0+0=0
\)
and it follows by induction that
\(
\begin{equation}
\left[\hat{M}^{2}, \hat{M}{ \pm}^{k}\right]=0 \quad \text { or } \quad \hat{M}^{2} \hat{M}{ \pm}^{k}=\hat{M}_{ \pm}^{k} \hat{M}^{2}, \quad k=0,1,2, \ldots \tag{5.124}
\end{equation}
\)
If we operate on (5.116) with $\hat{M}_{ \pm}^{k}$ and use (5.124), we get
\(
\begin{align}
\hat{M}{ \pm}^{k} \hat{M}^{2} Y & =\hat{M}{ \pm}^{k} c Y \
\hat{M}^{2}\left(\hat{M}{ \pm}^{k} Y\right) & =c\left(\hat{M}{ \pm}^{k} Y\right) \tag{5.125}
\end{align}
\)
which is what we wanted to prove.
Next we show that the set of eigenvalues of $\hat{M}{z}$ generated using the ladder operators must be bounded. For the particular eigenfunction $Y$ with $\hat{M}{z}$ eigenvalue $b$, we have
\(
\hat{M}_{z} Y=b Y
\)
and for the set of eigenfunctions and eigenvalues generated by the ladder operators, we have
\(
\begin{equation}
\hat{M}{z} Y{k}=b{k} Y{k} \tag{5.126}
\end{equation}
\)
where
\(
\begin{gather}
Y{k}=\hat{M}{ \pm}^{k} Y \tag{5.127}\
b_{k}=b \pm k \hbar \tag{5.128}
\end{gather}
\)
(Application of $\hat{M}{+}$or $\hat{M}{-}$destroys the normalization of $Y$, so $Y_{k}$ is not normalized. For the normalization constant, see Prob. 10.27.)
Operating on (5.126) with $\hat{M}_{z}$, we have
\(
\begin{align}
\hat{M}{z}^{2} Y{k} & =b{k} \hat{M}{z} Y{k} \
\hat{M}{z}^{2} Y{k} & =b{k}^{2} Y_{k} \tag{5.129}
\end{align}
\)
Now subtract (5.129) from (5.123), and use (5.127) and (5.108):
\(
\begin{align}
\hat{M}^{2} Y{k}-\hat{M}{z}^{2} Y{k} & =c Y{k}-b{k}^{2} Y{k} \
\left(\hat{M}{x}^{2}+\hat{M}{y}^{2}\right) Y{k} & =\left(c-b{k}^{2}\right) Y_{k} \tag{5.130}
\end{align}
\)
The operator $\hat{M}{x}^{2}+\hat{M}{y}^{2}$ corresponds to a nonnegative physical quantity and hence has nonnegative eigenvalues. (This is proved in Prob. 7.11.) Therefore, (5.130) implies that $c-b{k}^{2} \geq 0$ and $c^{1 / 2} \geq\left|b{k}\right|$. Thus
\(
\begin{equation}
c^{1 / 2} \geq b_{k} \geq-c^{1 / 2}, \quad k=0, \pm 1, \pm 2, \ldots \tag{5.131}
\end{equation}
\)
Since $c$ remains constant as $k$ varies, (5.131) shows that the set of eigenvalues $b{k}$ is bounded above and below. Let $b{\max }$ and $b{\text {min }}$ denote the maximum and minimum values of $b{k} . Y{\max }$ and $Y{\min }$ are the corresponding eigenfunctions:
\(
\begin{align}
\hat{M}{z} Y{\max } & =b{\max } Y{\max } \tag{5.132}\
\hat{M}{z} Y{\min } & =b{\min } Y{\min } \tag{5.133}
\end{align}
\)
Now operate on (5.132) with the raising operator and use (5.114):
\(
\begin{gather}
\hat{M}{+} \hat{M}{z} Y{\max }=b{\max } \hat{M}{+} Y{\max } \
\hat{M}{z}\left(\hat{M}{+} Y{\max }\right)=\left(b{\max }+\hbar\right)\left(\hat{M}{+} Y{\max }\right) \tag{5.134}
\end{gather}
\)
This last equation seems to contradict the statement that $b{\text {max }}$ is the largest eigenvalue of $\hat{M}{z}$, since it says that $\hat{M}{+} Y{\text {max }}$ is an eigenfunction of $\hat{M}{z}$ with eigenvalue $b{\text {max }}+\hbar$. The only way out of this contradiction is to have $\hat{M}{+} Y{\max }$ vanish. (We always reject zero as an eigenfunction on physical grounds.) Thus
\(
\begin{equation}
\hat{M}{+} Y{\max }=0 \tag{5.135}
\end{equation}
\)
Operating on (5.135) with the lowering operator and using (5.113), (5.132), and (5.116), we have
\(
\begin{gather}
0=\hat{M}{-} \hat{M}{+} Y{\max }=\left(\hat{M}^{2}-\hat{M}{z}^{2}-\hbar \hat{M}{z}\right) Y{\max }=\left(c-b{\max }^{2}-\hbar b{\max }\right) Y{\max } \
c-b{\max }^{2}-\hbar b{\max }=0 \
c=b{\max }^{2}+\hbar b_{\max } \tag{5.136}
\end{gather}
\)
A similar argument shows that
\(
\begin{equation}
\hat{M}{-} Y{\min }=0 \tag{5.137}
\end{equation}
\)
and by applying the raising operator to this equation and using (5.112), we find
\(
c=b{\min }^{2}-\hbar b{\min }
\)
Subtracting this last equation from (5.136), we have
\(
b{\max }^{2}+\hbar b{\max }+\left(\hbar b{\min }-b{\min }^{2}\right)=0
\)
This is a quadratic equation in the unknown $b_{\max }$, and using the usual formula (it still works in quantum mechanics), we find
\(
b{\max }=-b{\min }, \quad b{\max }=b{\min }-\hbar
\)
The second root is rejected, since it says that $b{\max }$ is less than $b{\text {min }}$. So
\(
\begin{equation}
b{\min }=-b{\max } \tag{5.138}
\end{equation}
\)
Moreover, (5.128) says that $b{\max }$ and $b{\text {min }}$ differ by an integral multiple of $\hbar$ :
\(
\begin{equation}
b{\max }-b{\min }=n \hbar, \quad n=0,1,2, \ldots \tag{5.139}
\end{equation}
\)
Substituting (5.138) in (5.139), we have for the $\hat{M}_{z}$ eigenvalues
\(
\begin{gather}
b{\max }=\frac{1}{2} n \hbar \
b{\max }=j \hbar, \quad j=0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots \tag{5.140}\
b_{\min }=-j \hbar \
b=-j \hbar,(-j+1) \hbar,(-j+2) \hbar, \ldots,(j-2) \hbar,(j-1) \hbar, j \hbar \tag{5.141}
\end{gather}
\)
and from (5.136) we find as the $\hat{M}^{2}$ eigenvalues
\(
\begin{equation}
c=j(j+1) \hbar^{2}, \quad j=0, \frac{1}{2}, 1, \frac{3}{2}, \ldots \tag{5.142}
\end{equation}
\)
Thus
\(
\begin{gather}
\hat{M}^{2} Y=j(j+1) \hbar^{2} Y, \quad j=0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots \tag{5.143}\
\hat{M}{z} Y=m{j} \hbar Y, \quad m_{j}=-j,-j+1, \ldots, j-1, j \tag{5.144}
\end{gather}
\)
We have found the eigenvalues of $\hat{M}^{2}$ and $\hat{M}_{z}$ using just the commutation relations. However, comparison of (5.143) and (5.144) with (5.104) and (5.105) shows that in addition to integral values for the angular-momentum quantum number $(l=0,1,2, \ldots)$ we now also have the possibility for half-integral values $\left(j=0, \frac{1}{2}, 1, \frac{3}{2}, \ldots\right)$. This perhaps suggests that there might be another kind of angular momentum besides orbital angular momentum. In Chapter 10 we shall see that spin angular momentum can have half-integral, as well as integral, quantum numbers. For orbital angular momentum, the boundary condition of single-valuedness of the $T(\phi)$ eigenfunctions [see the equation following (5.73)] eliminates the half-integral values of the angular-momentum quantum numbers. [Not everyone accepts single-valuedness as a valid boundary condition on wave functions, and many other reasons have been given for rejecting half-integral orbital-angular-momentum quantum numbers; see C. G. Gray, Am. J. Phys., 37, 559 (1969); M. L. Whippman, Am. J. Phys., 34, 656 (1966).]
The ladder-operator method can be used to solve other eigenvalue problems; see Prob. 5.36.