Eigenvalues of Orbital Angular Momentum

Suppose that the simultaneous eigenkets of $L^2$ and $L_z$ are completely specified by two (dimensionless) quantum numbers, $l$ and $m$ . These kets are denoted $\vert l, m\rangle$ . The quantum number $m$ is defined by

$$L_z \,\vert l, m\rangle = m\,\hbar\, \vert l, m\rangle.$$ (4.25)

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

We can write

$$L^2 \,\vert l, m\rangle = f(l,m)\, \hbar^{\,2}\,\vert l, m\rangle,$$ (4.26)

without loss of generality, where $f(l,m)$ is some real dimensionless function of $l$ and $m$ . Later on, we will show that $f(l,m) = l\,(l+1)$ . Now,

$$\langle l, m \vert\, L^2 - L_z^{\,2}\, \vert l, m\rangle =\langle... ... - m^{\,2}\, \hbar^{\,2}\, \vert l, m\rangle =[f(l,m) - m^{\,2}] \,\hbar^{\,2},$$ (4.27)

assuming that the $\vert l, m\rangle$ have unit norms. However,

$$\langle l, m \vert\,L^2 - L_z^{\,2}\,\vert l, m\rangle =\langle l... ...,L_x^{\,2}\,\vert l, m\rangle+ \langle l, m\vert\,L_y^{\,2}\,\vert l, m\rangle.$$ (4.28)

It is readily demonstrated that

$$\langle A\vert\,\xi^{\,2}\,\vert A\rangle\geq 0,$$ (4.29)

where $\vert A\rangle$ is a general ket, and $\xi$ an Hermitian operator. The proof follows from the observation that

$$\langle A\vert\,\xi^{\,2}\,\vert A\rangle = \langle A\vert\,\xi^{\dag }\, \xi\,\vert A\rangle = \langle B\vert B\rangle,$$ (4.30)

where $\vert B\rangle = \xi\, \vert A\rangle$ , plus the fact that $\langle B\vert B\rangle\geq 0$ for a general ket $\vert B\rangle$ . [See Equation (1.22).] It follows from Equations (4.27)-(4.29) that

$$m^{\,2} \leq f(l,m).$$ (4.31)

Consider the effect of the ladder operator $L^+$ on the eigenket $\vert l, m\rangle$ . It is easily demonstrated that

$$L^2\, (L^+ \vert l, m\rangle) = \hbar^{\,2}\, f(l,m)\, (L^+ \vert l,m\rangle),$$ (4.32)

where use has been made of Equation (4.26), plus the fact that $L^2$ and $L^+$ commute. It follows that the ket $L^+ \vert l,m\rangle$ is an eigenstate of $L^2$ corresponding to the same eigenvalue as the ket $\vert l, m\rangle$ . Thus, the ladder operator $L^+$ does not affect the magnitude of the angular momentum of any state that it acts upon. However,

$$L_z \,(L^+ \vert l, m\rangle) = (L^+ L_z + [L_z, L^+])\,\vert l,m\rangle = (L^+ L_z + \hbar\, L^+) \,\vert l,m\rangle$$

$$= (m+1)\,\hbar \,(L^+\vert l, m\rangle),$$ (4.33)

where use has been made of Equation (4.22). The previous equation implies that $L^+ \vert l,m\rangle$ is proportional to $\vert l, m+1\rangle$ . We can write

$$L^+ \vert l ,m\rangle = c^+_{l\,m}\, \hbar\,\vert l, m+1\rangle,$$ (4.34)

where $c^+_{l\, m}$ is a (dimensionless) number. It is clear that if the operator $L^+$ acts on a simultaneous eigenstate of $L^2$ and $L_z$ then the eigenvalue of $L^2$ remains unchanged, but the eigenvalue of $L_z$ is increased by $\hbar$ . For this reason, $L^+$ is called a raising operator.

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

$$L^-\, \vert l ,m\rangle = c^-_{l\,m}\,\hbar\, \vert l, m-1\rangle,$$ (4.35)

where $c^-_{l\,m}$ is a (dimensionless) number. Hence, $L^-$ is called a lowering operator.

The ladder operators, $L^+$ and $L^-$ , respectively step the value of $m$ up and down by unity each time they operate on one of the simultaneous eigenkets of $L^2$ and $L_z$ . It would appear, at first sight, that any value of $m$ 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 $m^{\,2}$ can take. This bound is determined by the eigenvalue of $L^2$ . [See Equation (4.26).] It follows that there is a maximum and a minimum possible value that $m$ can take. Suppose that we attempt to raise the value of $m$ above its maximum value, $m_{\rm max}$ . Because there is no state with $m> m_{\rm max}$ , we must have

$$L^+ \vert l, m_{\rm max}\rangle = \vert\rangle.$$ (4.36)

This implies that

$$L^-\, L^+ \vert l, m_{\rm max}\rangle = \vert\rangle.$$ (4.37)

However,

$$L^-\, L^+ = L_x^{\,2} + L_y^{\,2} + {\rm i}\,[L_x, L_y] = L^2 - L_z^{\,2} - \hbar \,L_z,$$ (4.38)

so Equation (4.37) yields

$$(L^2 - L_z^{\,2} - \hbar \,L_z) \,\vert l, m_{\rm max}\rangle = \vert\rangle.$$ (4.39)

The previous equation can be rearranged to give

$$L^2\, \vert l, m_{\rm max}\rangle = (L_z^{\,2} + \hbar \,L_z)\, \... ...e = m_{\rm max}\,(m_{\rm max} + 1) \,\hbar^{\,2}\, \vert l, m_{\rm max}\rangle.$$ (4.40)

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

$$f(l, m_{\rm max}) = m_{\rm max} (m_{\rm max} + 1).$$ (4.41)

But, when $L^-$ operates successively on $\vert n, m_{\rm max}\rangle$ it generates $\vert n, m_{\rm max}-1\rangle$ , $\vert n, m_{\rm max}-2\rangle$ , et cetera. Because the lowering operator does not change the eigenvalue of $L^2$ , all of these states must correspond to the same value of $f(l,m)$ ; namely, $m_{\rm max}\,(m_{\rm max} + 1)$ . Thus,

$$L^2 \,\vert l, m\rangle = m_{\rm max}\,(m_{\rm max} + 1)\,\hbar^{\,2}\, \vert l, m\rangle.$$ (4.42)

At this stage, we can give the unknown quantum number $l$ the value $m_{\rm max}$ , without loss of generality. We can also write the previous equation in the form

$$L^2 \,\vert l, m\rangle = l\,(l+1)\, \hbar^{\,2}\, \vert l, m\rangle.$$ (4.43)

It is easily seen that

$$L^- \,L^+ \,\vert l, m\rangle = (L^2 - L_z^{\,2}-\hbar\, L_z)\,\vert l, m \rangle = \hbar^{\,2} \,[l\,(l+1) - m\,(m+1)]\,\vert l,m\rangle.$$ (4.44)

Thus,

$$\langle l,m\vert\, L^- \,L^+\,\vert l,m\rangle =\hbar^{\,2} \,[l\,(l+1) - m\,(m+1)].$$ (4.45)

However, we also know that

$$\langle l,m\vert \,L^- \,L^+ \,\vert l,m\rangle = \langle l, m\ve... ...c^+_{l\,m}\, \vert l,m+1\rangle = \hbar^{\,2}\, c^+_{l\,m} \,c^{-}_{l\,\, m+1},$$ (4.46)

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

$$c^+_{l\,m}\, c^{-}_{l\,\,m+1} = l\,(l+1) - m\,(m+1).$$ (4.47)

Consider the following:

$$\langle l, m\vert\, L^-\, \vert l, m+1\rangle = \langle l, m\vert\,L_x\,\vert l,m+1 \rangle - {\rm i}\, \langle l, m\vert \,L_y\,\vert l,m+1 \rangle$$

$$= \langle l, m+1\vert \,L_x\,\vert l,m \rangle^\ast - {\rm i}\, \langle l, m+1\vert\, L_y\,\vert l,m \rangle^\ast$$

$$=( \langle l, m+1\vert\, L_x\,\vert l,m \rangle + {\rm i}\, \langle l, m+1\vert\, L_y\,\vert l,m \rangle)^\ast$$

$$= \langle l, m+1\vert\,L^+\,\vert l, m\rangle^\ast,$$ (4.48)

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

$$c^-_{l\,\, m+1} = (c_{l\,m}^+)^\ast$$ (4.49)

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

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

$$\vert c^+_{l\,m}\vert^{\,2} = l\,(l+1) - m \,(m+1).$$ (4.50)

The solution of the previous equation is

$$c_{l\,m}^+ = \sqrt{l\,(l+1)- m \,(m+1)}.$$ (4.51)

Note that $c_{l\,m}^+$ is undetermined to an arbitrary phase-factor [i.e., we can replace $c_{l\,m}^+$ , given previously, by $c_{l\, m}^+\exp(\,{\rm i}\,\gamma)$ , where $\gamma$ is real, and we still satisfy Equation (4.50)]. We have made the arbitrary, but convenient, choice that $c_{l\,m}^+$ is real and positive. This is equivalent to choosing the relative phases of the eigenkets $\vert l, m\rangle$ . According to Equation (4.49),

$$c_{l\, m}^- = (c_{l\, \,m-1}^+)^\ast = \sqrt{l\,(l+1)- m\, (m-1)}.$$ (4.52)

We have already seen that the inequality (4.31) implies that there is a maximum and a minimum possible value of $m$ . The maximum value of $m$ is denoted $l$ . What is the minimum value? Suppose that we try to lower the value of $m$ below its minimum value $m_{\rm min}$ . Because there is no state with $m<m_{\rm min}$ , we must have

$$L^- \,\vert l, m_{\rm min}\rangle = 0.$$ (4.53)

According to Equation (4.35), this implies that

$$c_{l\,\, m_{\rm min}}^- = 0.$$ (4.54)

It can be seen from Equation (4.52) that $m_{\rm min} = -l$ . We conclude that the quantum number $m$ can take a ``ladder'' of discrete values, each rung differing from its immediate neighbors by unity. The top rung is $l$ , and the bottom rung is $-l$ . There are only two possible choices for $l$ . Either it is an integer (e.g., $l=2$ , which allows $m$ to take the integer values $-2, -1, 0, 1, 2$ ), or it is a half-integer (e.g., $l=3/2$ , which allows $m$ to take the half-integer values $-3/2, -1/2, 1/2, 3/2$ ). In fact, we shall prove, in the next section, that an orbital angular momentum can only take integer values of $l$ .

In summary, just using the fundamental commutation relations (4.8)-(4.10), plus the fact that $L_x$ , $L_y$ , and $L_z$ are Hermitian operators, we have shown that the eigenvalues of $L^2 \equiv L_x^{\,2} + L_y^{\,2}+L_z^{\,2}$ can be written $l\,(l+1)\,\hbar^{\,2}$ , where $l$ is an integer, or a half-integer. Without loss of generality, we can assume that $l$ is non-negative. We have also demonstrated that the eigenvalues of $L_z$ can only take the values $m\,\hbar$ , where $m$ lies in the range $-l, -l+1,\cdots l-1, l$ . Finally, if $\vert l, m\rangle$ denotes a properly normalized simultaneous eigenket of $L^2$ and $L_z$ , belonging to the eigenvalues $l\,(l+1)\,\hbar^{\,2}$ and $m\,\hbar$ , respectively, then we have shown that

$$L^+ \,\vert l, m\rangle = \sqrt{l\,(l+1)-m\,(m+1)}\,\hbar\,\vert l, m+1\rangle$$ (4.55)

$$L^- \,\vert l,m \rangle = \sqrt{l\,(l+1)-m\,(m-1)}\,\hbar\,\vert l, m-1\rangle,$$ (4.56)

where $L^\pm = L_x \pm {\rm i} \,L_y$ are the so-called ladder operators.