Eigenvalues of Orbital Angular Momentum

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

$$L_z \,\vert l, m\rangle = m\,\hbar\, \vert l, m\rangle.$$ (314)

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

$$L^2 \,\vert l, m\rangle = f(l,m)\, \hbar^2\,\vert l, m\rangle,$$ (315)

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,

$$\langle l, m \vert\, L^2 - L_z^{\,2}\, \vert l, m\rangle =\langle... ...(l, m) \,\hbar^2 - m^2\, \hbar^2\, \vert l, m\rangle =[f(l,m) - m^2] \,\hbar^2,$$ (316)

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

$$\langle l, m \vert\,L^2 - L_z^{\,2}\,\vert l, m\rangle =\langle l... ...,L_x^{\,2}\,\vert l, m\rangle+ \langle l, m\vert\,L_y^{\,2}\,\vert l, m\rangle.$$ (317)

It is easily demonstrated that

$$\langle A\vert\,\xi^{\,2}\,\vert A\rangle\geq 0,$$ (318)

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

$$\langle A\vert\,\xi^{\,2}\,\vert A\rangle = \langle A\vert\,\xi^{\dag }\, \xi\,\vert A\rangle = \langle B\vert B\rangle,$$ (319)

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 (21)]. It follows from Equations (316)-(318) that

$$m^2 \leq f(l,m).$$ (320)

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

$$L^2 (L^+ \vert l, m\rangle) = \hbar^2\, f(l,m)\, (L^+ \vert l,m\rangle),$$ (321)

where use has been made of Equation (315), plus the fact that $L^2$ and $L^+$ commute. It follows that the ket $L^+ \vert l,m\rangle$ has the same eigenvalue of $L^2$ as the ket $\vert l, m\rangle$ . Thus, the shift operator $L^+$ does not affect the magnitude of the angular momentum of any eigenket it acts upon. However,

$$L_z \,L^+ \vert l, m\rangle = (L^+ L_z + [L_z, L^+])\,\vert l,m\r... ...^+ L_z + \hbar\, L^+) \,\vert l,m\rangle = (m+1)\,\hbar \,L^+\vert l, m\rangle,$$ (322)

where use has been made of Equation (311). The above equation implies that $L^+ \vert l,m\rangle$ is proportional to $\vert l, m+1\rangle$ . We can write

$$L^+ \vert l ,m\rangle = c^+_{l\,m}\, \hbar\,\vert l, m+1\rangle,$$ (323)

where $c^+_{l\, m}$ is a 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 given above, it is possible to demonstrate that

$$L^-\, \vert l ,m\rangle = c^-_{l\,m}\,\hbar\, \vert l, m-1\rangle.$$ (324)

Hence, $L^-$ is called a lowering operator.

The shift 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 (320), 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 (315)]. It follows that there is a maximum and a minimum possible value which $m$ can take. Suppose that we attempt to raise the value of $m$ above its maximum value $m_{\rm max}$ . Since there is no state with $m> m_{\rm max}$ , we must have

$$L^+ \vert l, m_{\rm max}\rangle = \vert\rangle.$$ (325)

This implies that

$$L^-\, L^+ \vert l, m_{\rm max}\rangle = \vert\rangle.$$ (326)

However,

$$L^-\, L^+ = L_x^{\,2} + L_y^{\,2} + {\rm i}\,[L_x, L_y] = L^2 - L_z^{\,2} - \hbar \,L_z,$$ (327)

so Equation (326) yields

$$(L^2 - L_z^{\,2} - \hbar \,L_z) \,\vert l, m_{\rm max}\rangle = \vert\rangle.$$ (328)

The above equation can be rearranged to give

$$L^2\, \vert l, m_{\rm max}\rangle = (L_z^{\,2} + \hbar \,L_z)\, \... ...\rangle = m_{\rm max}(m_{\rm max} + 1) \,\hbar^2\, \vert l, m_{\rm max}\rangle.$$ (329)

Comparison of this equation with Equation (315) yields the result

$$f(l, m_{\rm max}) = m_{\rm max} (m_{\rm max} + 1).$$ (330)

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

$$L^2 \,\vert l, m\rangle = m_{\rm max}(m_{\rm max} + 1)\,\hbar^2\, \vert l, m\rangle.$$ (331)

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 above equation in the form

$$L^2 \,\vert l, m\rangle = l\,(l+1)\, \hbar^2\, \vert l, m\rangle.$$ (332)

It is easily seen that

$$L^- \,L^+ \,\vert l, m\rangle = (L^2 - L_z^{\,2}-\hbar\, L_z)\,\vert l, m \rangle = \hbar^2 \,[l\,(l+1) - m\,(m+1)]\,\vert l,m\rangle.$$ (333)

Thus,

$$\langle l,m\vert\, L^- \,L^+\,\vert l,m\rangle =\hbar^2 \,[l\,(l+1) - m\,(m+1)].$$ (334)

However, we also know that

$$\langle l,m\vert L^- \,L^+ \vert l,m\rangle = \langle l, m\vert L... ...bar\, c^+_{l\,m} \vert l,m+1\rangle = \hbar^2\, c^+_{l\,m} \,c^{-}_{l\,\, m+1},$$ (335)

where use has been made of Equations (323) and (324). It follows that

$$c^+_{l\,m}\, c^{-}_{l\,\,m+1} = [l\,(l+1) - m\,(m+1)].$$ (336)

Consider the following:

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

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

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

$$= \langle l, m+1\vert\,L^+\,\vert l, m\rangle^\ast,$$ (337)

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

$$c^-_{l\,\, m+1} = (c_{l\,m}^+)^\ast$$ (338)

with the aid of Equations (323) and (324).

Equations (336) and (338) can be combined to give

$$\vert c^+_{l\,m}\vert^{\,2} = [l\,(l+1) - m \,(m+1)].$$ (339)

The solution of the above equation is

$$c_{l\,m}^+ = \sqrt{l\,(l+1)- m \,(m+1)}.$$ (340)

Note that $c_{l\,m}^+$ is undetermined to an arbitrary phase-factor [i.e., we can replace $c_{l\,m}^+$ , given above, by $c_{l\, m}^+\exp(\,{\rm i}\,\gamma)$ , where $\gamma$ is real, and we still satisfy Equation (339)]. 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 (338),

$$c_{l\, m}^- = (c_{l\, \,m-1}^+)^\ast = \sqrt{l\,(l+1)- m\, (m-1)}.$$ (341)

We have already seen that the inequality (320) 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

$$L^- \,\vert l, m_{\rm min}\rangle = 0.$$ (342)

According to Equation (324), this implies that

$$c_{l\,\, m_{\rm min}}^- = 0.$$ (343)

It can be seen from Equation (341) that $m_{\rm min} = -l$ . We conclude that $m$ 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 values $-2, -1, 0, 1, 2$ ), or it is a half-integer (e.g., $l=3/2$ , which allows $m$ to take the values $-3/2, -1/2, 1/2, 3/2$ ). We shall prove in the next section that an orbital angular momentum can only take integer values of $l$ .

In summary, using just the fundamental commutation relations (297)-(299), 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. We have also demonstrated that the eigenvalues of $L_z$ can only take the values $m\,\hbar$ , where $m$ lies in the range $-l, -l+1,\cdots l-1, l$ . Let $\vert l, m\rangle$ denote a properly normalized simultaneous eigenket of $L^2$ and $L_z$ , belonging to the eigenvalues $l\,(l+1)\,\hbar^2$ and $m\,\hbar$ , respectively. We have shown that

$$L^+ \,\vert l, m\rangle = \sqrt{l\,(l+1)-m\,(m+1)}\,\hbar\,\vert l, m+1\rangle$$ (344)

$$L^- \,\vert l,m \rangle = \sqrt{l\,(l+1)-m\,(m-1)}\,\hbar\,\vert l, m-1\rangle,$$ (345)

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