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4.2: 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

 

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

 

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

 

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

 

 

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

 

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

 

 

It is easily demonstrated that

 

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

 

 

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

 

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

 

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

 

 

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

 

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

 

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

 

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

 

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

 

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

 

 

This implies that

 

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

 

 

However,

 

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

 

 

so Equation (326) yields

 

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

 

 

The above equation can be rearranged to give

 

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

 

 

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

 

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

 

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

 

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

 

 

It is easily seen that

 

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

 

 

Thus,

 

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

 

 

However, we also know that

 

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

 

 

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

 

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

 

 

Consider the following:

 

$\displaystyle \langle l, m\vert\, L^-\, \vert l, m+1\rangle$ $\displaystyle = \langle l, m\vert\,L_x\,\vert l,m+1 \rangle - {\rm i}\, \langle l, m\vert \,L_y\,\vert l,m+1 \rangle$    
  $\displaystyle = \langle l, m+1\vert \,L_x\,\vert l,m \rangle^\ast - {\rm i}\, \langle l, m+1\vert\, L_y\,\vert l,m \rangle^\ast$    
  $\displaystyle =( \langle l, m+1\vert\, L_x\,\vert l,m \rangle + {\rm i}\, \langle l, m+1\vert\, L_y\,\vert l,m \rangle)^\ast$    
  $\displaystyle = \langle l, m+1\vert\,L^+\,\vert l, m\rangle^\ast,$ (337)

 

 

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

 

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

 

 

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

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

 

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

 

 

The solution of the above equation is

 

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

 

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

 

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

 

 

According to Equation (324), this implies that

 

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

 

$\displaystyle L^+ \,\vert l, m\rangle$ $\displaystyle = \sqrt{l\,(l+1)-m\,(m+1)}\,\hbar\,\vert l, m+1\rangle$ (344)
$\displaystyle L^- \,\vert l,m \rangle$ $\displaystyle = \sqrt{l\,(l+1)-m\,(m-1)}\,\hbar\,\vert l, m-1\rangle,$ (345)

 

 

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

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