next up previous
Next: Rotation Operators Up: Orbital Angular Momentum Previous: Orbital Angular Momentum

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

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

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

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

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

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

It is readily demonstrated that

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

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

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

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

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

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

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

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

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

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

This implies that

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


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

so Equation (4.37) yields

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

The previous equation can be rearranged to give

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

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

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

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

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

It is easily seen that

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


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

However, we also know that

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

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

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

Consider the following:

$\displaystyle \langle l, m\vert\, L^-\, \vert l, m+1\rangle$ $\displaystyle = \langle l, m\vert\,L_x\,\vert l,m+1 \rangle - {\rm i}\, \langle l, m\vert \,L_y\,\vert l,m+1 \rangle$    
  $\displaystyle = \langle l, m+1\vert \,L_x\,\vert l,m \rangle^\ast - {\rm i}\, \langle l, m+1\vert\, L_y\,\vert l,m \rangle^\ast$    
  $\displaystyle =( \langle l, m+1\vert\, L_x\,\vert l,m \rangle + {\rm i}\, \langle l, m+1\vert\, L_y\,\vert l,m \rangle)^\ast$    
  $\displaystyle = \langle l, m+1\vert\,L^+\,\vert l, m\rangle^\ast,$ (4.48)

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

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

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

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

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

The solution of the previous equation is

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

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

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

According to Equation (4.35), this implies that

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

$\displaystyle L^+ \,\vert l, m\rangle$ $\displaystyle = \sqrt{l\,(l+1)-m\,(m+1)}\,\hbar\,\vert l, m+1\rangle$ (4.55)
$\displaystyle L^- \,\vert l,m \rangle$ $\displaystyle = \sqrt{l\,(l+1)-m\,(m-1)}\,\hbar\,\vert l, m-1\rangle,$ (4.56)

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

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