Eigenvalues of 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

$$L_z \vert l, m\rangle = m \hbar \vert l, m\rangle.$$ (321)

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, since $L_z$ is an Hermitian operator.

We can write

$$L^2 \vert l, m\rangle = f(l,m) \hbar^2 \vert l, m\rangle,$$ (322)

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,

$$\langle l, m \vert L^2 - L_z^{ 2} \vert l, m\rangle =\langle... ...r^2 - m^2 \hbar^2 \vert l, m\rangle =[f(l,m) - m^2] \hbar^2,$$ (323)

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

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

$$= \langle l, m \vert L_x^{ 2}\vert l, m\rangle+ \langle l, m\vert L_y^{ 2}\vert l, m\rangle.$$ (324)

It is easily demonstrated that

$$\langle A\vert\xi^2\vert A\rangle\geq 0,$$ (325)

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

$$\langle A\vert\xi^2\vert A\rangle = \langle A\vert\xi^{\dag } \xi \vert A\rangle = \langle B\vert B\rangle,$$ (326)

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 Eq. (21)]. It follows from Eqs. (323)-(325) that

$$m^2 \leq f(l,m).$$ (327)

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

$$L^2 (L^+ \vert l, m\rangle) = \hbar^2 f(l,m) (L^+ \vert l,m\rangle),$$ (328)

where use has been made of Eq. (322), plus the fact that $L^2$ and $L_z$ 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. Note that

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

$$= (m+1) \hbar L^+\vert l, m\rangle,$$ (329)

where use has been made of Eq. (318). The above equation implies that $L^+ \vert l,m\rangle$ is proportional to $\vert l, m+1\rangle$. We can write

$$L^+ \vert l ,m\rangle = c^+_{l, m} \hbar \vert l, m+1\rangle,$$ (330)

where $c^+_{l, m}$ is a number. It is clear that when the operator $L^+$ acts on a simultaneous eigenstate of $L^2$ and $L_z$, 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

$$L^- \vert l ,m\rangle = c^-_{l, m} \hbar \vert l, m-1\rangle.$$ (331)

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

The shift operators 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 the shift operators a sufficient number of times. However, according to Eq. (327), 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 Eq. (322)]. 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

$$L^+ \vert l, m_{\rm max}\rangle = \vert\rangle.$$ (332)

This implies that

$$L^- L^+ \vert l, m_{\rm max}\rangle = \vert\rangle.$$ (333)

However,

$$L^- L^+ = L_x^{ 2} + L_y^{ 2} + {\rm i} [L_x, L_y] = L^2 - L_z^{ 2} - \hbar L_z,$$ (334)

so Eq. (333) yields

$$(L^2 - L_z^{ 2} - \hbar L_z) \vert l, m_{\rm max}\rangle = \vert\rangle.$$ (335)

The above equation can be rearranged to give

$$L^2 \vert l, m_{\rm max}\rangle = (L_z^{ 2} + \hbar L_z) \... ...max}(m_{\rm max} + 1) \hbar^2 \vert l, m_{\rm max}\rangle.$$ (336)

Comparison of this equation with Eq. (322) yields the result

$$f(l, m_{\rm max}) = m_{\rm max} (m_{\rm max} + 1).$$ (337)

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,

$$L^2 \vert l, m\rangle = m_{\rm max}(m_{\rm max} + 1) \hbar^2 \vert l, m\rangle.$$ (338)

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

$$L^2 \vert l, m\rangle = l (l+1) \hbar^2 \vert l, m\rangle.$$ (339)

It is easily seen that

$$L^- L^+ \vert l, m\rangle = (L^2 - L_z^{ 2}-\hbar L_z)\vert l, m \rangle = \hbar^2 [l (l+1) - m (m+1)]\vert l,m\rangle.$$ (340)

Thus,

$$\langle l,m\vert L^- L^+\vert l,m\rangle =\hbar^2 [l (l+1) - m (m+1)].$$ (341)

However, we also know that

$$\langle l,m\vert L^- L^+ \vert l,m\rangle = \langle l, m\v... ...,m} \vert l,m+1\rangle = \hbar^2 c^+_{l,m} c^{-}_{l, m+1},$$ (342)

where use has been made of Eqs. (330) and (331). It follows that

$$c^+_{l,m} c^{-}_{l, m+1} = [l (l+1) - m (m+1)].$$ (343)

Consider the following:

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

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

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

$$= \langle l, m+1\vert L^+\vert l, m\rangle^\ast,$$ (344)

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

$$c^-_{l, m+1} = (c_{l,m}^+)^\ast$$ (345)

with the aid of Eqs. (330) and (331).

Equations (343) and (345) can be combined to give

$$\vert c^+_{l,m}\vert^2 = [l (l+1) - m (m+1)].$$ (346)

The solution of the above equation is

$$c_{l, m}^+ = \sqrt{l (l+1)- m (m+1)}.$$ (347)

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 Eq. (346)]. 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 Eq. (345),

$$c_{l, m}^- = (c_{l, m-1}^+)^\ast = \sqrt{l (l+1)- m (m-1)}.$$ (348)

We have already seen that the inequality (327) 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 $m_{\rm min}$. Since there is no state with $m<m_{\rm min}$, we must have

$$L^- \vert l, m_{\rm min}\rangle = 0.$$ (349)

According to Eq. (331), this implies that

$$c_{l, m_{\rm min}}^- = 0.$$ (350)

It can be seen from Eq. (348) that $m_{\rm min} = -l$. We conclude that $m$ can take a ``ladder'' of discrete values, each rung differing from its immediate neighbours 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 will 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 (304)-(306), 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

$$L^+ \vert l, m\rangle = \sqrt{l (l+1)-m (m+1)} \hbar \vert l, m+1\rangle$$ (351)

$$L^- \vert l,m \rangle = \sqrt{l (l+1)-m (m-1)} \hbar \vert l, m-1\rangle,$$ (352)

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