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# 7.5: Eigenvalues of L²

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Consider the angular wavefunction $$\psi(\theta,\phi) = L_+\,Y_{l,m}(\theta,\phi)$$. We know that

$\oint \psi^\ast(\theta,\phi)\,\psi(\theta,\phi)\,d{\mit\Omega} \geq 0,$

because $$\psi^\ast\,\psi\equiv |\psi|^{\,2}$$ is a positive-definite real quantity. Hence, making use of Equations ([e5.48]) and ([e8.14]), we find that

\begin{aligned} \oint (L_+\,Y_{l,m})^\ast\,(L_+\,Y_{l,m})\,d{\mit\Omega} = \oint Y_{l,m}^{\,\ast}\,(L_+)^\dagger\,(L_+\,Y_{l,m})\,d{\mit\Omega}= \oint Y_{l,m}^{\,\ast}\,L_-\,L_+\,Y_{l,m}\,d{\mit\Omega}\geq 0.\end{aligned}

It follows from Equations ([e8.17]), and ([e8.29])–([e8.31]) that

\begin{aligned} \oint Y_{l,m}^{\,\ast}\,(L^2 -L_z^{\,2}-\hbar\,L_z)\,Y_{l,m}\,d{\mit\Omega} &= \oint Y_{l,m}^{\,\ast}\,\hbar^{\,2}\left[l\,(l+1) -m\,(m+1)\right]Y_{l,m}\,d{\mit\Omega}\nonumber\\[0.5ex] &= \hbar^{\,2}\left[l\,(l+1) -m\,(m+1)\right]\,\oint Y_{l,m}^{\,\ast}\,Y_{l,m}\,d{\mit\Omega}\nonumber\\[0.5ex] &=\hbar^{\,2}\left[l\,(l+1) -m\,(m+1)\right]\geq 0.\end{aligned}

We, thus, obtain the constraint $\label{e8.42} l\,(l+1) \geq m\,(m+1).$ Likewise, the inequality $\oint (L_-\,Y_{l,m})^\ast\,(L_-\,Y_{l,m})\,d{\mit\Omega} =\oint Y_{l,m}^{\,\ast}\,L_+\,L_-\,Y_{l,m}\,d{\mit\Omega}\geq 0$ leads to a second constraint:

$\label{e8.44} l\,(l+1) \geq m\,(m-1).$

Without loss of generality, we can assume that $$l\geq 0$$. This is reasonable, from a physical standpoint, because $$l\,(l+1)\,\hbar^{\,2}$$ is supposed to represent the magnitude squared of something, and should, therefore, only take non-negative values. If $$l$$ is non-negative then the constraints ([e8.42]) and ([e8.44]) are equivalent to the following constraint:

$-l \leq m \leq l.$

We, thus, conclude that the quantum number $$m$$ can only take a restricted range of integer values.

Now, if $$m$$ can only take a restricted range of integer values then there must exist a lowest possible value that it can take. Let us call this special value $$m_-$$, and let $$Y_{l,m_-}$$ be the corresponding eigenstate. Suppose we act on this eigenstate with the lowering operator $$L_-$$. According to Equation ([e8.32]), this will have the effect of converting the eigenstate into that of a state with a lower value of $$m$$. However, no such state exists. A non-existent state is represented in quantum mechanics by the null wavefunction, $$\psi=0$$. Thus, we must have $\label{e8.46} L_-\,Y_{l,m_-} = 0.$ From Equation ([e8.15]),

$L^2 = L_+\,L_-+L_z^{\,2} - \hbar\,L_z$ Hence, $L^2\,Y_{l,m_-} = (L_+\,L_-+L_z^{\,2} - \hbar\,L_z)\,Y_{l,m_-},$

or

$l\,(l+1)\,Y_{l,m_-} = m_-\,(m_- -1)\,Y_{l,m_-},$

where use has been made of ([e8.29]), ([e8.30]), and ([e8.46]). It follows that

$l\,(l+1) = m_-\,(m_--1).$

Assuming that $$m_-$$ is negative, the solution to the previous equation is $m_- = - l.$ We can similarly show that the largest possible value of $$m$$ is $m_+ =+ l.$ The previous two results imply that $$l$$ is an integer, because $$m_-$$ and $$m_+$$ are both constrained to be integers.

We can now formulate the rules that determine the allowed values of the quantum numbers $$l$$ and $$m$$. The quantum number $$l$$ takes the non-negative integer values $$0, 1, 2, 3,\cdots$$. Once $$l$$ is given, the quantum number $$m$$ can take any integer value in the range $-l,\,-l+1,\,\cdots\, 0, \,\cdots,l-1,\, l.$ Thus, if $$l=0$$ then $$m$$ can only take the value $$0$$, if $$l=1$$ then $$m$$ can take the values $$-1, 0, +1$$, if $$l=2$$ then $$m$$ can take the values $$-2,-1,0,+1,+2$$, and so on.

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