# 7.5: Eigenvalues of L²

- Page ID
- 15768

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.

## Contributors and Attributions

Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

\( \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}\) \(\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}\) \(\newcommand {\btau}{\mbox{\boldmath$\tau$}}\) \(\newcommand {\bmu}{\mbox{\boldmath$\mu$}}\) \(\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}\) \(\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}\) \(\newcommand {\bomega}{\mbox{\boldmath$\omega$}}\) \(\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}\)