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Physics LibreTexts

7.3: Eigenstates of Angular Momentum

  • Page ID
    15766
  • Let us find the simultaneous eigenstates of the angular momentum operators \(L_z\) and \(L^2\). Because both of these operators can be represented as purely angular differential operators, it stands to reason that their eigenstates only depend on the angular coordinates \(\theta\) and \(\phi\). Thus, we can write \[\begin{aligned} \label{e8.29} L_z\,Y_{l,m}(\theta,\phi) &= m\,\hbar\,Y_{l,m}(\theta,\phi),\\[0.5ex] L^2\,Y_{l,m}(\theta,\phi) &= l\,(l+1)\,\hbar^{\,2}\,Y_{l,m}(\theta,\phi).\label{e8.30}\end{aligned}\] Here, the \(Y_{l,m}(\theta,\phi)\) are the eigenstates in question, whereas the dimensionless quantities \(m\) and \(l\) parameterize the eigenvalues of \(L_z\) and \(L^2\), which are \(m\,\hbar\) and \(l\,(l+1)\,\hbar^{\,2}\), respectively. Of course, we expect the \(Y_{l,m}\) to be both mutually orthogonal and properly normalized (see Section [seig]), so that \[\label{e8.31} \oint Y^{\,\ast}_{l',m'}(\theta,\phi)\,Y_{l,m}(\theta,\phi)\,d{\mit\Omega }= \delta_{ll'}\,\delta_{mm'},\] where \(d{\mit\Omega} = \sin\theta\,d\theta\,d\phi\) is an element of solid angle, and the integral is over all solid angle.

    Now, \[\begin{aligned} L_z\,(L_+\,Y_{l,m}) &= (L_+\,L_z + [L_z, L_+])\,Y_{l,m}= (L_+\,L_z + \hbar\,L_+)\,Y_{l,m}\nonumber\\[0.5ex] &= (m+1)\,\hbar\,(L_+\,Y_{l,m}),\end{aligned}\] where use has been made of Equation ([e8.19]). We, thus, conclude that when the operator \(L_+\) operates on an eigenstate of \(L_z\) corresponding to the eigenvalue \(m\,\hbar\) it converts it to an eigenstate corresponding to the eigenvalue \((m+1)\,\hbar\). Hence, \(L_+\) is known as the raising operator (for \(L_z\)). It is also easily demonstrated that \[\label{e8.32} L_z\,(L_-\,Y_{l,m}) = (m-1)\,\hbar\,(L_-\,Y_{l,m}).\] In other words, when \(L_-\) operates on an eigenstate of \(L_z\) corresponding to the eigenvalue \(m\,\hbar\) it converts it to an eigenstate corresponding to the eigenvalue \((m-1)\,\hbar\). Hence, \(L_-\) is known as the lowering operator (for \(L_z\)).

    Writing \[\begin{aligned} L_+\,Y_{l,m} &= c_{l,m}^+\,Y_{l,m+1},\\[0.5ex] L_-\,Y_{l,m} &= c_{l,m}^-\,Y_{l,m-1},\end{aligned}\] we obtain \[L_-\,L_+\,Y_{l,m} = c^+_{l,m}\,c^-_{l,m+1}\,Y_{l,m} = [l\,(l+1)-m\,(m+1)]\,\hbar^2\,Y_{l,m},\] where use has been made of Equation ([e8.17]). Likewise, \[L_+\,L_-\,Y_{l,m} = c^+_{l,m-1}\,c^-_{l,m}\,Y_{l,m} = [l\,(l+1)-m\,(m-1)]\,\hbar^{\,2}\,Y_{l,m},\] where use has been made of Equation ([e8.15]). It follows that \[\begin{aligned} c^+_{l,m}\,c^-_{l,m+1}&= [l\,(l+1)-m\,(m+1)]\,\hbar^{\,2},\\[0.5ex] c^+_{l,m-1}\,c^-_{l,m}&= [l\,(l+1)-m\,(m-1)]\,\hbar^{\,2}.\end{aligned}\] These equations are satisfied when \[c^\pm_{l,m} = [l\,(l+l) - m\,(m\pm 1)]^{1/2}\,\hbar.\] Hence, we can write \[\begin{aligned} \label{eraise} L_+\,Y_{l,m} &= [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar\,Y_{l,m+1},\\[0.5ex] L_-\,Y_{l,m} &=[l\,(l+1)-m\,(m-1)]^{1/2}\,\hbar\,Y_{l,m-1}.\label{elow}\end{aligned}\]

    Contributors

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

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