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# 7.3: Eigenstates of Angular Momentum

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}
