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 $$\begin{aligned} \label{eraise} L_+\,Y_{l,m} &= [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar\,Y_{l,m+1},\\[0.5ex] L_-\,Y_{l,m} &=[l\,(l+1)-m\,(m-1)]^{1/2}\,\hbar\,Y_{l,m-1}.\label{elow}\end{aligned}$$

Contributors and Attributions

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

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