7.3: Eigenstates of Angular Momentum
( \newcommand{\kernel}{\mathrm{null}\,}\)
Let us find the simultaneous eigenstates of the angular momentum operators Lz and L2. 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 θ and ϕ. Thus, we can write LzYl,m(θ,ϕ)=mℏYl,m(θ,ϕ),L2Yl,m(θ,ϕ)=l(l+1)ℏ2Yl,m(θ,ϕ). Here, the Yl,m(θ,ϕ) are the eigenstates in question, whereas the dimensionless quantities m and l parameterize the eigenvalues of Lz and L2, which are mℏ and l(l+1)ℏ2, respectively. Of course, we expect the Yl,m to be both mutually orthogonal and properly normalized (see Section [seig]), so that ∮Y∗l′,m′(θ,ϕ)Yl,m(θ,ϕ)dΩ=δll′δmm′, where dΩ=sinθdθdϕ is an element of solid angle, and the integral is over all solid angle.
Now, Lz(L+Yl,m)=(L+Lz+[Lz,L+])Yl,m=(L+Lz+ℏL+)Yl,m=(m+1)ℏ(L+Yl,m), where use has been made of Equation ([e8.19]). We, thus, conclude that when the operator L+ operates on an eigenstate of Lz corresponding to the eigenvalue mℏ it converts it to an eigenstate corresponding to the eigenvalue (m+1)ℏ. Hence, L+ is known as the raising operator (for Lz). It is also easily demonstrated that Lz(L−Yl,m)=(m−1)ℏ(L−Yl,m). In other words, when L− operates on an eigenstate of Lz corresponding to the eigenvalue mℏ it converts it to an eigenstate corresponding to the eigenvalue (m−1)ℏ. Hence, L− is known as the lowering operator (for Lz).
Writing L+Yl,m=c+l,mYl,m+1,L−Yl,m=c−l,mYl,m−1, we obtain L−L+Yl,m=c+l,mc−l,m+1Yl,m=[l(l+1)−m(m+1)]ℏ2Yl,m, where use has been made of Equation ([e8.17]). Likewise, L+L−Yl,m=c+l,m−1c−l,mYl,m=[l(l+1)−m(m−1)]ℏ2Yl,m, where use has been made of Equation ([e8.15]). It follows that c+l,mc−l,m+1=[l(l+1)−m(m+1)]ℏ2,c+l,m−1c−l,m=[l(l+1)−m(m−1)]ℏ2. These equations are satisfied when c±l,m=[l(l+l)−m(m±1)]1/2ℏ. Hence, we can write L+Yl,m=[l(l+1)−m(m+1)]1/2ℏYl,m+1,L−Yl,m=[l(l+1)−m(m−1)]1/2ℏYl,m−1.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)