Angular Momentum Algebra: Raising and Lowering Operators

We have already derived the commutators of the angular momentum operators

\begin{eqnarray*}[L_x,L_y]&=&i\hbar L_z \\
{[L_i,L_j]}&=&i\hbar\epsilon_{ijk}L_k \\
{[L^2,L_i]}&=&0 .\\

We have shown that angular momentum is quantized for a rotor with a single angular variable. To progress toward the possible quantization of angular momentum variables in 3D, we define the operator \bgroup\color{black}$L_+$\egroup and its Hermitian conjugate \bgroup\color{black}$L_-$\egroup.

\begin{displaymath}\bgroup\color{black} L_\pm\equiv L_x\pm iL_y .\egroup\end{displaymath}

Since \bgroup\color{black}$L^2$\egroup commutes with \bgroup\color{black}$L_x$\egroup and \bgroup\color{black}$L_y$\egroup, it commutes with these operators.

\begin{displaymath}\bgroup\color{black} [L^2,L_\pm]=0 \egroup\end{displaymath}

The commutator with \bgroup\color{black}$L_z$\egroup is.

\begin{displaymath}\bgroup\color{black} [L_\pm,L_z]=[L_x,L_z]\pm i[L_y,L_z]=i\hbar(-L_y\pm iL_x)=\mp\hbar L_\pm .\egroup\end{displaymath}

From the commutators \bgroup\color{black}$[L^2,L_\pm]=0$\egroup and \bgroup\color{black}$[L_\pm,L_z]=\mp\hbar L_\pm$\egroup, we can derive the effect of the operators \bgroup\color{black}$L_\pm$\egroup on the eigenstates \bgroup\color{black}$Y_{\ell m}$\egroup, and in so doing, show that \bgroup\color{black}$\ell$\egroup is an integer greater than or equal to 0, and that \bgroup\color{black}$m$\egroup is also an integer

\bgroup\color{black}$\displaystyle \ell=0,1,2,...$\egroup
\bgroup\color{black}$\displaystyle -\ell\leq m\leq\ell$\egroup
\bgroup\color{black}$\displaystyle m=-\ell, -\ell+1,...,\ell$\egroup
\bgroup\color{black}$\displaystyle L_\pm Y_{\ell m}=\hbar\sqrt{\ell(\ell+1)-m(m\pm 1)}Y_{\ell(m\pm 1)}$\egroup
Therefore, \bgroup\color{black}$L_+$\egroup raises the \bgroup\color{black}$z$\egroup component of angular momentum by one unit of \bgroup\color{black}$\hbar$\egroup and \bgroup\color{black}$L_-$\egroup lowers it by one unit. The raising stops when \bgroup\color{black}$m=\ell$\egroup and the operation gives zero, \bgroup\color{black}$L_+Y_{\ell \ell}=0$\egroup. Similarly, the lowering stops because \bgroup\color{black}$L_-Y_{\ell -\ell}=0$\egroup.


Angular momentum is quantized. Any measurement of a component of angular momentum will give some integer times \bgroup\color{black}$\hbar$\egroup. Any measurement of the total angular momentum gives the somewhat curious result

\begin{displaymath}\bgroup\color{black} \vert L\vert=\sqrt{\ell(\ell+1)}\hbar \egroup\end{displaymath}

where \bgroup\color{black}$\ell$\egroup is an integer.

Note that we can easily write the components of angular momentum in terms of the raising and lowering operators.

L_x&=&{1\over 2}(L_+ +L_-) \\
L_y&=&{1\over 2i}(L_+ -L_-) \\

We will also find the following equations useful (and easy to compute).

\begin{eqnarray*}[L_+,L_-]&=&i[L_y,L_x]-i[L_x,L_y]=\hbar(L_z+L_z)=2\hbar L_z \\
L^2&=&L_+L_-+L_z^2-\hbar L_z .\\

* Example: What is the expectation value of $L_z$ in the state $\psi(\vec{r})=R(r)(\sqrt{2\over 3}Y_{11}(\theta,\phi)-i\sqrt{1\over 3}Y_{1-1}(\theta,\phi))$?*

* Example: What is the expectation value of $L_x$ in the state $\psi(\vec{r})=R(r)(\sqrt{2\over 3}Y_{11}(\theta,\phi)-\sqrt{1\over 3}Y_{10}(\theta,\phi))$?*

Jim Branson 2013-04-22