The Operators \bgroup\color{black}$L_\pm$\egroup

The next step is to figure out how the \bgroup\color{black}$L_\pm$\egroup operators change the eigenstate \bgroup\color{black}$Y_{\ell m}$\egroup. What eigenstates of \bgroup\color{black}$L^2$\egroup are generated when we operate with \bgroup\color{black}$L_+$\egroup or \bgroup\color{black}$L_-$\egroup?

\begin{displaymath}\bgroup\color{black} L^2(L_\pm Y_{\ell m})=L_\pm L^2 Y_{\ell m}=\ell(\ell+1)\hbar^2(L_\pm Y_{\ell m}) \egroup\end{displaymath}

Because \bgroup\color{black}$L^2$\egroup commutes with \bgroup\color{black}$L_\pm$\egroup, we see that we have the same \bgroup\color{black}$\ell(\ell+1)$\egroup after operation. This is also true for operations with \bgroup\color{black}$L_z$\egroup.

\begin{displaymath}\bgroup\color{black} L^2(L_z Y_{\ell m})=L_z L^2 Y_{\ell m}=\ell(\ell+1)\hbar^2(L_zY_{\ell m}) \egroup\end{displaymath}

The operators \bgroup\color{black}$L_+$\egroup, \bgroup\color{black}$L_-$\egroup and \bgroup\color{black}$L_Z$\egroup do not change \bgroup\color{black}$\ell$\egroup. That is, after we operate, the new state is still an eigenstate of \bgroup\color{black}$L^2$\egroup with the same eigenvalue, \bgroup\color{black}$\ell(\ell+1)$\egroup.

The eigenvalue of \bgroup\color{black}$L_z$\egroup is changed when we operate with \bgroup\color{black}$L_+$\egroup or \bgroup\color{black}$L_-$\egroup.

L_z(L_\pm Y_{\ell m})&=&L_\pm L_z Y_{\ell m}+[L_z,L_\pm]Y_{\e...\hbar L_\pm Y_{\ell m}
=(m\pm 1)\hbar (L_\pm Y_{\ell m}) \\

(This should remind you of the raising and lowering operators in the HO solution.)

From the above equation we can see that \bgroup\color{black}$(L_\pm Y_{\ell m})$\egroup is an eigenstate of \bgroup\color{black}$L_z$\egroup.

\begin{displaymath}\bgroup\color{black} L_\pm Y_{\ell m}=C_\pm(\ell,m)Y_{\ell (m\pm 1)} \egroup\end{displaymath}

These operators raise or lower the \bgroup\color{black}$z$\egroup component of angular momentum by one unit of \bgroup\color{black}$\hbar$\egroup.

Since \bgroup\color{black}$L_\pm^\dag =L_\mp$\egroup, its easy to show that the following is greater than zero.

\langle L_\pm Y_{\ell m}\vert L_\pm Y_{\ell m}\rangle\geq 0 \...
...\langle Y_{\ell m}\vert L_\mp L_\pm Y_{\ell m}\rangle\geq 0 \\

Writing \bgroup\color{black}$L_+L_-$\egroup in terms of our chosen operators,

L_\mp L_\pm&=&(L_x\mp iL_y)(L_x\pm iL_y)=L_x^2+L_y^2\pm iL_xL...
..._x \\
&=&L_x^2+L_y^2\pm i[L_x,L_y]=L^2-L_z^2\mp\hbar L_z \\

we can derive limits on the quantum numbers.

\langle Y_{\ell m}\vert(L^2-L_z^2\mp\hbar L_z) Y_{\ell m}\ran...
...l+1)-m^2\mp m)\hbar^2\geq 0 \\
\ell(\ell+1)\geq m(m\pm 1) \\

We know that the eigenvalue \bgroup\color{black}$\ell(\ell+1)\hbar^2$\egroup is greater than zero. We can assume that

\begin{displaymath}\bgroup\color{black} \ell\geq 0 \egroup\end{displaymath}

because negative values just repeat the same eigenvalues of \bgroup\color{black}$\ell(\ell+1)\hbar^2$\egroup.

The condition that \bgroup\color{black}$\ell(\ell+1)\geq m(m\pm 1)$\egroup then becomes a limit on \bgroup\color{black}$m$\egroup.

\begin{displaymath}\bgroup\color{black} -\ell\leq m\leq\ell \egroup\end{displaymath}

Now, \bgroup\color{black}$L_+$\egroup raises \bgroup\color{black}$m$\egroup by one and \bgroup\color{black}$L_-$\egroup lowers \bgroup\color{black}$m$\egroup by one, and does not change \bgroup\color{black}$\ell$\egroup. Since \bgroup\color{black}$m$\egroup is limited to be in the range \bgroup\color{black}$-\ell\leq m\leq\ell$\egroup, the raising and lowering must stop for \bgroup\color{black}$m=\pm\ell$\egroup,

L_- Y_{\ell (-\ell)}=0 \\
L_+ Y_{\ell \ell}=0 \\

The raising and lowering operators change \bgroup\color{black}$m$\egroup in integer steps, so, starting from \bgroup\color{black}$m=-\ell$\egroup, there will be states in integer steps up to \bgroup\color{black}$\ell$\egroup.

\begin{displaymath}\bgroup\color{black} m=-\ell, -\ell+1,...,\ell-1,\ell \egroup\end{displaymath}

Having the minimum at \bgroup\color{black}$-\ell$\egroup and the maximum at \bgroup\color{black}$+\ell$\egroup with integer steps only works if $\ell $ is an integer or a half-integer. There are \bgroup\color{black}$2\ell +1$\egroup states with the same \bgroup\color{black}$\ell$\egroup and different values of \bgroup\color{black}$m$\egroup. We now know what eigenstates are allowed.

The eigenstates of \bgroup\color{black}$L^2$\egroup and \bgroup\color{black}$L_z$\egroup should be normalized

\begin{displaymath}\bgroup\color{black} \langle Y_{\ell m}\vert Y_{\ell m}\rangle=1 .\egroup\end{displaymath}

The raising and lowering operators acting on \bgroup\color{black}$Y_{\ell m}$\egroup give

\begin{displaymath}\bgroup\color{black} L_\pm Y_{\ell m}=C_\pm(\ell,m)Y_{\ell (m\pm 1)} \egroup\end{displaymath}

The coefficient \bgroup\color{black}$C_\pm(\ell,m)$\egroup can be computed.

\langle L_\pm Y_{\ell m}\vert L_\pm Y_{\ell m}\rangle
...hbar^2 \\
C_\pm(l,m)&=&\hbar\sqrt{\ell(\ell+1)-m(m\pm 1)} \\

We now have the effect of the raising and lowering operators in terms of the normalized eigenstates.

\begin{displaymath}\bgroup\color{black} L_\pm Y_{\ell m}=\hbar\sqrt{\ell(\ell+1)-m(m\pm 1)}Y_{\ell(m\pm 1)}\egroup\end{displaymath}

Jim Branson 2013-04-22