The Operators $L_\pm$

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

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

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

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

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

The eigenvalue of $L_z$ is changed when we operate with $L_+$ or $L_-$.

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

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

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

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

These operators raise or lower the $z$ component of angular momentum by one unit of $\hbar$.

Since $L_\pm^\dag =L_\mp$, its easy to show that the following is greater than zero.

$$\begin{eqnarray*} \langle L_\pm Y_{\ell m}\vert L_\pm Y_{\ell m}\rangle\geq 0 \... ...\langle Y_{\ell m}\vert L_\mp L_\pm Y_{\ell m}\rangle\geq 0 \\ \end{eqnarray*}$$

Writing $L_+L_-$ in terms of our chosen operators,

$$\begin{eqnarray*} L_\mp L_\pm&=&(L_x\mp iL_y)(L_x\pm iL_y)=L_x^2+L_y^2\pm iL_xL... ..._x \\ &=&L_x^2+L_y^2\pm i[L_x,L_y]=L^2-L_z^2\mp\hbar L_z \\ \end{eqnarray*}$$

we can derive limits on the quantum numbers.

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

We know that the eigenvalue $\ell(\ell+1)\hbar^2$ is greater than zero. We can assume that

$$\bgroup\color{black} \ell\geq 0 \egroup$$

because negative values just repeat the same eigenvalues of $\ell(\ell+1)\hbar^2$.

The condition that $\ell(\ell+1)\geq m(m\pm 1)$ then becomes a limit on $m$.

$$\bgroup\color{black} -\ell\leq m\leq\ell \egroup$$

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

$$\begin{eqnarray*} L_- Y_{\ell (-\ell)}=0 \\ L_+ Y_{\ell \ell}=0 \\ \end{eqnarray*}$$

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

$$\bgroup\color{black} m=-\ell, -\ell+1,...,\ell-1,\ell \egroup$$

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

The eigenstates of $L^2$ and $L_z$ should be normalized

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

The raising and lowering operators acting on $Y_{\ell m}$ give

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

The coefficient $C_\pm(\ell,m)$ can be computed.

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

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

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