Exploring Quantum Physics - Week 6 Bonus Question

Question:

Prove that $-l \le m \le l$ for $|lm\rangle$ is both an eigenvector of $L^2$ and $L_3$.

Solution:

Thanks Gang Xu for the ladder operator hint.

$\begin{eqnarray*} & & L_3 L_\pm|lm\rangle \\ &=& ( L_\pm L_3 - L_\pm L_3 + L_3 L_\pm)|lm\rangle \\ &=& ( L_\pm L_3 + L_3 L_\pm - L_\pm L_3)|lm\rangle \\ &=& ( L_\pm L_3 + [L_3, L_\pm])|lm\rangle \\ &=& ( L_\pm L_3 + \pm\hbar L_\pm)|lm\rangle \\ &=& ( L_\pm \hbar m + \pm\hbar L_\pm)|lm\rangle \\ &=& \hbar (m \pm 1) L_\pm |lm\rangle \\ \end{eqnarray*}$

To proceed, we need to know one more identity

$\begin{eqnarray*} & & L_- L_+ \\ &=& (L_1 - i L_2)(L_1 + i L_2) \\ &=& (L_1^2 - i L_2 L_1 + i L_1 L_2 + L_2^2) \\ &=& (L_1^2 + L_2^2) + i L_1 L_2 - i L_2 L_1 \\ &=& (L_1^2 + L_2^2) + i [L_1, L_2] \\ &=& (L_1^2 + L_2^2) + i (i\hbar L_3) \\ &=& (L_1^2 + L_2^2) - \hbar L_3 \\ &=& L^2 - L_3^2 - \hbar L_3 \\ \end{eqnarray*}$

So we can combine the two identities into a single one $L_\pm L_\mp = L^2 - L_3^2 \pm \hbar L_3$. Now we can efficiently do both at the same time

$\begin{eqnarray*} & & L^2 L_\pm | lm\rangle \\ &=& (L_\pm L_\mp + L_3^2 \mp \hbar L_3) L_\pm | lm\rangle \\ &=& (L_\pm L_\mp L_\pm + L_3^2 L_\pm \mp \hbar L_3 L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + L_3^2 L_\pm \mp \hbar L_3 L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + L_3^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + L_3 \hbar (m \pm 1) L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + \hbar (m \pm 1) L_3 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3 \hbar m \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - \hbar m L_3 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - \hbar^2 m^2 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (\hbar^2 l (l+1) - \hbar^2 m^2 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& ((\hbar^2 l (l+1) - \hbar^2 m^2 \mp \hbar^2 m) L_\pm + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (\hbar^2 l (l+1) - \hbar^2 m^2 \mp \hbar^2 m + \hbar^2 (m \pm 1)^2 \mp \hbar^2 (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1) - m^2 \mp m + (m \pm 1)^2 \mp (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1) - m^2 \mp m + m^2 \pm 2m + 1 \mp (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1) \mp m \pm 2m + 1 \mp (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1)) L_\pm | lm\rangle \\ \end{eqnarray*}$

The last equality comes from just plus or minus sign case study - they will all cancel in either case. In retrospect - this result is not surprising, as $L_2$ is supposed to commute with $L_1$ and $L_2$. That is so far I can go.