As usual let $|l,m\rangle$ denote an eigenstate of $\vec{L}^2$ and $L_z$. I know that
\begin{align}
\vec{L}^2 |l,m\rangle &= \hbar^2 l(l+1) |l,m\rangle, \\
L_z |l,m\rangle &= \hbar m |l,m\rangle.
\end{align}
Consider $|1,1\rangle$.
Then, I can write $|1,1\rangle$ in terms of $|1,m_x\rangle$ with $m_x = \pm 1,0$, where $|1,m_x\rangle$ denotes an eigenbasis of $L_x$, as
\begin{align}
|1,1\rangle_z = c_1\cdot |1,-1\rangle_x + c_2 \cdot |1,0\rangle_x + c_3 \cdot |1,1\rangle_x
\end{align}
with $|c_1|^2 + |c_2|^2 + |c_3|^2 = 1$.
I want to compute the coefficients $c_1, c_2, c_3$ working with
\begin{align}
\Delta^2L_x := \langle (L_x – \langle L_x \rangle)^2\rangle = \frac{1}{2} \cdot \left( \langle L^2\rangle – \langle L_z^2\rangle\right) = \frac{\hbar^2}{2}\cdot\left(l(l+1) – m^2\right)
\end{align}
as in this case (for $|1,1\rangle$) we have $\Delta^2L_x = \frac{\hbar^2}{2}$. However, I don't understand how this can be done. Can someone please explain it?
Best Answer
Your requested hint: $$ L_x (c_1 |1,-1\rangle_x + c_2 |1,0\rangle_x + c_3 |1,1\rangle_x)=\hbar(-c_1 |1,-1\rangle_x + c_3 |1,1\rangle_x). $$ Can you take it from here? It is straightforward linear algebra.