Note: NO, this is not at all a homework question. I struggle to understand the solution our professor gave us to a problem, and that's why I am asking here. Also, given the nature of the problem, I truly believe this could be helpful to a broader audience.
Consider an experiment, where we can prepare a hydrogen atom in any state we want and then measure both magnitude $(L^2)$ and component along any axis $i$ (by e. g. rotating the detector) of the orbital angular momentum $(L_i)$. In an attempt to obtain information about both $L_z$ and $L_x$ for a hydrogen atom in the $2p$ state, we rotate our measurement an angle $\theta = 45°$ between the $z$ and the $x$-axis, so that we measure the eigenvalues of :
$$\hat{L}_{z'}=\frac{1}{\sqrt{2}}(\hat{L}_{x}+\hat{L}_{z})= \frac{1}{\sqrt{2}}(\frac{\hat{L}_{+}+\hat{L}_{-}}{2}+\hat{L}_{z})$$
as can be shown in the figure below:
From the figure, we can estimate the uncertainty as $m_l=\langle \hat{L}_z \rangle \pm \Delta L_z$.
Your task is to find $\Delta L_z, \Delta L_x, \Delta L_y \ $and $ \Delta L_z'$. And after that, verify that the Heisenberg Uncertainty relations
$\Delta_x \Delta_y \geq \frac{\hbar}{2} \langle \hat{L}_z \rangle \ $, $\Delta_z \Delta_x \geq \frac{\hbar}{2} \langle \hat{L}_y \rangle \ $, $\Delta_y \Delta_z \geq \frac{\hbar}{2} \langle \hat{L}_x \rangle$
for angular momentum hold.
Hint: The action of the $L_{\pm}$ operators on an $L_z$ eigenstate is
$\hat{L}_{\pm}Y_{l}^{m}=\hbar \sqrt{l(l+1)-m(m\pm 1)}Y_{l}^{m\pm 1}$
Solution:
From simple geometry, we find that:
$\Delta L_{z'}=0$
$\Delta L_{x}=\Delta L_{z}=\frac{\hbar}{2}$
$\Delta L_{y}=\frac{\hbar}{\sqrt{2}}$
(...)
My question is:
"from simple geometry", how do they find that $\Delta L_{x}=\Delta L_{z}=\frac{\hbar}{2}$ and $\Delta L_{y}=\frac{\hbar}{\sqrt{2}}$ ?
It's clear that if $\Delta L_{z'}=0$, the Uncertainty Principle applies to $\Delta L_{x}, \Delta L_{y}$ and $\Delta L_{z}$, and thus we cannot measure them with total accuracy because $\Delta L_{z'}$ already is. But how did they find out those values? How does $\ \cos{45°}=\sin{45°}=\frac{1}{\sqrt{2}}$ relate to that (The angle of $45°$ is obviously of use)?
Any help/ hint would be appreciated. Thanks for your help !