Why does $\Delta L_{x}=\Delta L_{z}=\frac{\hbar}{2}$ and $\Delta L_{y}=\frac{\hbar}{\sqrt{2}}$ in this experiment?

Question

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.

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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

$\hat{L}_{\pm}Y_{l}^{m}=\hbar \sqrt{l(l+1)-m(m\pm 1)}Y_{l}^{m\pm 1}$

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Solution:

From simple geometry, we find that:

$\Delta L_{z'}=0$
$\Delta L_{x}=\Delta L_{z}=\frac{\hbar}{2}$
$\Delta L_{y}=\frac{\hbar}{\sqrt{2}}$

(...)

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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 !

Answer 1

Review the vector model of angular momentum. Absorbing $\hbar$ into L, one has now non-dimensionalized.

On the z' axis, the 3 states with eigenvalues -1,0,1 for $\hat L_{z'}$, respectively, are the two back -to back cones and a disc in the middle. You are just projecting the +1 eigenstate cone in your figures. Since $\ell =1$, the length of the side of the cone is $| {\bf L}|=\sqrt{1(1+1)}=\sqrt 2$, so both the height and the radius of the base of the cone are =1.

Project this cone on the three axes, x,y,z.

The projection on both the x and z axes yields $$ \langle L_x \rangle =\langle L_z \rangle = \Delta L_x = \Delta L_z =\frac{1}{\sqrt 2} ~, $$ from trigonometry. (Had you looked at the middle, 0 eigenvalue state, a collapsed cone, so a disc of radius $\sqrt 2$, you'd have $\langle L_x \rangle =\langle L_z \rangle=0$ and $\Delta L_x = \Delta L_z=1$.)

The projection on the y axis has to be $\pm$ symmetric, so the center of the cone base projects to the origin of y. $$ \langle L_y \rangle = 0, \qquad \Delta L_y =1. $$ For the 0 eigenvalue state, the y-axis would go through it, so $\Delta L_y=\sqrt 2$.

They all satisfy the Robertson inequalities of the UP.

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Full disclosure: a footnote from hell. _{I am not a physical chemist, so I may be missing fine print in the lore of the vector model. In particular, I took your picture in good faith, accepting its radius of your cone base as some type of $\Delta L_i$. Your official proffered answers, then, would agree with these here if only their $\Delta L_i$ were $1/\sqrt 2$ of mine, here: the ratios coincide already. There may be an unstated such normalization in the uncertainty in your picture. Why am I wondering about that? Because if one does the brute QM calculation, one actually finds that for the top eigenstate of $\hat L_{z'}$ with eigenvalue 1, that is the state $|\psi\rangle= (|0\rangle +|1\rangle/(2-\sqrt{2}) + |-1\rangle /(2+\sqrt{2}) ~ )/2$, where $|m\rangle$ denotes the eigenstates of $\hat L_z$, we have: $\langle L_z\rangle=1/\sqrt 2$ but also $\langle L_z^2\rangle-\langle L_z\rangle^2=1/4$, so $\Delta L_z= 1/2$, your questioner's value, not my geometrical one. Moreover, $\langle L_y\rangle=0$ and $\langle L_y^2\rangle=1/2=(\Delta L_y)^2$; surely this cannot be a coincidence! It is therefore possible that the geometrical vector picture, as labelled, normalizes the error slightly differently than the conventional QM picture--people do play games with variances and errors this way.}

Answer 2

Aside of that, there is another assignment for this problem. I would like to reword the solution of this assignment here in my own words because maybe it can help someone else to understand. Also, if I ever forget how they got those results, I can simply look back at this answer. I'm sure it can be of use to the site and that other users can profit of it.

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Problem:

Remember that

$\hat{L}_{z'}=\frac{1}{\sqrt{2}}(\hat{L}_{x}+\hat{L}_{z})= \frac{1}{\sqrt{2}}(\frac{\hat{L}_{+}+\hat{L}_{-}}{2}+\hat{L}_{z})$

and that the action of the $L_{\pm}$ operators on an $L_z$ eigenstate is

$\hat{L}_{\pm}Y_{l}^{m}=\hbar \sqrt{l(l+1)-m(m\pm 1)}Y_{l}^{m\pm 1}$

Suppose we measured a value of $+ \hbar$ for the angular momentum in this basis (eigenbasis to $L_{z'}$ ) . Express the state of the system in the $L_z$ eigenbasis.

Hint: You need to solve the eigenvalue equation

$\hat{L}_{z'}\psi ' = \hat{L}_{z'}(a Y_1^{-1}+bY_1^0 + cY_1^1)=\hbar \psi '$

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My explanation of the above equation in my own terms:

1)The above equation means that if we apply $\hat{L}_{z'}$ on $\psi '$, we get the same $\psi '$ back multiplied by $\hbar$. That's why we talk about the $\hat{L}_{z'}$ eigenbasis. We get the same function back, multiplied by some number, here $\hbar$. In this context, $\hbar$ is the eigenvalue of the eigenfunction $\psi '$ in the eigenbasis $\hat{L}_{z'}$ (or $\hat{L}_{z}$).

2) The atom is in the $2p$ state, thus the azimuthal quantum number $l=1$, and consequently the magnetic quantum number $m=-l, ... ,l = -1, 0, 1$. Also, in spherical coordinates $(r, \theta, \phi)$, $\hat{L}_{z} $ and $ \hat{L}_{z'}$ only depend on $\phi$, i.e. only affect the angular part $Y_l^m$ of the wave-function $\psi '$. That implies that we can write $\hat{L}_{z'} \psi ' = \hat{L}_{z'}(aY_1^{-1} + bY_1^0 + cY_1^1)$, where $a, b, c$ are coefficients.

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Reworded solution:

Writing the equation in the hint with $L_{z'}$ explicitly, we find (using the fact that $L_{\pm}Y_1^m= \hbar \sqrt{2} Y_1^{m\pm 1}$ or zero, and $\hat{L}_z Y_1^m=\hbar m Y_1^m$)

$$\hat{L}_{z'} \psi ' = \frac{1}{\sqrt{2}}(\frac{\hat{L}_{+}+\hat{L}_{-}}{2}+\hat{L}_{z}) (a Y_1^{-1}+bY_1^0 + cY_1^1)=$$ $$ \frac{\hbar}{\sqrt{2}}(\frac{1}{\sqrt{2}}(aY_1^0 + bY_1^1 + 0) + (0 + bY_1^{-1} + cY_1^0)) + ( (-1)a Y_1^{-1} + 0b Y_1^{0} + 1c Y_1^1)) =$$ $$\frac{\hbar}{\sqrt{2}}(Y_1^{-1}(\frac{b}{\sqrt{2}} - a) + Y_1^0 (\frac{a}{\sqrt{2}} + \frac{c}{\sqrt{2}}) + Y_1^1(\frac{b}{\sqrt{2}} + c)) \stackrel{!}{=}$$ $$\hbar(aY_1^{-1} + bY_1^0 + cY_1^1)$$

Setting all coefficients equal to each other, we get:
$\frac{1}{\sqrt{2}}(\frac{b}{\sqrt{2}}-a) = a$
$\frac{1}{\sqrt{2}}(\frac{a}{\sqrt{2}}+\frac{c}{\sqrt{2}}) = b$
$\frac{1}{\sqrt{2}}(\frac{b}{\sqrt{2}}+c) = c$
and the normalization condition $\mid {a} \mid^2+\mid b\mid^2+\mid c \mid^2 = 1$

Thus
$b= (2+\sqrt{2})a $
$c= 2b-a=(3+2\sqrt{2})a$

Plugging those two equations into $\mid {a} \mid^2+\mid b\mid^2+\mid c \mid^2 = 1$, we obtain $a=0.1465$

This gives $\psi ' = 0.1465 Y_1^{-1} + 0.5Y_1^{0} + 0.854Y_1^1$