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:

enter image description here

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

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


From simple geometry, we find that:

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

enter image description here


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 !

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    $\begingroup$ Simple geometry is a terrible way of computing $\Delta L_y$ since the angular momentum vector does not have a well-defined position. It might work in some specific cases, and it may provide some intuition, but it should not be trusted for exact calculations. $\endgroup$ Nov 10, 2019 at 22:53
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    $\begingroup$ @CosmasZachos I've already seen that before. How should that help me ? $\endgroup$
    – user213816
    Nov 11, 2019 at 10:47
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    $\begingroup$ You are drawing the state with eigenvalue 1 of $L_{z'}$, so | L |=$\sqrt 2$, hence the height and the radius of your cone are both 1. I have absorbed the pestiferous $\hbar$ in the normalization of the L s. Their x and z projections are hence all $1/\sqrt 2$. The projections on the y axis, however, are 0 and 1, respectively, so $\langle L_y\rangle=0$ and $\Delta L_y=1$. This last value differs from the wrong entry in your question. They trivially satisfy the Robertson inequality. $\endgroup$ Nov 11, 2019 at 15:55
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    $\begingroup$ Of course. In this case |L| is $\sqrt2 $ times the $L_z'$ component, and you took that to be 1 here. $\endgroup$ Nov 11, 2019 at 16:08
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    $\begingroup$ NO, the cone is 3D, and the base projects to the entire [-1,1] interval on the y axis. That is why I wanted to ensure you are comfortable with the WP language. As far as the y-axis is concerned, you have the flat disc in-between the +1 and -1 cones. $\endgroup$ Nov 11, 2019 at 16:48

2 Answers 2


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.

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.

  • $\begingroup$ I read your answer since the vey moment you posted it, but I still do not understand some parts of it. Why would $\Delta L_x = \Delta L_y = \sqrt{2}$ for the $0$ eigenvalue state ? The cone, be it in the $-1,0$ or $1$ eigenvalue state, is after all still "along" the $z'$ axis, not the $z$ axis. So even the disc of radius $\sqrt{2}$ has the $z'$ axis that goes through its center. And second question: does "ys" stand for y symmetric ? $\endgroup$
    – user213816
    Nov 11, 2019 at 19:40
  • $\begingroup$ By the way, Thank You so much for the support you gave until now, I really appreciate it. More than 4 days ago, I wrote an email to the person in charge of the exercises, but for now I got no response at all. That's why I'm asking on this website. $\endgroup$
    – user213816
    Nov 11, 2019 at 19:45
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    $\begingroup$ First was a typo, y s was plural of y, but generic singular is as good, when not confusing. $\endgroup$ Nov 11, 2019 at 19:52
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    $\begingroup$ Yes, it would be $\sqrt 2$: the y-axis would be a diameter of the disc. But, mea culpa!, the other two variances would be 1, the projections of that disc. Corrected the answer. $\endgroup$ Nov 11, 2019 at 22:37
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    $\begingroup$ p means $\ell=1$. s is 0. $\endgroup$ Nov 12, 2019 at 11:05

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.


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

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

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.

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 $\mid {a} \mid^2+\mid b\mid^2+\mid c \mid^2 = 1$

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

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    $\begingroup$ Indeed, obviously $\psi ' =\langle \theta,\phi |\psi '\rangle $ of the other answer, which is of course how it was found. $\endgroup$ Nov 18, 2019 at 17:41
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    $\begingroup$ Exactly as in exactly : spherical harmonics are a redundant representation of the three states $|m\rangle$ for l =1, m =-1,0,1 in the $\theta,\phi$ representation, which is, however, redundant, as the angles are never used. The eigenstates are automatically normalized. This is my point about the brute force QM calculation in the footnote of my answer. I did not mean to detract from the pleasure of seeing the point. The problem first suggests the brute QM calculation, and then expedites the answer by geometry, albeit with the misbegotten normalization the teacher was careless in fixing. $\endgroup$ Nov 18, 2019 at 17:55

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