# Uncertainty relation for $S_z$ eigenstates [closed]

I am content with the method of finding the uncertainty relation for $L_z$ eigenstates in a spin-1/2 system where $|\uparrow\rangle=|m=1/2\rangle$ and $|\downarrow\rangle=|m=-1/2\rangle$. I have used the fact that $[L_x,L_y]=i\hbar{L_z}$ and $L_z|\psi\rangle=m\hbar|\psi\rangle$

$$\Delta{L_x}\Delta{L_y}\geqslant \frac{1}{2}|\langle{\psi}|[L_x,L_y]|\psi\rangle|\qquad \hbox{with}\qquad \textstyle\frac{1}{2}|\langle{\psi}|[L_x,L_y]|\psi\rangle|=\frac{\hbar}{2}|\langle\psi|L_z|\psi\rangle|\, .$$

For $|\psi\rangle=|nlm\rangle$ we have \begin{align}\frac{\hbar}{2}|\langle\psi|L_z|\psi\rangle|&=\frac{m\hbar^2}{2}\langle{nlm}|nlm\rangle \tag{1}\, ,\\ \Delta{L_x}\Delta{L_y}&\geqslant\frac{m\hbar^2}{2}\langle{nlm}|nlm\rangle=\frac{m\hbar^2}{2} \tag{2} \end{align} and clearly for m=1/2 we have $\Delta{L_x}\Delta{L_y}\geqslant\frac{\hbar^2}{4}$ and for m=-1/2 we have $\Delta{L_x}\Delta{L_y}\geqslant-\frac{\hbar^2}{4}$

I understand how this works however for unknown reasons I am struggling to do the same for $\Delta{S_x}\Delta{S_y}$ where $i\hbar{S_z}=[S_x,S_y]$

some guidance would be helpful, is the process the same?

## closed as off-topic by Emilio Pisanty, Buzz, Kyle Kanos, Martin, FGSUZJan 9 at 0:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, Martin, FGSUZ
If this question can be reworded to fit the rules in the help center, please edit the question.

• The commutation relations for spin and angular momentum are exactly the same. Here $i\hbar S_z=[S_x, S_y]$. Also note that "$L_z=m\hbar$" is something of an abuse of notation. – Adomas Baliuka Apr 14 '17 at 11:58
• sorry should it be $L_z=m_l\hbar$ ? – Sam Apr 14 '17 at 12:03
• No, $L_z$ is an operator and operators are not equal to their eigenvalues (which are numbers). You could instead write $L_z|m\rangle=\hbar m|m\rangle$. – Adomas Baliuka Apr 14 '17 at 12:06
• oh I see of course I will fix that now – Sam Apr 14 '17 at 12:07
• does $S_z|\psi\rangle=m\hbar|\psi\rangle$ ? – Sam Apr 14 '17 at 12:09

1. You need to make sure you keep the absolute value in $\frac{1}{2}\vert\langle \psi\vert [A,B]\vert\psi\rangle\vert$. In the specific case of your question the right hand sides of (1) and (2) should respectively be \begin{align} &\vert m\vert\frac{\hbar^2}{2}\langle n\ell m\vert n\ell m\rangle\, ,\qquad \hbox{and}\qquad \vert m\vert \frac{\hbar^2}{2}\, . \end{align} so the product $\Delta L_x\Delta L_y$, which is the product of two positive numbers, remains $\ge +\frac{\hbar^2}{4}$ even for the $m=-1/2$ state, not $-\frac{\hbar^2}{4}$ as you have it.
2. For kets denoted by $\vert n\ell m\rangle$, $\ell$ is often an integer so only integer values of $m$ can occur. Thus a more appropriate version of your $\langle n\ell m\vert n\ell m\rangle$ would be $\langle\uparrow\vert\uparrow\rangle$ for the $m=1/2$ state, and $\langle\downarrow\vert\downarrow\rangle$ for the $m=-1/2$ state. In both cases, $\Delta L_x\Delta L_y\ge \frac{\vert m\vert\hbar^2}{2}$.
3. The derivation you have provided for $L_x,L_y,L_z$ depends only on the commutations relations and so it holds for the spin operators $S_x,S_y$ and $S_z$ as they have the same commutation relations as the angular momentum operators $L_x,L_y,L_z$.