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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?

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  • $\begingroup$ 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. $\endgroup$ – Adomas Baliuka Apr 14 '17 at 11:58
  • $\begingroup$ sorry should it be $L_z=m_l\hbar$ ? $\endgroup$ – Sam Apr 14 '17 at 12:03
  • $\begingroup$ No, $L_z$ is an operator and operators are not equal to their eigenvalues (which are numbers). You could instead write $L_z|m\rangle=\hbar m|m\rangle$. $\endgroup$ – Adomas Baliuka Apr 14 '17 at 12:06
  • $\begingroup$ oh I see of course I will fix that now $\endgroup$ – Sam Apr 14 '17 at 12:07
  • $\begingroup$ does $S_z|\psi\rangle=m\hbar|\psi\rangle$ ? $\endgroup$ – Sam Apr 14 '17 at 12:09
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  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 $L_x,L_y,L_z$.
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