HUP for spin seems violated: 0 x sy > sz/2...?

[JBlue]

TL;DR Summary

After measuring Sx, we know its state exactly. This seems to violate Heisenberg's Uncertainty Principle (HUP).

HUP for spins reads
$$\langle\sigma_z^2\rangle\langle\sigma_x^2\rangle \ge \frac{1}{4}|\langle\sigma_y\rangle|^2$$
Right after measuring $\sigma_z$, we know it exactly, and so $\langle\sigma_z^2\rangle=0$.
However, HUP then implies that $\langle\sigma_y^2\rangle=\infty$

Even if we say that the uncertainty isn't $0$ but some small $\epsilon$, we still get a $\langle\sigma_y^2\rangle$ that is very large, though it shouldn't be greater than $\hbar$.

What am I missing?

 

[Morbert]

Hmm, considering a state $|z^+\rangle$, I get the following quantities

$\sigma_x=1$
$\sigma_z=0$
$\sigma_x\sigma_z = 0$
$\frac{1}{2}|\langle\left[S_x,S_z\right]\rangle| = 0$
$\therefore \sigma_x\sigma_z \geq \frac{1}{2}|\langle\left[S_x,S_z\right]\rangle| = 0$

[edit] - So this is the the Robertson uncertainty relation. I also tested it for the schroedinger uncertainty relation and I also get $0\geq0$

 

[PeterDonis]

Right after measuring $\sigma_z$, we know it exactly, and so $\langle\sigma_z^2\rangle=0$.

No, it isn't. See here:

https://phys.libretexts.org/Bookshe...inty/11.3.2:_Uncertainty_in_Quantum_Mechanics

Discrete observables like spin are counterintuitive in this respect.

 

[vanhees71]

TL;DR Summary: After measuring Sx, we know its state exactly. This seems to violate Heisenberg's Uncertainty Principle (HUP).

HUP for spins reads
$$\langle\sigma_z^2\rangle\langle\sigma_x^2\rangle \ge \frac{1}{4}|\langle\sigma_y\rangle|^2$$
Right after measuring $\sigma_z$, we know it exactly, and so $\langle\sigma_z^2\rangle=0$.
However, HUP then implies that $\langle\sigma_y^2\rangle=\infty$

Even if we say that the uncertainty isn't $0$ but some small $\epsilon$, we still get a $\langle\sigma_y^2\rangle$ that is very large, though it shouldn't be greater than $\hbar$.

What am I missing?

Let's see. You prepared the system in an eigenstate of $s_x$, say in the eigenstate with $\sigma_x=\hbar/2$. In the usual $\hat{s}_z$ eigenbasis, $|\pm 1/2 \rangle$, it's
$$|\sigma_x=\hbar/2 \rangle=\frac{1}{\sqrt{2}}(|1/2 \rangle +|-1/2 \rangle).$$
This gives
$$\langle \sigma_z \rangle=\langle \sigma_x =1/2|\hat{s}_z|\sigma_x=1/2 \rangle=0,$$
and
$$\langle \sigma_z^2 \rangle = \langle \sigma_x =1/2|\hat{s}_z^2|\sigma_x=1/2 \rangle=\hbar^2/4,$$
i.e.,
$$\Delta \sigma_z=\hbar/2.$$
In the same way you also get
$$\Delta \sigma_y=\hbar/2.$$
The HUP states that
$$\delta \sigma_y \Delta \sigma_z \geq \frac{\hbar}{2} |\langle (-\mathrm{i}) [\hat{s}_y,\hat{s}_z] \rangle|.$$
Now $[\hat{s}_y,\hat{s}_z]=\mathrm{i} \hat{s}_x$. The expectation value on the right-hand side of our HUP thus is $\hbar/2$ and thus the right-hand side gives $\hbar^2/4$, i.e., the HUP is valid with the equality sign.

If you want to apply the HUP to $s_x$ and $s_z$ you get $0$ on the left-hand side of the HUP, because you prepared an eigenstate of $s_x$ and thus $\Delta s_z=0$. Then the HUP is of course always fulfilled. Of course it's not $\langle s_z^2 \rangle$ you have to use on the lefthand side of the HUP but the standard deviation, $\Delta s_z$, which is defined as
$$\Delta s_z^2=\langle s_z^2 \rangle-\langle s_z \rangle^2.$$

 

[JBlue]

Thanks, vanhees71, for walking through the example.
I had a silly confusion that is now perfectly cleared up!