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!