What does an uncertainity of 0 mean?

[CrosisBH]

TL;DR

$\Delta S_x = 0$ for all quantum state vectors. What exactly does it mean for spin in the x direction to have 0 uncertainty?

I'm just starting my undergraduate Quantum Mechanics course. I had a homework problem to show that \Delta S_x = \sqrt{\langle S_x^2 \rangle - \langle S_x \rangle ^2} = 0, S_x being the spin in the x direction. I managed to solve it, but the physical interpretation is confusing me. If I remember my stats course correctly, the uncertainty is pretty much the standard deviation mathematically, and if standard deviation is 0, that means all the data points are the same, which would seem to suggest we have certainty of S_x.

This feels wrong to me. The math seems to suggest that to me, but I also happen to know that S_x can be \pm \frac{\hbar}{2} each with probability \frac{1}{2}, so I'm guessing my idea of uncertainty is wrong, so that's why I'm asking here. What does a state having 0 uncertainty mean in the context of Quantum Mechanics?

Thank you!

 

[anuttarasammyak]

Are you sure $\triangle S_x = 0$ for all quantum state vectors ? Your counter example says $\triangle S_x =\frac{\hbar}{2}> 0$.

 

[PeroK]

Summary:: \Delta S_x = 0 for all quantum state vectors. What exactly does it mean for spin in the x direction to have 0 uncertainty?

I'm just starting my undergraduate Quantum Mechanics course. I had a homework problem to show that \Delta S_x = \sqrt{\langle S_x^2 \rangle - \langle S_x \rangle ^2} = 0, S_x being the spin in the x direction. I managed to solve it, but the physical interpretation is confusing me. If I remember my stats course correctly, the uncertainty is pretty much the standard deviation mathematically, and if standard deviation is 0, that means all the data points are the same, which would seem to suggest we have certainty of S_x.

This feels wrong to me. The math seems to suggest that to me, but I also happen to know that S_x can be \pm \frac{\hbar}{2} each with probability \frac{1}{2}, so I'm guessing my idea of uncertainty is wrong, so that's why I'm asking here. What does a state having 0 uncertainty mean in the context of Quantum Mechanics?

Thank you!

You are seriously misunderstanding something here. It's difficult to help you without seeing what you think you've calculated.

There are certain states where a particular observable will only have a single value if measured (these are called eigenstates of that observable). There are, for example, eigenstates of $S_x$: there are two of them, which can be labelled $|+x \rangle$ and $|-x \rangle$. In the $|+x \rangle$ state a measurement of $S_x$ will yield $+\frac {\hbar} 2$ with certainty.

For an eigenstate of an observable that observable has zero uncertainty, which statistically means the measurements (in that particular state) have zero variance.

When you say "all quantum state vectors" have $\Delta S_x = 0$ you have clearly misunderstood something fundamental.

 

[PeroK]

This feels wrong to me. The math seems to suggest that to me, but I also happen to know that S_x can be \pm \frac{\hbar}{2} each with probability \frac{1}{2},

PS S_x can be \pm \frac{\hbar}{2} but the probability depends on the state. In general, any state can be expressed (in the x-basis) as: $$|\chi \rangle = \alpha |+x \rangle + \beta |-x \rangle$$
Where the probability of measuring $\frac {\hbar}{2}$ is $|\alpha|^2$ and the probability of measuring $-\frac {\hbar}{2}$ is $|\beta|^2$. Hence the probability of measuring $\frac {\hbar}{2}$ can be anything from $0$ to $1$, depending on the state.

And, if the probability is either $0$ or $1$, then the state is an eigenstate of $S_x$, with eigenvalues $\pm \frac {\hbar}{2}$.

 

[atyy]

Summary:: \Delta S_x = 0 for all quantum state vectors.

That is not true. For some quantum states, when you measure $\hat{S}_x$, even though the state and the measurement are the same, you can get different results each time you do the measurement.

I'm just starting my undergraduate Quantum Mechanics course. I had a homework problem to show that \Delta S_x = \sqrt{\langle S_x^2 \rangle - \langle S_x \rangle ^2} = 0, S_x being the spin in the x direction. I managed to solve it, but the physical interpretation is confusing me. If I remember my stats course correctly, the uncertainty is pretty much the standard deviation mathematically, and if standard deviation is 0, that means all the data points are the same, which would seem to suggest we have certainty of S_x.

Yes, in this context the uncertainty does mean the standard deviation. And if you have certainty, then the standard deviation is 0. However, when one makes a measurement of $\hat{S}_{x}$, it is only for particular quantum states for which one obtains certain results. There are 2 eigenstates of $\hat{S}_{x}$, which we can denote $|+\rangle$ and $|-\rangle$. If the quantum state is either one of the eigenstates of $\hat{S}_{x}$, then measuring $\hat{S}_{x}$ will give a certain result.

This feels wrong to me. The math seems to suggest that to me, but I also happen to know that S_x can be \pm \frac{\hbar}{2} each with probability \frac{1}{2}, so I'm guessing my idea of uncertainty is wrong, so that's why I'm asking here. What does a state having 0 uncertainty mean in the context of Quantum Mechanics?

If the quantum state is a superposition of eigenstates $\frac{1}{\sqrt{2}}|+\rangle +\frac{1}{\sqrt{2}} |-\rangle$, then measuring $\hat{S}_x$ will give you \pm \frac{\hbar}{2} each with probability \frac{1}{2}, and the standard deviation or uncertainty will not be 0.