- #1
CrosisBH
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- TL;DR Summary
- [itex] \Delta S_x = 0[/itex] 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 [itex] \Delta S_x = \sqrt{\langle S_x^2 \rangle - \langle S_x \rangle ^2} = 0 [/itex], [itex] S_x [/itex] 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 [itex] S_x [/itex].
This feels wrong to me. The math seems to suggest that to me, but I also happen to know that [itex] S_x [/itex] can be [itex] \pm \frac{\hbar}{2} [/itex] each with probability [itex] \frac{1}{2} [/itex], 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!
This feels wrong to me. The math seems to suggest that to me, but I also happen to know that [itex] S_x [/itex] can be [itex] \pm \frac{\hbar}{2} [/itex] each with probability [itex] \frac{1}{2} [/itex], 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!