- #1
etotheipi
Hey, sorry if this is a dumb question but I wondered if someone could clarify. If you have two Cartesian coordinate systems ##\mathcal{F}## and ##\mathcal{F}'## whose origins coincide except their ##x## axes point in opposite directions (i.e. ##\hat{x} = -\hat{x}'##), then the spin operators along their respective positive ##x## directions would seem to be related via. ##\hat{S}_x = - \hat{S}_{x'}##. AFAIK they're compatible because ##[\hat{S}_x, \hat{S}_{x'}] = 0##.
Does that mean that if you measure the observable corresponding to ##\hat{S}_x## to be a definite ##S_x##, then this automatically fixes ##S_{x'} = -S_x##? That's to say, does a single measurement cut it, or do you still need to actually measure the spin in each coordinate system individually?
Sorry if this doesn't make sense![Winking face with tongue :stuck_out_tongue_winking_eye: 😜](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61c.png)
Does that mean that if you measure the observable corresponding to ##\hat{S}_x## to be a definite ##S_x##, then this automatically fixes ##S_{x'} = -S_x##? That's to say, does a single measurement cut it, or do you still need to actually measure the spin in each coordinate system individually?
Sorry if this doesn't make sense
![Winking face with tongue :stuck_out_tongue_winking_eye: 😜](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61c.png)