2
$\begingroup$

If we consider the spins of two, far apart, entangled electrons, what will happen if we make a measurement on both spins at exactly the same time (let's assume time is not discrete)? I see it as a problem because if you measure one spin the other one pops instantaneously in the opposite spin state, depending on what the observed spin of the other electron is. But that's the same the other way round. Isn't the system in some state of not to "know" (I don't know how to phrase it differently) which spin pair will emerge? Is this a paradox? Do the measurements adapt to each other?

I can't see a problem if you consider the wavefunction as a pilot wave, in which case the particles have well-defined positions, energies, and spins (which are always opposite and have already before the measurement well defined opposite values). By the way, I see that the uncertainty principle has nothing to do with this (as I presumed in the earlier formulation of this question). The locations of the two particles are not spooky connected.

$\endgroup$

closed as unclear what you're asking by ZeroTheHero, Yashas, Jon Custer, sammy gerbil, Michael Seifert Jun 9 '17 at 14:01

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ In principle, it could be done. But what is the exciting part about that? $\endgroup$ – ABCD Jun 7 '17 at 10:33
  • $\begingroup$ I mean, which of the measurements is responsible for the change in the total wave function? If you measure the wave function of the spins at the same time, will that have the same effect as doing one spin measurement on one electron? The moment you make a measurement on one spin will decide the final combination of the two spins, but if you make a measurement on both spins at the same time, won't this have an effect on the measurement of the other spin, and vice-versa? You can't say you make a measurement of a combination of spins the electrons are already in. Isn't there some kind of "loop"? $\endgroup$ – descheleschilder Jun 7 '17 at 11:50
  • $\begingroup$ You also misunderstand energy-time uncertainty. The $\Delta t$ in the energy-time uncertainty inequality refers to the time it takes for an individual energy-transferring process to take place, not the time between two measurements. $\endgroup$ – probably_someone Jun 7 '17 at 15:30
  • $\begingroup$ Bipartite entanglement is the statement that the whole is in a definite state whereas a part is not. So, in principle, you cannot assign a state vector(or a rep in pos.space which one calls the wavefunction) to either of these. Also, entanglement has nothing to do with the measurement process in the way you have put in. Please read Nielsen and Chuang's book for more details on the basics of entanglement. $\endgroup$ – ABCD Jun 7 '17 at 15:48
  • $\begingroup$ This question as worded is doesn't make much sense: 1. measuring spins has nothing to do with time-energy uncertainty, 2. time-energy uncertainty at best requires careful definition of what is $\Delta t$ (see physics.stackexchange.com/q/53802/36194), something not done (or doable?) in the question. $\endgroup$ – ZeroTheHero Jun 7 '17 at 15:57
3
$\begingroup$

Nothing special happens. What you end up is making two measurements, which confirm that one particle spins one way and the other particle spins the other, just like you expected to see.

There's no paradox here unless you presume a naive version of causality. If you are worried about whether Measurement 1 caused the results of Measurement 2, or whether Measurement 2 caused the results of measurement 1, you could end up in a bit of a bind. However, what you find is that no paradox occurs because the observable results of the two measurements is unaffected by this existential crisis. If necessary, you can break this issue down by saying "The simultaneous measurements of Measurement 1 and Measurement 2 caused the particle to be in a particular state."

$\endgroup$
1
$\begingroup$

First problem you have with that is, what do you mean by exactly the same time? Both observers in the same rest frame? So lets exclude different rest frames.

There is still a problem with pilot wave theory. If one of the observers changes which direction he wants to measure (entanglement is only interesting if you allow for different non compatible measurements), then something has to change immediately at the other observers place (if you believe in realism, which seems you do). This is because Bell-inequalities (and CHSH) show that local realism is not compatible with experimentally confirmed QM. So how does a superluminal change of the pilot wave help there in any way?

$\endgroup$
  • $\begingroup$ In pilot wave theory, the two electrons have at any moment in time a correlation of spins (up-down), which is already present before you make the measurements. So if you make the measurements at exactly the same time (both electrons in the same inertial frame) you always get as the outcome of a spin measurement on one electron the spin it has at that moment, while the measurement at exactly the same time of the spin of the other electron will be always the opposite. $\endgroup$ – descheleschilder Jun 9 '17 at 17:29
  • $\begingroup$ The difference lies in the fact that in the pilot wave theory the state of the particles is already there before the measurement, which is not the case in the probability (Copenhagen) interpretation. The Bell inequalities do not exclude real non-local hidden variables, which are of importance in this case. $\endgroup$ – descheleschilder Jun 9 '17 at 17:29
  • $\begingroup$ @descheleschilder that you always get the opposite result is not true. Only for certain kinds of measurement and not generally for the ones needed to show chsh. $\endgroup$ – lalala Jun 9 '17 at 19:05
  • $\begingroup$ Well, I'm only talking about these certain kinds of measurements, and not about the ones needed to show the chsh inequality. $\endgroup$ – descheleschilder Jun 10 '17 at 6:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.