To separate the two spins of electron of a particular direction, we use Stern-Gerlach apparatuses to apply a non-uniform Magnetic field. Suppose we have two identical electrons and we know their spin adds to 0, and we know their initial spin-state (for concreteness, let's say they are in the singlet configuration, $\lvert\psi\rangle=\frac{1}{\sqrt2}(\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle)$.

Now if we send both electrons in the same direction and measure one of them, then immediately we know the spin of the other.

My questions are:

  1. If we send both electrons at 90 degrees to each other (assume the first electron is always sent in the direction of the known state), then there is no entanglement effect, since one component of spin does not affect another. Is this correct?

  2. If we send both electrons at an arbitrary angle $\theta$, then there is some entanglement effect. Is this correct?

  3. If the last statement is correct, how do we determine the correlation in terms of the angle and initial entangled state?

  4. If the initial state is the triplet state instead of single, what effect does that have on the system?

  • $\begingroup$ Presumably here you mean the relative angles of the Stern-Gerlach apparatus? At the moment it seems like you are talking about the directions in which the spins move in space. The position or velocity of the spins has no effect on the correlations. $\endgroup$ Commented May 18, 2015 at 14:12
  • $\begingroup$ Since your initial state is a scalar, the angle simply cannot affect anything. Yes, you've written it w.r.t to the z-axis single particle spin eigenstates, but to make them w.r.t the x-axis, all you have to is say "they are w.r.t to the x-axis". $\endgroup$
    – JEB
    Commented Apr 20, 2018 at 1:57

1 Answer 1


The entanglement does not depend on the velocity vectors of the particles. The SG apparatus will determine the spin state of the measured particle, not necessarily the whole system.

By performing a measurement on one particle in a known entangled state you would know what the state of the other particle is as well. You do not need to measure it.

  • $\begingroup$ You are right that the particle's velocity doesn't have anything to do with it, but if I measure each particle's spin from a different angle, it will. Maybe I'm not being clear enough with that. $\endgroup$ Commented May 22, 2015 at 18:30
  • $\begingroup$ Why would that matter? $\endgroup$
    – Yogi DMT
    Commented Nov 4, 2016 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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