Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say we start with an entangled electron--positron pair and we separate them. I want to take the entangled electron and bind it to a proton or the outer shell of an atom.

Is it possible to do this while having the entangled electron still remain entangled with the positron? Would the electromagnetic interaction with the proton "make a measurement" of the entangled electron's spin because of its magnetic field?

Assuming my above example is possible, say we then measure the spin of the positron that's far away. Whatever it may be, the entangled electron bound to the atom will now assume the opposite spin state. The total angular momentum of the atom will now be at a different value than what it was before, and therefore something must happen to conserve this, emit a photon?

Again, if my above example has a measurable effect whether it emits a photon or changes in some way, then this suggests to me that if we have an ensemble of these special atoms, we will be able to send information via quantum entanglement. By having these special atoms in one station and the entangled positrons in another station, we can measure or not measure, 1 or 0, the spin of the entangled positron and thus produce a measurable effect at the other station via the special atoms.

share|cite|improve this question
up vote 4 down vote accepted

You say:

I want to take the entangled electron and bound it to a proton or the outer shell of an atom.

The act of binding the electron with a proton, or an outer shell of an atom, means that some energy in the form of electromagnetic radiation will be given up by the electron in order to fall in an available energy level. A photon takes away spin 1 and any mathematical relationship that your electron's spin had with the world outside the atom world is lost.

So the answer is no.

share|cite|improve this answer

I'm not going to answer your electron<->atom+photon interaction question because my atomic physics is veeery rusty..

But regardless, suppose it works like you write, what you are proposing is simply a spin-measurement device. By measuring the positron in the "other end", you collapse the wavefunction consisting of the positron and electron+atom+photon system. Likewise if your atom ejects a photon which you detect.

To answer your first question, the electron would involve the atom in the spin superposition at first, but if it includes the emission and non-emission of a photon as well, this is where the fun ends since the photon emission and detection is a "collapsing" act.

So you are not better off than simply measuring the electron spin the same way you measure the positron spin, with equally zero chance of information transmission (as have doubtlessly been discussed in many other threads here).

share|cite|improve this answer

I'm assuming you mean spin entanglement (for entanglement including location the answer would clearly be "no" because to bind it to the proton or atom you would have to localize it there).

As soon as the energy depends on the spin of the electron (be it through spin-orbit interaction in a many-electron atom, or through interaction with the nuclear spin), the energy of the photon emitted on binding depends on the spin state of the electron, that is, it contains information about that state, and therefore the electron and positron are no longer entangled (the system consisting of electron, rest of the atom, positron and emitted photon is still entangled, though).

Note that if the energy doesn't depend on the electron's state, the entanglement may be conserved (at least partially) because the spin of the photon may be taken from the orbit angular momentum of the electron. Only processes which involve an electron spin flip will reduce the entanglement in that case.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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