# Quantum entanglement thought experiment

One can "filter" a collection of electron pairs generated in an entangled (fixed-direction) spin state (e.g. singlet $\left|\uparrow \downarrow \right\rangle - \left|\downarrow \uparrow\right\rangle$) by discarding all pairs where the left particle has spin down. This can be accomplished by separating each pair, running the "left" particle through the magnet of a Stern-Gerlach apparatus that only has a "down" detector, and discarding its "right" partner when detected; the "up" fraction is allowed to continue, along with its partner. Up to this point, the left survivor hasn't actually been measured, so the wave function of the pair is still intact. [EDIT: This conclusion is wrong, see answers.]

The spin of the "right" particles should therefore still have expectation 0, right?

• Note that as soon as you perform the measurement on the left particle (by running it through the apparatus) the two particles are disentangled, regardless of the outcome. So it is not true to say that "the wave function of the pair is still intact". – Harry Johnston Aug 27 '15 at 23:16
• Makes sense.... – suissidle Aug 27 '15 at 23:26
• By not detecting the left particle with your spin down detector, you have effectively measured it as spin up. So in the Copenhagen picture, the right particle's state collapses to down as required by the entanglement. – adipy Aug 27 '15 at 23:34

## 2 Answers

Imagine you have a pair of coins. Whenever you throw them, each of them is fully random, but their outcomes are opposite.

Now imagine you throw the two coins. You look at the left coin. When it is head, you discard both coins and start again, when it is tail, you keep it. Since you have never looked at the right coin, it should still be completely random.

But this is not correct: You discard the right coin when it is tail -- while you did not look at it, you know it is tail, since the results of the two coins are opposite.

So the answer is no. You have postselected your state -- you have discarded the right electron dependent on the state of the left electron, which carries information about the right electron's state as well. Thus, you affect the state of the remaining electrons on the right.

• I thought the whole point of QM was that you cannot use "coin" logic. I'm trying to understand what's special about entanglement at the quantum level. I hear that before any spin measurement on either the "left" or "right" particle, the state is still a superposition, and each particle behaves randomly (spin-wise). But this is not substantive, as without any knowledge about either left or right, the spins really are random. – suissidle Aug 27 '15 at 23:23
• @suissidle As long as you measure the spin in a single fixed direction (e.g. up/down), it is coin logic, and it does indeed miss the point of QM. You might want to read up on Bell inequalities or the CHSH inequality. You should find references e.g. here on physics.se or on Wikipedia. – Norbert Schuch Aug 27 '15 at 23:25
• In that case, I don't understand what the measurable effect of the entangled superposition of states is in general, beyond the mere lack of knowledge on the part of the observer. I understand Bell's inequality and how it breaks in some special situations. – suissidle Aug 27 '15 at 23:37
• Entanglement is "stronger" than classical correlations. When you measure the two spins on different axes, the results are more correlated than is possible in a classical system, resulting in violations of Bell's inequality. – adipy Aug 27 '15 at 23:47
• @suissidle Everything interesting with entanglement happens when you have incompatible operators act on the two different particle. Otherwise it is quite boring. – Timaeus Aug 28 '15 at 1:16

No.

The particles have a joint spin state, $\left|\uparrow \downarrow \right\rangle - \left|\downarrow \uparrow\right\rangle.$

When you send the left particle through a Stern-Gerlach device then the device changes from $\left|0\right\rangle$ to $\left|1\right\rangle$ when the left particle is detected as down.

So $\left(\left|\uparrow \downarrow \right\rangle - \left|\downarrow \uparrow\right\rangle\right)\otimes \left|0\right\rangle$ evolves into

$\left|\uparrow \downarrow \right\rangle\otimes \left|0\right\rangle- \left|\downarrow \uparrow\right\rangle\otimes \left|1\right\rangle$

At this point you can people that see the detector having gone off can ignore their particles but their particles are already particles with a definite spin. And the people that don't see the detector go off also already have particles with a definite spin state.

Up to this point, the left survivor hasn't actually been measured, so the wave function of the pair is still intact.

The particles have spatial degrees of freedom and a joint spin state and the detectors have states. And they are all evolving. The Stern-Gerlach device isn't magic, it evolves the particles spin states and spatial degrees of freedom by subjecting them to an inhomogeneous magnetic field. Inhomogeneous magnetic fields change the spin state. Since they have a joint spin state, changing the spin state of one of them changes the spin state of both of them.

So sending one particle through a Stern-Gerlach changes the spin state of both. And your claim that you didn't even affect the particle you sent through is even more wrong.

The fact that you could detect one and not the other is because the spatial degrees of freedom of particle one changed. And the spin changed to so that the spatial degree of freedom became entangled with the spin. And then your detector evolved to have the detectors state become entangled with the spatial degree of freedom.

For are changing your particles and the only reason you think you haven't is because people like to call some things measurements and detectors. Really you are evolving things. And sometimes that evolution gets the different parts so separate that they will never interact again. That when the things interacting with them can pretend like just one outcome happened.

But you always evolve things. And a joint spin state means when you evolve the spin state of one you evolve the spin state of the other. And only when you have interacted differently with so many different things that it is impossible to get them to all get back to the same state together that is when you can take one branch as it's own.

If you to see it. Look at $\left(\left|\uparrow \downarrow \right\rangle - \left|\downarrow \uparrow\right\rangle\right)\otimes \left|C\right\rangle\otimes \left|0\right\rangle\otimes \left|B\right\rangle$

Where the $C$ means the spatial degree of freedom of particle 1 isin the center position,and the $B$ means the human is bored.

This evolves into $\left(\left|\uparrow \downarrow \right\rangle\otimes \left|U\right\rangle - \left|\downarrow \uparrow\right\rangle\otimes \left|D\right\rangle\right)\otimes \left|0\right\rangle\otimes \left|B\right\rangle$

Where the $U$ means it was spatially deflected up and the $D$ means it was spatially deflected down. Then we only placed the detector where the left particle is down. So it becomes $\left(\left|\uparrow \downarrow \right\rangle\otimes \left|U\right\rangle\otimes \left|0\right\rangle - \left|\downarrow \uparrow\right\rangle\otimes \left|D\right\rangle\otimes \left|1\right\rangle \right)\otimes \left|B\right\rangle$ and then the sector interacts with the human $\left|\uparrow \downarrow \right\rangle\otimes \left|U\right\rangle\otimes \left|0\right\rangle\otimes \left|B\right\rangle - \left|\downarrow \uparrow\right\rangle\otimes \left|D\right\rangle\otimes \left|1\right\rangle\otimes \left|E\right\rangle.$ Where $E$ isn't bored. So originally the spins were entangled and notnrslted to anything else. Now we have two whole worlds.

One world with bored humans and detectors that didn't go off and a left particles that was deflected the direction that spin up particles go and the left particle has spin up and the right particle has spin down.

And a second world with excited humans, detectors that are in a fired state and a beam that where the left particles was deflected the direction that spin down particles go and the left particle has spin down and the right particle has spin up.

And to get these worlds to affect each other is impossible because there are actually too many particles with different states in these two worlds you can't get them to overlap. In a double slit you have to the beam from one slit to overlap the beam from the other slit. But the overlap is actually in configuration space so you need every single particle in the universe to be in the same configuration but have gone through those two histories.

Back when all you did was deflect the beam it was possible to deflect them to the same place. But now you'd have to deflect those beams to the same place and undo the changes to he detector and the human and undo those in a way where undoing hem doesn't change anything else. That is hopeless. It is not going to happen.

So the two worlds will not interfere ever again. So they act like a world onto their own. So you either have a world where ...

There are bored humans and detectors that didn't go off and a left particle that was deflected the direction that spin up particles go and the left particle has spin up and the right particle has spin down.

Or else you have a world where there are excited humans, detectors that are in a fired state and a beam that where the left particles was deflected the direction that spin down particles go and the left particle has spin down and the right particle has spin up.

It is those two worlds that are going to throw away something or start messing with particles. They are already as separate as they are going to be. You've done a measurement. But it started back when the beams of particles were deflected different directions by a magnetic field that also polarized the spin. But things became irreversible some time after that and even after that is way too late.

spin of the "right" particles should therefore still have expectation 0, right?

Nope. The right particle has a totally defined spin state and that happened even before the detector went off, it happened when the left particle went through the Stern-Gerlach device.

• Thank you for taking the time for this detailed answer. I'm now in a superposition of Copenhagen and many-worlds. – suissidle Aug 27 '15 at 23:48
• @suissidle There isn't much wiggle room. All I did was use the Schrödinger equation for the actual experimental setup. And that can be tested by for instance placing extra detectors at any earlier stage and seeing that yes the beam continuously spilt into two directions and the spin continuously polarized etc. – Timaeus Aug 28 '15 at 0:00
• "the two worlds will not interfere ever again": You mean nothing like a breaking of Bell's inequality would be possible for this superposition of worlds? – suissidle Aug 28 '15 at 9:58
• @suissidle I can't tell what you mean. Each world can safely act as if it is the entire world. So each perceives that it lives in a world that isn't entangled with the other world. It can't tell the universe is in a superposition of the two worlds. And Bell's inequality is about the correlations in results if you prepare many systems. For that you need to wrote the wave function for many copies of the system (and the wave function of the aggregator that measures correlations) or else cheat and use Born rule. This is one system and one measurement. One isn't many. So they are unrelated. – Timaeus Aug 28 '15 at 15:20