# Prepared states and quantum entanglement [duplicate]

Imagine two entangled quantum particles in the singlet state, one held by Alice and the other by Bob. Alice and Bob are both widely separated. Bob measures his spin in some axis and finds it to be up (say). And that puts his particle now into the prepared state of spin up. Entanglement means Alice, if she were to do a similar measurement would find spin down. But assume Alice does not actually do any measurement. So does Bob doing his measurement mean that Alice's particle has immediately and spontaneously collapsed into a prepared state of spin down? Or is the state of her unmeasured particle unchanged?

The literature, and popular press, are both vague and inconsistent on this seemingly important point.

• How do you want to find out what happens? By measuring it? ;) Sep 24, 2016 at 0:30
• If you want it explained by a Nobel Prize winner : quantamagazine.org/20160428-entanglement-made-simple now that's not a chance you get every day, it's a good article, imo
– user108787
Sep 24, 2016 at 0:33

Let's say the Hilbert space of one spin, what Bob can measure, is $\mathcal{H}_s$ (spanned by $|\uparrow\rangle$ and $|\downarrow\rangle$). The Hilbert space of the rest of the world is $\mathcal{H}_w$. The total Hilbert space is $\mathcal{H}_s\otimes \mathcal{H}_w$. A general state in this this system is $|\psi\rangle=k_1|\uparrow\rangle\otimes|a\rangle +k_2|\downarrow\rangle\otimes|b\rangle$. The question boils down to how to correctly apply the Born rule to this state when you can only measure $\mathcal{H}_s$ and can't measure $\mathcal{H}_w$.

So Bob doesn't care about the rest of the world. If he looks in his detector and measures the state $|\uparrow\rangle$, he must now project the wavefunction onto this state. $P_s=|\uparrow\rangle\langle\uparrow|$ is the projection operator we want to use on $\mathcal{H}_s$. Bob can't have any physical interaction with $\mathcal{H}_w$, so we act with the identity operator there. $P_w=I$. Acting on psi:

\begin{align*} (P_s\otimes P_w)|\psi\rangle&= k_1|\uparrow\rangle\langle\uparrow|\uparrow\rangle\otimes|a\rangle +k_2|\uparrow\rangle\langle\uparrow|\downarrow\rangle\otimes|b\rangle \\ &=k_1|\uparrow\rangle\otimes|a\rangle\\ \end{align*}

Of course this has to be normalized.

So yes, by doing absolutely no physics/observation on $\mathcal{H}_w$, we still manage to pick out state $|a\rangle$ over state $|b\rangle$. We did learn something about Alice, but of course we had a huge amount of information inside the wavefunction $|\psi\rangle$ to begin with. To get consistent physics and to predict the probabilities of making a future measurement (perhaps the future measurement is Alice's reaction when they meet back up and say "our spins are opposite, how weird is that?"), Bob must calculate the unitary time evolution of this new state, $|\uparrow\rangle\otimes|a\rangle$, and apply Born's rule again.

If this were a classical probability distribution, this wouldn't be surprising at all. Imagine I take a pair of shoes and perfectly randomly put them in separate boxes. If Bob knows how shoes work, that there is one left shoe and one right shoe (ie, if he knows the probability distribution), then once he opens the box he knows that Alice has the opposite kind of shoe. It's not surprising that he can tell this, because the information was in the probability distribution (which he knew) all along. (I credit John McGreevy for teaching me about classical shoe physics)

A much more convincing demonstration of quantum weirdness is that of "quantum pseudo-telepathy" (that's what Wikipedia calls it anyways, I've never heard that exact phrase before), demonstrating 'success rates' which would be impossible in classical physics.

Quantum pseudo-telepathy is a phenomenon in quantum game theory resulting in anomalously high success rates in coordination games between separated players.

The particles prepared in this way, have opposite spin in any angle you measure them as long as the angle of measurement is same in both the measurements. Let us call this phenomena "anti correlation"

This happens due to law of conservation of angular momentum. It doers not matter when and where you measure the two particles, so distance is not a factor. It is much more (filthy) functionally rich version of pair of shoes.

If it does not make sense, think again. Anti correlation in this case is a guaranteed outcome. Probabilities do not give guaranteed outcomes.

If you say probability of 1 gives a guaranteed outcome, then it is not probability, it is a law, and that law is law of conservation of angular momentum.

To explain entanglement, anti correlation and statistical correlation need to be explained separately and independent of one another. The whole mystery of entanglement is a result of mixing the two and then trying to explain them.

Same angle measurement outcomes of particles of the same pair, can be explained with the quantum function, which is assigned due to conservation laws.

But statistical correlation (when two particles are measured at different angles) is a correlation between outcomes of various pairs and that needs independent scrutiny.

• I've deleted some off-topic comments. Sep 24, 2016 at 15:48