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Problem with past explanations is they emphasize the need for random choice of measurement angle, but my understanding is that was only necessary in experiments seeking to remove any possible loop-holes from "hidden variable" approaches. So, you emit entangled articles to Alice and Bob. Alice measures her particle at a farther distance from the source than Bob and always measures if spin is down. Bob measures either up or at 60 degrees. Assuming that Alice can't detect any difference in her measurements depending on Bob's choice, why not? (Shouldn't there be a change in the correlation between particles and so in the share of particle Alice measures as up?)

(P.s., if the answer is that entanglement ends once Bob measures, am I wrong in assuming the state of Alice's particle is then no longer described by the wave function?)

Qmechanic
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John Strider
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1 Answers1

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Suppose Alice and Bob measure entangled spins in the state \begin{align} \left|\psi\right> = \frac{1}{2}\left(\left|\uparrow\downarrow\right> + \left|\downarrow\uparrow\right> \right) \end{align} Alice always measures $S_z$. Bob measures either $S_z$ or \begin{align} S_{B} = \frac{1}{2}S_x + \frac{\sqrt{3}}{2}S_z \end{align} If Bob chooses to measure $S_z$, then Alice will detect spin up or spin down each with 50% probability. (I take it you accept this claim without contention.) She will detect the same thing if Bob instead chooses to measure $S_B$. In this case, Bob will detect one of the eigenstates of $S_B$ each with 50% probability. If Bob measures spin "up" along the $60^{\circ}$ axis, Alice will measure spin up or spin down along the $z$ axis with probabilities $P(\uparrow|S_{B,+})\approx 93.3\%$ and $P(\downarrow|S_{B,+})\approx 6.7\%$, respectively. If Bob measures spin "down" along his axis, the probabilities are the opposite: Alice gets spin up with probability $P(\uparrow|S_{B,-})\approx 6.7\%$ and spin down with probability $P(\downarrow|S_{B,-})\approx 93.3\%$. Alice's probabilities to detect either spin up or spin down are \begin{align} P(\uparrow) &= P(\uparrow|S_{B,+})P(S_{B,+}) + P(\uparrow|S_{B,-})P(S_{B,-})\\ &= (0.933)(0.5) + (0.067)(0.5)\\ &= 0.5 \end{align} or \begin{align} P(\downarrow) &= P(\downarrow|S_{B,+})P(S_{B,+}) + P(\downarrow|S_{B,-})P(S_{B,-})\\ &= (0.067)(0.5) + (0.933)(0.5)\\ &= 0.5 \end{align} respectively. So the outcome for Alice is the same regardless of Bob's choice of axis.

d_b
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    I think this answers it for me but if you wouldn't mind clarifying one issue. My understanding is that If Bob measures "up" at 60%, then Alice's particle is "set" to down at 60%. Then the probability you mention of 93.3% is that of Alice's measurement returning a yes or no when she measures "up" for a particle that is actually down at 60%. Is my understanding roughly on the mark? (Bonus: is it fair to say (at a high, abstract level) these probabilities have more to do with the Uncertainty Principle as oppose to the wave function or entanglement per se. – John Strider Jan 07 '23 at 21:07