Suppose Alice and Bob each have a spin-$\frac{1}{2}$ particle which are entangled in the spin-$0$ state

$ \lvert 0, 0 \rangle = \frac{1}{\sqrt{2}} \lvert +z, -z \rangle + \frac{1}{\sqrt{2}} \lvert -z, +z \rangle $

Since their clocks are synchronized they can perform the following steps in order:

  1. Alice measures her particle's spin $\hat{S}_{1z}$ and collapses the state into either $\lvert +z, -z \rangle$ or $\lvert -z, +z \rangle$

  2. Bob measures his particle's spin $\hat{S}_{2z}$ and (because Alice has already measured her particle) gets the same state $\lvert +z, -z \rangle$ or $\lvert -z, +z \rangle$, respectively

  3. Alice measures her particle's spin in a different basis $\hat{S}_{1x}$ and the state is now either $\lvert +x, -x \rangle$ or $\lvert -x, +x \rangle$

  4. Bob measures his particle's spin $\hat{S}_{2z}$ again

What will be the outcome of Bob's measurement? My guess is that since Alice measured her particle and got $\lvert +x \rangle _{1}$ or $\lvert -x \rangle _{1}$, then Bob's must be in $\lvert -x \rangle _{2}$ or $\lvert +x \rangle _{2}$, respectively. Since $\lvert -x \rangle _{2}$ or $\lvert +x \rangle _{2}$ has an amplitude for being in $\lvert +z \rangle _{2}$ or $\lvert -z \rangle _{2}$, there is a probability $p$ of Bob's second measurement (step 4) being different from his first (step 2).

This doesn't seem right as it implies that they could set up a protocol where if Alice wants to send a bit of data to Bob, they could agree on the following:

  • To send a 0, Alice skips her second measurement (step 3) and Bob is guaranteed both his measurements (steps 2 and 4) are the same.

  • To send a 1, Alice performs her second measurement (step 3) and Bob has a nonzero probability $p$ of seeing $\hat{S}_{2z}$ give different consecutive measurements (steps 2 and 4).

To achieve arbitrary certainty, Alice and Bob could simply repeat this N times and the probability of all N iterations yielding false negatives (Bob's measurement not changing from $\lvert +z \rangle _{2}$ to $\lvert -z \rangle _{2}$ or vice-versa) becomes $(1-p)^N$. Since FTL communication can't be possible, there must be an error in my reasoning but I can't seem to see it.


1 Answer 1


After the first measurement, their states are longer entangled. So measurement 3 has no effect on Bob's particle- measurement 4 will be the same as measurement 2 regardless.


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