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The question relates to this post.

As shown in Peskin and Schroeder's introduction to quantum field theory p. 28.,

$$[\phi(x),\phi(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0$$,

which implies the measurements between two spacelike places do not affect each other. However, in the EPR paradox, even two experiments are spacelike separated, once experimentalist $A$ obtained $+$ result of $S_z$, $B$ has to obtain $-$.

Though at this stage of EPR, no information is delivered. It still seems to me, actually, that the two measurements affect each other. Is there any further reasoning to reconcile this inconsistency?

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I think I got answer myself. The vanish commutator and EPR paradox are not correlated.

The vanishing commutator simply says, once a measurement at $x$ was done, the obtained state will not bring uncertainty for measurement of $y$, by the simutaneous eigenstate property. Not like one measures momentum, the state becomes $|p \rangle$, then measure position will have full uncertainty in state $| p \rangle$. In this sense, the measurement of A has no effect of B.

The EPR paradox is $| + - \rangle + | - + \rangle \rightarrow | + - \rangle$ or $| - + \rangle$. Either I obtain $+$ or $-$ for the spin of the first particle, the spin of the second particle is a different state than the first particle, e.g. $ | + \rangle | - \rangle$ or $ | - \rangle | + \rangle$.

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  • Doesn't the vanish commutator contradict with EPR paradox? The vanish commutator says measurement A=(spin at x) has no effect of measurement B=(spin at y), but EPR paradox says there are relationship. – Harry Feb 12 '17 at 05:23