EPR measurement + local unitary operation

Question

Consider an EPR pair $\frac{1}{\sqrt2}\left( |\uparrow \rangle_A \otimes |\downarrow \rangle_B - | \downarrow \rangle_A \otimes | \uparrow \rangle_B \right)$ and the standard conventions (notation) for up/down $1/2-$spin, and $x$, $y$, and $z$ axes. Suppose we can perform a local unitary operation on the first (qubit), $A$, and a subsequent "hard measurement" on the pair, in a time much shorter than what it takes for a light signal to travel between the two particles (parties, Alice, Bob) $A$ and $B$. Is the action of the local unitary "instantaneous" on the EPR pair, or the effect is propagated with a finite speed? More specifically, you can think party $A$ acting with Pauli $\sigma_x \otimes I$, with the effect of modifying the original EPR state to $\frac{1}{\sqrt2}\left( |\downarrow \rangle_A \otimes |\downarrow \rangle_B - | \uparrow \rangle_A \otimes | \uparrow \rangle_B \right)$, then rapidly doing the spin measurement on its end, with party $B$ doing the spin measurement on its own end. I would think we'd expect that the spin measurement outcomes between $A$ and $B$ should be the same (100% correlated) in this instance -- as opposed to anti-correlated, as would have been the case with the original EPR pair. Is it correct?

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Clarification after the original post: All spin measurements, by both $A$ and $B$, are along the $z$ axis.

The question can also be summarized as follows: Is the effect of a local unitary operation on an entangled state instantaneous, or it propagates with finite speed? I would think the former is the case. (As a side note, observe that this can not modify the partial trace / local density matrix for either party.)

Answer 1

I think that the principle of non-locality naturally extends to your example.

In other words, even if your operation acts "locally", the entanglement says that you are performing also something non-local; which depends on the state and the operation.

This is generally how quantum mechanics works. Perhaps, a more intuitive example is when the entanglement is based on non-distinguishable particles.

Assume two photons are distributed between A and B, this time, the EPR describes the fact that you don't know what photon A and B have respectively:

$$\frac{1}{\sqrt{2}}(|p_1,p_2\rangle_{AB} + |p_2,p_1\rangle_{AB})$$

Even by applying a local unitary, you don't know what photon you are operating on. Hence in this case doesn't make much sense to think of the operation as something that propagates.

Answer 2

After acting with the Pauli X, the correlation between measurements of A and B will depend on the basis. The Y and Z measurement anticommutes with X, so the correlation will be flipped (i.e., correlated). The X measurement, on the other hand, commutes with X, so it will still be anticorrelated.

This has nothing to do with "instantaneous change" of any kind: It is a local transformation done by A, and one way to think of it is that A effectively measures in a different basis -- nothing which B has to be concerned about.