1
$\begingroup$

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?


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.)

$\endgroup$
0

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Sorry, I wasn't clear in the original question; "hard measurements" of spin by A and B are in the z-axis. Hence, it is undoubtedly true that A and B will observe the same spin measurement outcome if the experiment is performed sufficiently long after applying the local unitary by A. So, before performing the measurement along the z-axis, party $A$ applies unitary to flip the first "qubit". $\endgroup$ Dec 10, 2022 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.