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If two people (Alice and Bob) are given a state $|\phi^+\rangle = \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ from a list of states, then Alice can read the first qubit, see if it is 0 or 1, and send the measurement to Bob who can measure the second qubit to see that they have $|\phi^+\rangle$.

However, what if they are given a state like $\frac{1}{\sqrt2}(i|00\rangle + |11\rangle)$ or $\frac{1}{\sqrt2}(|00\rangle + i|11\rangle)$? I do not think that Alice and Bob can tell the difference between this state and $|\phi^+\rangle$, since all they get is a qubit, and not the value in front of the qubit. They won't see the $i$ and so they can't tell the difference. Do I have the right idea here?

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    $\begingroup$ Note that questions of the form "Am I right?" typically don't make good Q&A here because the answer (yes or no) isn't long enough to be a valid answer. Consider rephrasing your question to be a bit more open to an explanation. $\endgroup$ – Kyle Kanos Sep 29 '15 at 12:01
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Not quite. Your typical qubit is a spin-$\frac12$ system, where the spin around any axis in 3D will be measured as either "up" or "down" along that axis. The observable matrices are the Pauli matrices; the "computational basis" $|0\rangle,|1\rangle$ is therefore the $z$-axis.

Now if you wrote $i|00\rangle + i|11\rangle$, that would be an "overall phase factor" and that is indeed not observable, but since you wrote $i$ for one and $+1$ for the other, you can, just by measuring about a different axis. The $z$-axis conveniently erases the entanglement information.

But you can confirm that this state is equivalent, up to an overall phase of $i$, to $\sqrt{\frac12} |00\rangle - i\sqrt{\frac12} |11\rangle,$ which can be written as $$\sqrt{\frac18}\big(|0\rangle -|1\rangle\big)\otimes\big(|0\rangle + i |1\rangle\big) + \sqrt{\frac18}\big(|0\rangle + |1\rangle\big)\otimes\big(|0\rangle - i |1\rangle\big).$$In terms of the eigenvectors $|x_\pm\rangle \propto |0\rangle \pm |1\rangle$ and $|y_\pm\rangle\propto|0\rangle \pm i|1\rangle,$ this is the state$$\sqrt{\frac12}|x_-~y_+\rangle + \sqrt{\frac12}|x_+~y_-\rangle.$$So the entangled state that you just gave can be distinguished by Alice making a bunch of measurements on the $x$-axis and Bob making a bunch of measurements on the $y$-axis, or vice versa. If they do not have the prefactor $i$ on the $|00\rangle$ state, then Alice's $x_\pm$ will not correlate at all with Bob's $y_\pm$ but if they do, then they will correlate perfectly.

Edit: per @NorbertSchuch, it is also important to see how this varies from the $|00\rangle + |11\rangle$ case, which can be written as $|x_+~x_+\rangle + |x_-~x_-\rangle.$

To distinguish them, Alice measures in the $x$-direction and Bob measures in the $y$-direction; in the $-i$ case they strictly anti-correlate whereas in the $+$ case Bob measures a straight 50/50 split no matter what Alice measures.

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  • $\begingroup$ The first time a 1/8 appears your state has a squared norm of 5/8 not 1 and doesn't give both spins equally often in the z basis, how could it possibly be a phase factor different than the triplet of spin 0? $\endgroup$ – Timaeus Sep 29 '15 at 6:48
  • $\begingroup$ You also have to check what happens to the other entangled state under this basis transformation! $\endgroup$ – Norbert Schuch Sep 29 '15 at 10:14
  • $\begingroup$ @Timaeus: good catch; I had written the \sqrt{\frac12} factors then went back to edit them; I must have been too overzealous and accidentally corrected three of them to \sqrt{\frac18} rather than just the two of them. $\endgroup$ – CR Drost Sep 29 '15 at 13:49
  • $\begingroup$ @NorbertSchuch I mean, there is no transformation; it is just a rewriting. With that rewriting you can see that if you rotate the first qubit so that $+x$ becomes $+z$ and the second qubit so that $+y$ becomes $+z$, the result is $\sqrt{\frac12}|\uparrow\downarrow\rangle + \sqrt{\frac12}|\downarrow\uparrow\rangle,$$ but even without these transforms we can still surely write the above state in the above way. $\endgroup$ – CR Drost Sep 29 '15 at 13:51
  • $\begingroup$ Yes, but you still need to rewrite both states in the new basis; it seems you only rewrite one of them. $\endgroup$ – Norbert Schuch Sep 29 '15 at 17:06

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