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Michael A. Nielsen & Isaac L. Chuang, Quantum Computation and Quantum Information, 10th Anniversary Edition p.113, Box 2.7 states that "if a measurement of $\vec v\cdot\vec\sigma$ is performed on both qubits [of $\psi:=\frac{|ab\rangle-|ba\rangle}{\sqrt 2}$ ] , then we can see that a result of $+1 (−1)$ on the first qubit implies a result of $−1 (+1)$ on the second qubit," where $|a\rangle$ and $|b\rangle$ are the eigenstates with eigenvalue $+1$ and $-1$ respectively of $\vec v\cdot\vec\sigma$ where $\vec v$ is a unit $R^3$ vector and $\vec\sigma$ is the vector for the $3$ Pauli matrices.

What does it mean precisely in terms of operator actions by "a measurement of $\vec v\cdot\vec\sigma$ is performed on both qubits, then see a result of +1"? Is it the following? $$\vec v\cdot\vec\sigma\otimes I\ |\psi\rangle = \frac{|ab\rangle+|ba\rangle}{\sqrt2}.$$

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The probability that both parties get the +1 measurement result is calculated by $$ \frac{1}{4}\langle\psi|(I+\vec{v}\cdot\vec{\sigma})\otimes (I+\vec{v}\cdot\vec{\sigma})|\psi\rangle $$ where $|\psi\rangle$ is the state that you're measuring. This is because the projector onto the $+$ solution is $$ \frac12(I+\vec{v}\cdot\vec{\sigma}). $$

Similarly, if I wanted to know the probability of Alice getting +1 and Bob getting -1, I'd evaluate $$ \frac{1}{4}\langle\psi|(I+\vec{v}\cdot\vec{\sigma})\otimes (I-\vec{v}\cdot\vec{\sigma})|\psi\rangle $$

Now, what you're really interested in is the probability that both parties get different answers: $$ \frac{1}{4}\langle\psi|(I+\vec{v}\cdot\vec{\sigma})\otimes (I-\vec{v}\cdot\vec{\sigma})|\psi\rangle+\frac{1}{4}\langle\psi|(I-\vec{v}\cdot\vec{\sigma})\otimes (I+\vec{v}\cdot\vec{\sigma})|\psi\rangle. $$ If you expand this, it's the same as $$ \frac{1}{2}\langle\psi|(I\otimes I-\vec{v}\cdot\vec{\sigma}\otimes \vec{v}\cdot\vec{\sigma})|\psi\rangle $$ In other words, you're after the $-1$ outcome of the measurement $\vec{v}\cdot\vec{\sigma}\otimes \vec{v}\cdot\vec{\sigma}$.

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  • $\begingroup$ Yes. The reason is that $\vec v\cdot\vec\sigma=|a\rangle\langle a|-|b\rangle\langle b|$ and $I=|a\rangle\langle a|+|b\rangle\langle b|$, right? $\endgroup$
    – Hans
    Sep 22 at 8:18
  • $\begingroup$ Yes, that's right. $\endgroup$
    – DaftWullie
    Sep 22 at 12:11
  • $\begingroup$ Then don't you think the phrase "a measurement of $\vec v\cdot\vec\sigma$" is incorrect? It should rather be "a measurement of $ \frac12(I+\vec{v}\cdot\vec{\sigma})$". Do you agree? $\endgroup$
    – Hans
    Sep 22 at 19:08
  • $\begingroup$ There are two different ways of defining a measurement. You can either use projectors, in which case you have to give all projectors: $\{|a\rangle\langle a|,|b\rangle\langle b|\}|$ or, you can specify an operator such as $\vec{v}\cdot\vec{\sigma}$. This implicitly defines the set of projectors as being the projectors onto the different eigenspaces. It is very common, for example, to say "perform a $Z$ measurement" because it's easier than talking about projecting onto the $|0\rangle/|1\rangle$ basis. $\endgroup$
    – DaftWullie
    Sep 23 at 6:55
  • $\begingroup$ You misunderstood me. I understand the equivalence between these two forms of the operator. I have no objections to using operators composed of $\vec v$ and $\vec\sigma$ but to the required operator being $\vec{v}\cdot\vec{\sigma}$ rather than $\frac12(I+\vec{v}\cdot\vec{\sigma})$. Do you agree that the correct operator should be the latter rather than the former? $\endgroup$
    – Hans
    Sep 23 at 8:50

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