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I'm trying to understand the section "EPR and the Bell inequality" from the book Quantum Computation and Quantum Information by Nielson and Chuang. While going through the box titled "Anti-correlations in the EPR experiment," I have trouble understanding the following line:

It turns out that no matter what choice of $v$ we make, the results of the two measurements are always opposite to one another. That is, if the measurement on the first qubit yields $+1$, then the measurement on the second qubit will yield $−1$, and vice versa.

While the authors go on explaining why this is true, I doesn't quite fit with my understanding by performing computations.

Specifically, as the authors show, first we show that $$\frac{|01\rangle-|10\rangle}{\sqrt{2}} = \frac{|ab\rangle-|ba\rangle}{\sqrt{2}},$$ up to an unobservable global phase factor, where $|a\rangle$ and $|b\rangle$ are the eigenstates of $v.\sigma$. Now, the text says

...if a measurement of $v.\sigma$ is performed on both qubits, then we can see that a result of $+1$ ($−1$) on the first qubit implies a result of $−1$ ($+1$) on the second qubit.

That means, if we apply $v.\sigma\otimes v.\sigma$ on the state we get $$(v.\sigma\otimes v.\sigma)\frac{|ab\rangle-|ba\rangle}{\sqrt{2}} = -\frac{|ab\rangle-|ba\rangle}{\sqrt{2}}.$$ How does this justify the above statement?

Also, on the same page of the book, it's mentioned that

Alice performs a measurement of spin along the $v$ axis, that is, she measures the observable $v.\sigma$.

Shouldn't then we be using the operator $v.\sigma\otimes I$ instead of $v.\sigma \otimes v.\sigma$?

May be I'm misunderstanding something. Please help me sort this out. Thanks!

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To measure the a single-qubit observable, you rotate the qubit to align that observable's axis with the computational basis (e.g. align it along the Z axis) then do a measurement. In other words, for any single-qubit observable $O$ you measure $M_O$ by instead performing a single-qubit operation $u$ then a Z-basis measurement $M_Z$. For all $M_O$, there exists a $u$ such that $M_Z \cdot u \equiv M_O$.

Now consider the case where two parties are measuring $M_O$ on their respective parts of a singlet state. They perform $M_O$ by applying $u$ and then measuring $M_Z$. However, the singlet state is not changed by both parties applying $u$, i.e. $(u \otimes u) \cdot (|01\rangle - |10\rangle)$ gives $|01\rangle - |10\rangle$ again (up to global phase). Therefore you can just drop those operations without changing the expected outcome:

$$\begin{align} (M_O \otimes M_O) \cdot (|01\rangle - |10\rangle) &= (M_Z \cdot u) \otimes (M_Z \cdot u) \cdot (|01\rangle - |10\rangle) \\ &= (M_Z \otimes M_Z) \cdot (u \otimes u) \cdot (|01\rangle - |10\rangle) \\ &\propto (M_Z \otimes M_Z) \cdot (|01\rangle - |10\rangle) \\ &\rightarrow \text{measurements give opposite answers} \end{align}$$

Because the singlet state gives opposite answers in the computation basis, and the singlet state is not affected by doing a basis change to both qubits, it must give opposite answers in every basis.

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