# Question regarding the measurement of Pauli matrices on a Bell state

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? $$\vec v\cdot\vec\sigma\otimes I\ |\psi\rangle = \frac{|ab\rangle+|ba\rangle}{\sqrt2}.$$

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 $$\vec{v}\cdot\vec{\sigma}\otimes \vec{v}\cdot\vec{\sigma}$$.

• 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?
– Hans
Sep 22 at 8:18
• Yes, that's right. Sep 22 at 12:11
• 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?
– Hans
Sep 22 at 19:08
• 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. Sep 23 at 6:55
• 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?
– Hans
Sep 23 at 8:50