### Question 10.

This may be a very stupid question, but in the notes we talk about a Bell measurement which can determine which Bell state two qubits are in, and these are given by the projection operators: \[P_{00} = | \phi^{+} \rangle \langle \phi^{+} | \] …etc. In the notes it claims that a Bell Measurement leaves the state unchanged, but surely that’s only the case if the Bell state of the two qubits matches that in the projector, otherwise you have two orthogonal states and the result of applying the projection operator is zero?

I don’t think that’s a stupid question! But I’m not entirely sure which bit of the notes you’re referring to. Definitely in general, measurements change the state. In the case of a projective measurement \(\{P_j\}_{j\in J}\) (with \(\sum_j P_j = I\)) on a state \(\rho\), if an outcome \(k\) is obtained, then after the measurement the state of the system will be \[
\frac{P_k \rho P_k}{\operatorname{tr}[P_k \rho P_k]}.
\] This is only the same as \(\rho\) if \(P_k \rho P_k\) is proportional to \(\rho\), which is equivalent to the support of \(\rho\) being contained in the subspace \(P_k \mathcal{H}\). For the case of \(\rho\) being a pure state and the projectors being rank-1 (like they are with the Bell measurement), then the subspace \(P_k\mathcal{H}\) is one-dimensional and the support of \(\rho\) is one-dimensional, so \(\rho\) being proportional to \(P_k\rho P_k\) is actually equivalent to \(\rho = P_k\). That is, the state can only be unchanged if it is one of the projectors to start with, like you said. In that case though, it will always be unchanged, since you will *always* get outcome \(k\): the probability of getting outcome \(j\neq k\) is \(\operatorname{tr}[P_j \rho] = \operatorname{tr}[P_j P_k] = 0\), since the projectors are orthogonal. So if the projectors are all rank-1, and the state you are measuring is a pure state, then it’s unchanged iff it is one of the projectors.