### Question 13.

In Notes 11, on encoding two classical bits as a Bell state it says: ‘We can encode two classical bits in the state of the two qubit system. This information can be recovered by performing a joint measurement (on the two qubits) which projects onto the Bell basis. This is called a Bell measurement and is a measurement in which the measurement operators are the Bell state projectors’.

Could you outline how this works? Surely if you make a measurement with one of the Bell projectors you can only tell whether the two qubits are in that Bell state or if they aren’t, after which you aren’t able to perform a second measurement and the state has collapsed to the post measurement state. So then you only have a 1 in 4 chance of recovering the information which is the equivalent of just guessing?

Ah, good to clarify this point. The measurement isn’t a two outcome measurement for each projector, asking “is it in this state or not”. Instead, it’s a four outcome measurement. The measurement is described by the set of four projectors \[ \{| \Phi^+ \rangle\langle \Phi^+ |_{AB}, | \Phi^- \rangle\langle \Phi^- |_{AB}, | \Psi^+ \rangle\langle \Psi^+ |_{AB}, | \Psi^- \rangle\langle \Psi^- |_{AB}\} \] and the four measurement outcomes correspond to the four Bell states. This measurement which will pick out the right state, without modifying it, iff the state being measured is indeed one of the four Bell states, as discussed in Question 10. This is a nonlocal measurement because the projectors involve both the \(A\) and \(B\) system (and in fact are entangled). So to perform such a measurement, one needs access to both parts of the state.

Maybe the following statements will help clarify things, in case this seems murky or maybe like cheating:

- if you can perform arbitrary measurements on the two qubits, the tensor product structure doesn’t really matter anymore. The fact that these states are entangled, etc, doesn’t play a role; you are simply trying to determine which of four
*orthogonal*states you have, which can you can do with 100% success rate (as long as the state you are measuring is indeed one of those four states). - if you only have access to one of the two systems, say Alice’s system \(A\), then what you have access to is the
*reduced density matrices*resulting from tracing out \(B\)^{1}. But for all four of these states, the reduced density matrix is the same: \(\frac{1}{2}I_A\). Therefore there is nothing you can do*with access to \(A\) alone*to determine which of the four states you have. - If Alice and Bob can each perform local measurements and communicate classically, then it’s a bit more complicated. Ghosh, Kar, Roy, Sen(De) and Sen showed in 2001 that the four Bell states cannot be perfectly distinguished by such a protocol.
- if the possible states you could have are not all orthognal to each other, then you can’t perfectly distinguish them even with arbitrary measurements. Note that this doesn’t have anything to do with the tensor product structure (so in particular it doesn’t have to do with entanglement); this is true even for only one system. If you restrict to projective measurements, you will sometimes make an error; the measurement will sometimes tell you the state is \(| \psi \rangle\) when really the state is \(| \chi \rangle\), if \(\langle \psi|\chi \rangle\neq 0\). If you allow POVMs then you might be able to do better, like in Exercise 10 of Example Sheet 2; in that scheme, sometimes you can get “no result,” but you never get the wrong answer.

Why? One way to see this is by equation (2.180) of Nielsen and Chaung: \(\operatorname{tr}[M \rho_A] = \operatorname{tr}[ (M\otimes I)\rho_{AB}]\) for any observable (Hermitian matrix) \(M\). An observable \(M\) on \(A\) alone has expectation value \(\operatorname{tr}[ (M\otimes I)\rho_{AB}]\), but we see these values are described exactly by the reduced density matrix. In fact, you can take this as the definition of the partial trace, since it is the unique function \(f\) from density matrices on \(AB\) to density matrices on \(A\) such that \(\operatorname{tr}( M f(\rho_{AB})) = \operatorname{tr}[ (M\otimes I) \rho_{AB}]\), as described in Box 2.6 of Nilsen and Chaung.↩