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**Two things** (6 June 2018):

- Question 7 was updated (someone found a good way to do it), and Question 9 was corrected.
- Ask me your questions
**before 7pm tonight**if you want them answered before the exam! I’ll try to answer them all by 9 or 10pm. Feel free to keep asking them after 7pm, but those I’ll answer tomorrow (so it would be for your interest/knowledge etc, not as preparation for the exam).

Uhlmann’s theorem says that \(F(\rho, \sigma) = \max_{\text{all purifications}} | \langle \phi_\rho | \phi_\sigma \rangle |\). Why is the absolute value there? Is it just to make complex numbers into reals, so that we can compare them? So technically the formula would be correct even if there was \(\text{Re} \langle \phi_\rho | \phi_\sigma \rangle\) or \(\text{Im} \langle \phi_\rho | \phi_\sigma \rangle\) instead? Am I right to think that we can always adjust the complex phase of \(\langle \phi_\rho | \phi_\sigma \rangle\) by considering \(|\phi_\rho \rangle \rightarrow |\psi_\rho\rangle = |\phi_\rho \rangle |0\rangle\) and \(|\phi_\sigma \rangle \rightarrow |\psi_\sigma\rangle = |\phi_\sigma \rangle[\cos(\theta) + i\sin(\theta)] |0\rangle\)?

This way \(| \langle \phi_\rho | \phi_\sigma \rangle | = | \langle \psi_\rho | \psi_\sigma \rangle |\), but the complex phase of \(\langle \psi_\rho | \psi_\sigma \rangle\) is arbitrary (depending on how we choose \(\theta\)).

I’m asking because in 2014 exam question 1 iv) (proving joint concavity of the fidelity), it would help to be able to assume \(| \langle \phi_\rho | \phi_\sigma \rangle | = \langle \phi_\rho | \phi_\sigma \rangle\) (because of some absolute value manipulations).

Yeah, the absolute value is just to make them all real to compare them. You are right that you can find purifications so that the inner product has whatever phase you want but the same absolute value (and in fact could just maximize over the real or imaginary part). As for the joint concavity of the fidelity: that was actually Example Sheet 3, Exercise 4, and I did just as you suggested (assume \(| \langle \phi_\rho | \phi_\sigma \rangle | = \langle \phi_\rho | \phi_\sigma \rangle\)) in the solutions!

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.↩

When we talk about capacities in a completely classical context, Shannon’s Source Coding Theorem gives that any rate \(R>H(U)\) is reliable. Whereas in the quantum case the HSW Theorem states that a rate \(R\) is reliable if \(R< \chi^*(\Lambda)\) . But in both contexts means roughly the same thing. For the classical case: the number of bits of codeword/number of uses of the source. And for quantum: ‘number of bits of classical message that is transmitted per use of the channel’. So why in the clasical case do we appear to want to minimize \(R\) and in the quantum case want to maximise it?

Ah, good question. It’s actually not a classical vs. quantum difference you’re encountering here; it’s a difference between source coding and channel coding.

- In source coding (i.e. data compression), we don’t control the messages being sent out from the source; we are just trying to compress them. Here, the rate is the number of bits of storage space per message that we need. We want to minimize the rate since that means we’re compressing the most.
- In channel coding (i.e. data transmission), we have messages we wish to send over the channel in a way that’s robust to the noisyness of the channel. We are thinking of the channel as a resource we are using, and we want to minimize the number of times we have to use the channel. Here, the rate is the number of bits of information we can transmit per use of the channel. We want to maximize the rate in order to transmit as much information as possible per use of the channel.

So in Shannon’s Source Coding Theorem (classical) and in Schumacher’s Theorem (quantum), we are compressing data and want to minimize the rate (compress the most). In Shannon’s Noisy Coding Theorem (classical) and in the HSW Theorem (quantum), we are sending information over a channel, and want to maximize the rate (send the most info per use of the channel).