Questions and Answers

This is an attempt to have a LaTeX-friendly way to ask me questions.

  1. Write your question in the box below. The “Edit question” button should pop-out a more full-featured editor (via StackEdit), but it might not work on mobile.
  2. When you’re done, click the “x” in the top-left to leave the editor, and press “Submit”.

That will email me the question via a service called formspree. I am not affiliated with them or with StackEdit, so please don’t type in anything secret. If you want me to respond privately, please say so and include an email address. Otherwise I’ll (try to) answer below, which could help other people with similar questions. The questions may be edited for clarity.

Two things (6 June 2018):

  1. Question 7 was updated (someone found a good way to do it), and Question 9 was corrected.
  2. 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).

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Some questions answered

Question 14.

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!


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:

  1. 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.


Question 12.

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.

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).


Ask me - Eric Hanson