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Suppose you know that a qubit is either is in state $|+\rangle$ with probability $p$ or in state $|-\rangle$ with probability $1-p$. If this is the best you know about the qubit's state, where in the Bloch sphere would you represent this qubit?

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In your first sentence, are you specifying that the qubit is in a pure state? See response and comments below. –  joshphysics Apr 5 '13 at 16:20
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Since this is a homework question, here's a bit to get you started: I'll assume you mean that the qubit is in the pure state $$ |\psi\rangle = \sqrt{p}|+\rangle + \sqrt{1-p}|-\rangle $$ The bloch sphere representation of pure states can be written as follows: $$ |\psi\rangle = e^{i\phi}\sin(\theta/2)|+\rangle + \cos(\theta/2)|-\rangle $$ Post more of your work if you're still stuck and need guidance.

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That's not what the question says -- if I'm not mistaken, it says $\rho = p\vert+\rangle\langle+\vert + (1-p)\vert-\rangle\langle-\vert$. –  Norbert Schuch Apr 5 '13 at 15:51
    
@NorbertSchuch Normalization demands $\langle \psi|\psi\rangle = 1$. Given my expression for the state, one has $\langle \psi|\psi\rangle = \sqrt p^2 + \sqrt{1-p}^2 = p + 1-p=1$. –  joshphysics Apr 5 '13 at 16:12
    
Yes, but the way the question is phrased it talks about a mixed state rather than a pure one. –  Norbert Schuch Apr 5 '13 at 16:13
    
@Norbert: the way the question is phrased, it is simply ambiguous. –  Abdelmalek Abdesselam Apr 5 '13 at 16:17
    
@NorbertSchuch You may be right; I'll ask for clarification from the user... –  joshphysics Apr 5 '13 at 16:18
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