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I want to compute the POVM $E_{(\theta, \phi)}$ of the measure which gives the probability of a qubit state lying over the surface of Bloch sphere, with angles $\theta, \phi$. How can I handle this?

Dividing the surface area through the volume of Bloch sphere seems not to be a solution because it gives and absurd result of 3.

How should I start?

_____EDIT____________

For clarification, I will write down the full statement of the problem:

Find the POVM $E_{(\theta, \phi)}$ of the measure which determines the probability that the state of a qubit is on the surface of the Bloch sphere with polar angle $\theta$ and azimuthal $\phi,$ the relation of completeness being fulfilled in this case: $$\int d^{2} \Omega E_{(\theta, \phi)}=\mathbb{I}$$ Is that a projective measure? Prove also that: $$p(\theta, \phi)=\operatorname{Tr}\left[E_{(\theta, \phi)} \rho\right]=\frac{1}{4 \pi}(1+\boldsymbol{r} \cdot \boldsymbol{\Omega})$$ with $r$ being the state's Bloch vector and $\Omega=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$.

Explain why the only determination of the maximum of $p(\theta, \phi)$ allows tomography the state.

I was trying to use the pure state $\rho = \frac{1}{2}\left( \mathbb{I}+\vec{\Omega } \cdot \vec{\sigma} \right)$ as a operator measure for the mixed state $\rho '$ but can't obtain the factor $\frac{1}{4\pi}$.

I hope this update clears up your doubts

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    $\begingroup$ Isn't that simply the differential element of the surface - $d\theta\,\sin\theta\,d\phi$, times $1/4\pi$? $\endgroup$
    – Norbert Schuch
    Jan 5, 2020 at 17:08
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    $\begingroup$ What distribution are you drawing your states from? If it's a uniformly distribution over the volume of the Bloch sphere, then the probability is 0. $\endgroup$
    – Craig Gidney
    Jan 5, 2020 at 17:39
  • $\begingroup$ Yes it might be an uniform distribution, but how can I then compute the POVM $E_{(\theta, \phi)}$? Cause they must satisfy the completeness relation $\int d^\Omega E_{(\theta, \phi)}=\mathbb{I}$ $\endgroup$
    – Dani
    Jan 5, 2020 at 18:36
  • $\begingroup$ $dE_{\theta,\phi} = d\theta \sin\theta d\phi/4\pi$. $dE$ is a measure over which you have to integrate. $\endgroup$
    – Norbert Schuch
    Jan 5, 2020 at 22:35
  • $\begingroup$ Where does this question come from? $\endgroup$
    – DaftWullie
    Jan 6, 2020 at 10:06

2 Answers 2

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For a single qubit that's in a pure state (no decoherence) the probability that the state is somewhere on the surface of the sphere is 1. You don't need to measure this.

If you're going to do a POVM to learn a bit of information about the state, you have to choose an axis of measurement, and you will only gain information about the qubit with respect to that axis. This means that you can learn about $\theta$ or $\phi$, but not both.

Many measurements along different axes will give you full information about the qubit, but that involves more than one POVM. Look up "state tomography".

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I find the wording and notation of the question you've been given a bit odd. I think you've taken the correct strategy, and the factor of $4\pi$ in the question doesn't make any sense to me either. I guess it should just be a 2.

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  • $\begingroup$ yes I agree I think it should be 2, but maybe it could be related to integration over solid angle in some sense? $\endgroup$
    – Dani
    Jan 6, 2020 at 11:02
  • $\begingroup$ The integration is part of what I find perturbing. If it really means an integration as the notation actually implies, then the stated integral is not a completeness relation. To make it consistent, I think what you want is $E_{\tilde\Omega}=\delta_{\tilde\Omega=\pm\Omega}(I+\tilde\Omega\cdot\vec{\sigma})/2$, but then you need to introduce the dummy integration variables $\tilde \Omega$. $\endgroup$
    – DaftWullie
    Jan 6, 2020 at 11:11
  • $\begingroup$ Indeed we can do: $$\operatorname{tr}(\rho^2 \rho' )= \operatorname{tr}(\rho \rho')= \operatorname{tr}\left[\frac{1}{4}(1+\vec{\Omega} \cdot \vec{\sigma})(1+\vec{r} \cdot \vec{\sigma})\right]= \operatorname{tr}\left[\frac{1}{4}(1+\vec{r} \cdot \vec{\sigma}+\vec{\Omega} \cdot \vec{\sigma}+(\vec{\Omega} \cdot \vec{\sigma})(\vec{r} \cdot \vec{\sigma}))\right]=\operatorname{tr}\left[\frac{1}{4}(1+\vec{r} \cdot \vec{\sigma}+\vec{\Omega} \cdot \vec{\sigma}+(\vec{r} \cdot \vec{\Omega}) 1+i(\vec{\Omega} \times \vec{r}) \cdot \vec{\sigma})\right]=\frac{1}{2}(1+\vec{r}\cdot\vec{\Omega})$$ $\endgroup$
    – Dani
    Jan 6, 2020 at 11:24
  • $\begingroup$ and maybe defining $\rho = \frac{1}{4\pi}(\mathbb{I}+\vec{\Omega}\cdot\vec{\sigma})$ instead of $\rho = \frac{1}{2}(\mathbb{I}+\vec{\Omega}\cdot\vec{\sigma})$? In this way the integral gives the identity, as expected, but then $\rho$ is not a projector, so it won't be a projective measure. Is this righ? $\endgroup$
    – Dani
    Jan 6, 2020 at 11:48
  • $\begingroup$ I don't think so. As I say, I think the question is a bit suspect. $\endgroup$
    – DaftWullie
    Jan 6, 2020 at 12:03

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