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Quantum Computing

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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? $\endgroup$ Commented 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 $\tilde \Omega$. $\endgroup$ Commented Jan 6, 2020 at 11:11
  • $\begingroup$ Indeed we can do: $$\operatorname{tr}(\rho^2 \rho' )= \operatorname{tr}(\rho \rho')= \operatorname{tr}\left[\frac{1}{4}(1+\vec{\Omega} \cdot \vec{\sigma})(1+\vec{r} \cdot \vec{\sigma})\right]= \operatorname{tr}\left[\frac{1}{4}(1+\vec{r} \cdot \vec{\sigma}+\vec{\Omega} \cdot \vec{\sigma}+(\vec{\Omega} \cdot \vec{\sigma})(\vec{r} \cdot \vec{\sigma}))\right]=\operatorname{tr}\left[\frac{1}{4}(1+\vec{r} \cdot \vec{\sigma}+\vec{\Omega} \cdot \vec{\sigma}+(\vec{r} \cdot \vec{\Omega}) 1+i(\vec{\Omega} \times \vec{r}) \cdot \vec{\sigma})\right]=\frac{1}{2}(1+\vec{r}\cdot\vec{\Omega})$$ $\endgroup$ Commented 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? $\endgroup$ Commented Jan 6, 2020 at 11:48
  • $\begingroup$ I don't think so. As I say, I think the question is a bit suspect. $\endgroup$ Commented Jan 6, 2020 at 12:03