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roshoka
  • Member for 8 years, 6 months
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Thinking of a linear operator as a (1,1) tensor
I see, thanks for the information
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Thinking of a linear operator as a (1,1) tensor
It seems like before the components part the important thing was to find $\mathcal{I}^{-1}$. What is the significance of that? Find the inverse in the first place?
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Thinking of a linear operator as a (1,1) tensor
So why is it justified to use $A$ instead of $\alpha$ in what I wrote? Is it just abuse of notation?
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If $\rho_{AB}$ is at most rank 2, why is $\operatorname{det}(\rho_{AB})=0$?
So according to Wikipedia: "The rank of [a matrix] equals the number of non-zero singular values". Since the density matrices are 4x4, a rank of 2 would mean that two of the eigenvalues would be zero, making the determinant zero. Right?
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If $\rho_{AB}$ is at most rank 2, why is $\operatorname{det}(\rho_{AB})=0$?
I actually don't know what either of those mean for this, but I'll try to figure it out.
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