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On page 47 of Quantum Machine Learning you can read that:

Interference terms-the off-diagonal elements-are present in the density matrix of a pure state (Equation 3.10), but they are absent in a mixed state (Equation 3.11)

Where 3.10 is a pure state on $|\psi> = a|0>+b|1>$

$\rho = |\psi><\psi|= \begin{bmatrix}|a|^2&ab^*\\a^*b&|b|^2\end{bmatrix}$

and 3.11, described as a sum of projectors:

$\rho_{mixed} = |a|^2 |0><0| + |b|^2 |1><1| = \begin{bmatrix}|a|^2&0\\0&|b|^2\end{bmatrix}$

He is saying that the matrices for mixed states are purely diagonal. I do not agree with what he said, especially because looking at the definition of mixed state (on all the books I found), to build one, I can use any old state that I want, and thus on off diagonal elements I can have non zero values.

Can anyone clarify better what is written there, and why? If I’m not wrong, that quote is true only if you put in the density matrix of the mixed states only states that are basis for that space (so you have only diagonal matrices in the summation, as in the example).

Thank you!

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    $\begingroup$ could you also include equations 3.10 and 3.11? $\endgroup$ – glS Oct 25 '16 at 16:35
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    $\begingroup$ Please make your post self-contained. Don't expect anyone to go that other site to get your whole query. $\endgroup$ – user36790 Oct 25 '16 at 16:35
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    $\begingroup$ The statement refers to a very specific example for which it is correct (indeed, if the density matrix of a pure state is diagonal, then it cannot have more than one nonzero element on the diagonal). However, I find this section very convoluted and not very enlightening. $\endgroup$ – Martin Oct 25 '16 at 16:42
  • $\begingroup$ The way to express bra-ket notation in latex (in the absence of the "physics" package) is \langle ... | ... \rangle. $\endgroup$ – Adomas Baliuka Oct 25 '16 at 20:14

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