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I have a mixed state $\rho$ and its hamiltonian $H$. Firstly, I find the eigenvalues $\{p_j\}$ of $\rho$, and orthonormal basis of $H$. I write $\rho$ in terms of $H$'s eigenstates and $\rho$'s eigenvalues as:

$$ \rho = \sum_{j} p_j|n_j\rangle\langle n_j| \hspace{2em} (1) $$

My question is: is equation (1) a valid construction for any mixed states? Any help would be appreciated, thank you in advance.

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    $\begingroup$ What do you mean by a mixed state and its Hamiltonian"? Generically, a Hamiltonian and the state it acts upon are initially not associated, and there's no reason to think that $\rho$ should be diagonal in the basis of $H$. $\endgroup$
    – DaftWullie
    Commented Apr 24 at 6:38
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    $\begingroup$ Yes, indeed, but that's an important aspect of your question. $\endgroup$
    – DaftWullie
    Commented Apr 24 at 8:13

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The question is, how do you perform the diagonalisation routine? A typical way to achieve this is to have a set of unitaries $\{U_i\}$ for which the $|n_j\rangle$ are eigenstates with $\pm 1$ eigenvalues. If, for example, there's a pair $j,k$ of them with opposite eigenvalues for $U_i$, then choosing to apply $U_i$ 50:50 at random cancels the off-diagonal term $|n_j\rangle\!\langle n_k|$. Setting aside any question of finding a modestly sized set of $\{U_i\}$ and implementing them, whatever your protocol, you're going to need some randomness in there. A typical way of thinking about this is to have an extra qubit. Whether you prepare it in a $|+\rangle$ state or a maximally mixed state doesn't matter, but you then do a controlled-$U_i$ off that qubit.

What I'm saying is that, ultimately, to perform the diagonalisation step, you're going to have to introduce an ancilla system.

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  • $\begingroup$ I haven't thought about how to do diagonalisation using unitaries. Do you have some papers about this technique? thank you very much!! $\endgroup$ Commented Apr 24 at 8:59
  • $\begingroup$ I've never previously thought about it as generally as I stated it in my answer (so have no idea where it might have appeared in the literature in this form) but it's a common technique, starting from the fact that dephasing noise diagaonalises (in the computational basis) a density matrix. But it also appears in magic state distillation and studies of graph-diagonal states (for example, you make an entanglement witness for a graph diagonal state, and you can equally apply it to any non-diagonal state). Section 2.E of arxiv.org/abs/quant-ph/0405045 might help. $\endgroup$
    – DaftWullie
    Commented Apr 24 at 11:52
  • $\begingroup$ I have a question like this: can the operation of converting from pure state to density matrix $|\psi\rangle -> |\psi\rangle \langle \psi|$ be understood as a quantum operation? Currently, I have to measure the state $|\psi\rangle = \sum_{j} \sqrt{p_j}|n_j\rangle$ to attain a probability vector so that I can calculate equation 1. My opinion is that when I do a measurement, it will stop being quantum (the final state can't exist on quantum computers). Is that right? Whereas in double systems, you have partial trace operations which are basically quantum observables. $\endgroup$ Commented Apr 24 at 13:03
  • $\begingroup$ There is no operation that converts $|\psi\rangle\rightarrow|\psi\rangle\langle\psi|$. They are the same thing just written using different formalisms. $\endgroup$
    – DaftWullie
    Commented Apr 25 at 6:16
  • $\begingroup$ I suspect you're thinking about destructive, projective measurements where there's a number of projectors equal to the dimension of the Hilbert space. There are also partial measurements where not all projectors are rank 1. Performed in a non-destructive way, they can achieve some of what you're looking for. $\endgroup$
    – DaftWullie
    Commented Apr 25 at 6:18

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