Hi, I am trying to familiarize myself with the quantum mechanical trace distance and hit a brick wall. Thus, I would appreciate your help with the matter!(adsbygoogle = window.adsbygoogle || []).push({});

I am reading up on trace distance using Nielsen, Chunang - Quantum Computation and Quantum Information and Bengtsson, Zyczkowski - Geometry of quantum states; an introduction to quantum entanglement.

Unfortunately, I can't seem to wrap my head around the prove of Theorem 9.1 in Nielsen, Chuang.

It states:

Let ##\{E_m\}## be a set of POVMs, with ##p_m:=tr(\rho E_m)## and ##q_m:=tr(\sigma E_m)## as the probabilities of obtaining a measurement outcome labeled by ##m##. Then ##D(\rho,\sigma) = max_{\{E_m\}} D(q_m, q_m)##, where the maximization is over all POVMs.

The prove of this theorem states:

Note that ##D(p_m,q_m ) =\frac{1}{2} \sum_m |tr(E_m(\rho-\sigma))|##.

Using the spectral decomposition, we may write ##\sigma - \rho =Q-S##, where ##Q## and ##S## are positive operators with orthogonal support. Thus ##|\rho-\sigma| = Q+S##.

Here ##|A|:=\sqrt{|A^\dagger A|}##

How exactly does one obtain this last relation?

I have tried writing ##\rho-\sigma## as ##UDU^\dagger## and split the diagonal matrix into positive and negative parts.

This yields:

[itex] \rho-\sigma= UQU^\dagger + USU^\dagger[/itex]

[itex]\begin{align*}|\rho-\sigma|&=\sqrt{\big( UDU^\dagger\big)^\dagger \big( UDU^\dagger\big)}\\

&=\sqrt{\big( UD^\dagger U^\dagger\big) \big( UDU^\dagger\big)}\\

&=UDU^\dagger = Q-S\end{align*}[/itex] since ##D## is diagonal and real, right?

However that doesn't look like it's correct at all.

Could you guys help me out? Thanks!

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# Trace Distance in quantum mechanics

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