Trace Distance in quantum mechanics

[Emil_M]

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!

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:

$\rho-\sigma= UQU^\dagger + USU^\dagger$

$\begin{align*}|\rho-\sigma|&=\sqrt{\big( UDU^\dagger\big)^\dagger \big( UDU^\dagger\big)}\\<br /> &=\sqrt{\big( UD^\dagger U^\dagger\big) \big( UDU^\dagger\big)}\\<br /> &=UDU^\dagger = Q-S\end{align*}$ 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!

 

[Truecrimson]

The OP might already solved the problem, but here we go.

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

The keyword is "orthogonal". $\rho - \sigma$ is Hermitian so by the spectral decomposition,
$$\rho - \sigma = \sum_i \lambda_i P_i,$$
where the $\lambda_i$'s are eigenvalues with the corresponding orthogonal projections $P_i$ onto the eigenvectors. Both $$\sum_{\substack{i \\ \lambda_i \ge 0}} \lambda_i P_i$$ and $$- \sum_{\substack{i \\ \lambda_i < 0}} \lambda_i P_i$$ are positive operators. Call them $Q$ and $S$ respectively so that $$\rho - \sigma = Q - S.$$
Then $| \rho - \sigma | = Q + S$ because $Q$ and $S$ are orthogonal i.e. $QS = SQ = 0$.