# Trace Distance in quantum mechanics

1. Dec 5, 2015

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

2. Dec 10, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Dec 14, 2015

### Truecrimson

The OP might already solved the problem, but here we go.
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. $QS = SQ = 0$.