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I was thinking about the trace distance and what it means for the evolution of two density matrices that are close together in the trace distance.

Specifically, suppose $\rho$ and $\sigma$ are two density matrices such that $\lVert \rho-\sigma\rVert_{Tr}<\varepsilon$, what can we say about their distance after evaluating a quantum gate $U$ on both, or performing a measurement or projecting onto a certain state $|x\rangle$?

  • As unitaries preserve the norm, I suspect that $\lVert U(\rho-\sigma)U^{\dagger}\rVert_{Tr}<\varepsilon$ as well.
  • How about a projection onto one of the computational basis states? What can we say about: $\lVert |x\rangle\langle x|(\rho-\sigma)\rVert_{Tr}<\varepsilon$?
nippon
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1 Answers1

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Your argument for unitary evolutions is correct. More generally, for every evolution (i.e., every completely positive and trace-preserving map $\Phi$) one has $\|\Phi(\rho-\sigma)\|_1\leq\|\rho-\sigma\|_1<\varepsilon$ because every such operation is trace-norm contractive.

The key to your second point is the inequality $$ \|ABC\|_{\rm tr}\leq\|A\|_\infty\|B\|_{\rm tr}\|C\|_\infty $$ with $\|\cdot\|_{\rm op}$ the usual operator norm = largest singular value which holds for all $A,B,C$. (NB: this also follows from the duality between trace norm and operator norm). In your case this yields $$ \|\,|x\rangle\langle x|(\rho-\sigma)\|_{\rm tr}\leq\|\,|x\rangle\langle x|\,\|_\infty\|\rho-\sigma\|_{\rm tr}=\|\rho-\sigma\|_{\rm tr}<\varepsilon\,. $$ Notably, this idea transfers to the case of measurement probabilities: given a POVM $\{M_i\}_i$, i.e., $M_i\geq 0$ and $\sum_i M_i={\bf1}$, the difference of the corresponding probabilities for all $i$ satisfies $$ \big|{\rm tr}(\rho M_i)-{\rm tr}(\sigma M_i)\big|\leq\|(\rho-\sigma)M_i\|_1\leq\|\rho-\sigma\|_1\|M_i\|_\infty\leq \|\rho-\sigma\|_1<\varepsilon\,. $$ In the first step we used that the trace is upper bounded by the trace norm, and in the third step we used that $M_i\leq{\bf1}$ (because ${\bf1}-M_i=\sum_{j\neq i}M_j\geq 0$).

Frederik vom Ende
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