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We are given two states $|\psi_1\rangle, |\psi_2\rangle \in \mathbb{C}^2 \otimes \mathbb{C}^2$ with trace distance $\leq \varepsilon$, so they are very close to each other. Now, assume we measure the first qubit of each state in the standard basis and say we get outcome $m \in \{0,1\}$ in both cases. Then the two states collapse to $|m\rangle \otimes |\psi_1'\rangle$ and $|m\rangle \otimes |\psi_2'\rangle$.

What can we say about the trace distance of the post-measurement states $|\psi_1'\rangle$ and $|\psi_2'\rangle$?

If we trace out the first qubit, the trace distance stays under $\leq \varepsilon$ as the trace distance is contractive under CPTP maps. Is there a similar result for measurements?

Background: In measurement-based quantum computation (MBQC) we can't work with perfect qubits, hence there are always small errors. The question is aimed at how the error evolves over time.

Blau
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2 Answers2

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You can't say anything. Consider for example $$\begin{align} |\psi_1\rangle = \sqrt{\epsilon}|00\rangle + \sqrt{1-\epsilon}|11\rangle \\ |\psi_2\rangle = \sqrt{\epsilon}|01\rangle + \sqrt{1-\epsilon}|11\rangle \end{align} $$ The trace distance between them is quite small (well, not $\epsilon$, but some reasonable function of $\epsilon$), but if you measure the first qubit and obtain 0, the resulting states will be $|00\rangle$ and $|01\rangle$, which have maximal trace distance.

Mateus Araújo
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If you could find a way to make close states become consistently far apart after measurement, you could use the deferred measurement principle to transform that into a process that did it with only unitary gates. So you can't.

Craig Gidney
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