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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.

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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.

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  • $\begingroup$ Thanks, I had a similar example, but I was hoping for something like: taking the probability of getting $m=0$ into account to still get a nice inequality (maybe probability of outcome times trace distance of post-measurement state is still small?). $\endgroup$
    – Blau
    Commented Jun 26 at 13:39
  • $\begingroup$ @Blau if you want some sort of averaging over the measurement outcomes, that's a pretty big detail to leave out of the question. It basically equates to not measuring the first system at all, and just tracing it out. $\endgroup$
    – DaftWullie
    Commented Jun 26 at 13:43
  • $\begingroup$ @DaftWullie I am not really sure if this is the same. I really want to measure and don't forget about the measurement outcome $m$. If we trace out things, then everything is fine (as I explained in my post). But as you can see if we measure, stupid things happen. I need a formula how these things can be quantified taking the measurement outcome into account :S $\endgroup$
    – Blau
    Commented Jun 26 at 13:51
  • $\begingroup$ @DaftWullie In MBQC we also do not trace out. We measure and use that measurement outcome for further measurements. $\endgroup$
    – Blau
    Commented Jun 26 at 13:51
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    $\begingroup$ If you don't renormalize after measurement then you do get the statement you wanted from Hölder's inequality: $\|\Pi(|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|)\|_1 \le \|\Pi\|_\infty \|(|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|)\|_1 = \|(|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|)\|_1$. $\endgroup$
    – Mateus Araújo
    Commented Jun 26 at 14:21
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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.

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