# Are close states still close after measurement (regarding trace distance)?

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.

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.

• 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?).
– Blau
Commented Jun 26 at 13:39
• @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. Commented Jun 26 at 13:43
• @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
– Blau
Commented Jun 26 at 13:51
• @DaftWullie In MBQC we also do not trace out. We measure and use that measurement outcome for further measurements.
– Blau
Commented Jun 26 at 13:51
• If you don't renormalize after measurement then you do get the statement you wanted from Hölder's inequality: $\|\Pi(|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|)\|_1 \le \|\Pi\|_\infty \|(|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|)\|_1 = \|(|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|)\|_1$. Commented Jun 26 at 14:21

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.