# Fault-tolerant correlated measurement on logical qubits

So I have an encoded Bell pair $$(|\overline{0}\overline{0}\rangle+|\overline{1}\overline{1}\rangle)$$ in CSS code(e.g. Steane code) and I hope to perform stabilizer measurement of $$\overline{XX}$$ non-destructively. Can I simply perform Shor's style measurement on each side and combine their results? i.e. on each side, prepare ancillary $$|0\rangle^{\otimes n}+|1\rangle^{\otimes n}$$ and then do transversal CZ and measure in $$\overline{X}$$ basis? of course, I do this measurement 3 times to ensure fault-tolerance in the case of Steane code.\ I have doubt because this measurement is like, correlated, so maybe they should share an entangled ancillary state to begin with as well? Or is this independent measurement fine?

• The two things that you mark as "cat" should be one big, common, cat state. Otherwise, you end up doing two separate $X$ measurements, which is more destructive than $X\otimes X$. Commented Mar 25 at 9:21

This approach will get you the correct measurement result, but will also collapse the encoded state.

Consider first the case with no underlying stabilizer code. You start with your Bell pair $$\left|00\right\rangle + \left|11\right\rangle$$ and measure $$X_1$$ and $$X_2$$ separately.

These measurement outcomes will be correlated, they will both give both +1 (or -1), which will collapse the state to $$\left|+\right\rangle\otimes\left|+\right\rangle$$ (or $$\left|-\right\rangle\otimes\left|-\right\rangle$$). However, you can reconstruct the measurement outcome of $$X_1X_2$$ by multiplying the two measurement outcomes of $$X_1$$ and $$X_2$$. This product gives +1 in all cases, which is not surprising since $$X_1X_2$$ was a stabilizer of your initial state.

In general, you can infer the measurement outcome of any Pauli operator that is a product of commuting measured Pauli operators.

Note that measuring $$X_1$$ and $$X_2$$ separately has a side effect: you initial state stabilized by $$X_1X_2$$ and $$Z_1Z_2$$ is now stabilized by $$X_1X_2$$ and $$\pm X_1$$ (and equivalently $$\pm X_2$$). This is probably not what you want, i.e. you should probably only measure $$X_1X_2$$ in one go.

The analysis is the same for the encoded case, your measurements are just more involved.

As mentionned in a comment, if you entangle your cat states in a big $$\left|00\right\rangle^{\otimes n} + \left|11\right\rangle^{\otimes n}$$, proceed with your Shor measurement and measure the $$2n$$ ancillary qubits at once, you should only be measuring your stabilizer operator.

• I see, thanks this is quite clear. What about say the two parties have two independent states, but they want to perform a correlated $X\otimes X$ measurement? In this case do they need to share an entangled ancilla? Commented Mar 25 at 14:36
• @AndyLiuin After a correlated $X\otimes X$ measurement, your states are stabilized by $X_1X_2$, but not necessarily by $X_1$ or $X_2$. This cannot be achieved without an entangled ancilla. By measuring your two states independently in the $X$ basis, you lose all entanglement between them.
– AG47
Commented Mar 25 at 14:59
• Maybe I did not explain my follow-up question very clearly. I'm saying, if instead of the Bell pair, I start with a product state $|\phi\rangle\otimes|psi\rangle$, and I still want to perform non-destructive $X\otimes X$ measurement. Do I need an entangled ancilla? If I now measure separately will this disturb the original state? Commented Mar 25 at 16:01
• @AndyLiuin You will almost certainly need an entangled ancilla. I tried to convey the intuition that the $X \otimes X$ measurement introduces entanglement between your states, which cannot be done by a procedure only acting independently on the states, unless you consume some entangled resources doing so.
– AG47
Commented Mar 25 at 16:21
• @AndyLiuin Feel free to ask a follow up question on the website.
– AG47
Commented Mar 25 at 16:22