Timeline for Why does measuring the $X$ stabilisers at the boundary for rough merging give $X_{L}X_{L}$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 30 at 4:34 | comment | added | Navneeth Ramakrishnan | Yes, you're correct. | |
| Aug 28 at 13:26 | comment | added | GaussStrife | So the new joint state of the two patches of the planar code is in a $(-1)^{S_{1},S_{2},S_{3},S_{4}}$ eigenstate of $X_{L}X_{L}$, given the product of the stabilisers. This much makes sense to me. But it will also be an eigenstate of the 4 new stabilisers, with the results the measurments returned, yes? As in those 3 qubits are now part of the planar code state as well, and we have to adjsut the stabilisers using the measurement results, but the overall state is still an eigenstate of those product logical operators. | |
| Aug 27 at 8:13 | comment | added | Navneeth Ramakrishnan | If you know the outcomes of the observable $X_1X_2$ and the observable $X_2X_3$, you also know the outcome of the product of those two observables which is $X_1X_3$. You have "ignored" the second qubit here because $X_2^2 = I$. Similarly here, you know the outcomes of the stabilizer measurements so you also know the outcome of the observable constructed by their products, which is $X_LX_L$. | |
| Aug 26 at 7:58 | comment | added | GaussStrife | They are still part of the overall surface code, correct? I mean the new merged surface will have them present where the old boundary was, so why would we be able to ignore them? I'm not sure what notation you have used to express the outcome of the projector that is applied tbh, as now, looking at it, I fail to see how we couldn't just ignore a large amount of qubits in a stabiliser code, as we could derive projectors that would ignore many of them taking the approach above, ie any qubits the stabiliser measurments share. | |
| Aug 26 at 7:50 | comment | added | GaussStrife | OK so I understand that, since the application of these 4 stabilisers gives the logical state of the surface code weighted by $(-1)^{s_{1}s_{2}s_{3}s_{4}}$, which means that, given $X_{L}X_{L}$ is equal to the product of these stabilisers, the logical state is a $(-1)^{s_{1}s_{2}s_{3}s_{4}}$ eigenstate of $X_{L}X_{L}$ as well. But what is that measurement formalism you used that shows we can ignore the middle qubits? | |
| Aug 23 at 8:23 | vote | accept | GaussStrife | ||
| Aug 19 at 11:43 | comment | added | Navneeth Ramakrishnan | As for your first comment, consider measuring two Pauli observables $P_1, P_2$ and obtaining $+1$ for both or $-1$ for both (i.e. the product is $+1$). This means that you apply the projector $\frac{1+P_1}{2}\frac{1+P_2}{2} + \frac{1-P_1}{2}\frac{1-P_2}{2} = \frac{1 + P_1P_2}{2}$ on the state. Replace $P_i$ with the stabilizers here to see why you can ignore the $X$ operations on the pink qubits. | |
| Aug 19 at 10:22 | comment | added | Navneeth Ramakrishnan | You get the four individual outcomes of $\pm 1$ for each stabilizer (these are indeed random) but the product of those four outcomes is the same as the outcome of the observable $X_LX_L$ which may or may not be random depending on the joint state of the two patches. | |
| Aug 14 at 7:40 | comment | added | GaussStrife | ie if I measure those 4 stabilisers, it will collapse the two surfaces into an eigenstate of the 4 of them. But that wouldn't be the same as just measuring the logical operators, and the eigenstate in question wouldn't even be a +1 eigenstate, correct? | |
| Aug 14 at 7:38 | comment | added | GaussStrife | Thank you for your answer. Ok so it was my first guess. But how does the measurement actually give us $X_{L}X_{L}$? As in my question, I can clearly see that the product of the operators of the stabilisers will indeed give the product of those two logical operators. But the measurement of stabilisers doesn't cancel overlapping operators in the same way the product does. | |
| Aug 14 at 6:32 | history | edited | Navneeth Ramakrishnan | CC BY-SA 4.0 |
added 9 characters in body
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| Aug 14 at 6:27 | history | answered | Navneeth Ramakrishnan | CC BY-SA 4.0 |