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I learned that logical $|0\rangle$ of surface code, is an eigenstate, where all stabilizers are +1 value, and since the z-stabilizer is enforcing an even amount of edged in each node, and the x-stab is forcing an equal superposition,

then the final state is an equal superposition of all circles which is very making sense in this diagram as one of the terms :

enter image description here

But, it is never the diagram! actually, the diagrams look like this:

enter image description here

Or:

enter image description here

Which are letting open circles that are connected to the boundary!

I will be happy if anyone can explain more about the topological aspects of the eigenstates of surface code. is it a superposition of open circles? and what is the topolgical difference between $|0L\rangle$ and $|1L\rangle$

Thank you

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2 Answers 2

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All those three cases are different quantum error correction codes.

First toric code has periodic boundary condition, but second one which is surface code has boundaries. Lastly third one is rotated surface code which also have different stabilizer group.

Normally, we define logical state with stabilizer operators within surface code used n data qubit

$Z_i$ : i th Z stabilizer $X_i$ : i th X stabilizer

$ | 0_L \rangle = \prod{(\hat{I}+ \hat{X}_j )} |0^{\otimes n} \rangle $ , $ | 1_L \rangle = \hat{X}_L | 0_L \rangle = \prod{(\hat{I}+ \hat{X}_j )} |1^{\otimes n} \rangle $

We can define $\hat{X}_L$ as a logical X operator in surface code which meets above equations. And simply $\hat{X}_L$ can be found from series of data qubit X operators. Note that $\hat{X}_L$ should remain all the stabilizer's eigenvalue +1.

enter image description here

Likewise, Logical Z operator can be also understood in this way.

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  • $\begingroup$ Thank you, I learned new things from your answer. So, what about circles? the eigenstate is not a superposition of circles? but of open circles? $\endgroup$
    – Ron Cohen
    Apr 10, 2022 at 13:50
  • $\begingroup$ Those semi-circles also represent stabilizers. So the eigenstate is a superposition of semi-circles. Those are not opened but closed, just image shows semi-circles within the plaquettes. $\endgroup$ Apr 10, 2022 at 14:02
  • $\begingroup$ But it seems from the structure of the stabilizer, that they are forming semi-circles, and not circles $\endgroup$
    – Ron Cohen
    Apr 10, 2022 at 14:16
  • $\begingroup$ Because that is a right stabilizers, in order to detect and correct errors in rotated surface code structure. $\endgroup$ Apr 10, 2022 at 14:18
  • $\begingroup$ I wonder why thoae boundary vondition is necessary. Why are we enforcing semicircles and not complete circles? How is it helpful for the code? Does it matter? $\endgroup$
    – Ron Cohen
    Apr 10, 2022 at 18:58
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First, the reason that bottom and left(also top&left, top&right, bottom&right) are not in the same boundary type (X/Z) is that if they were, then a very short accidental chain of X errors from left to bottom will be interpreted as a logical-X-operation (new purple chain):

enter image description here

And the reason for cutting the boundary from 4-qubits stabilizers, to 3-qubits stabilizers, is that we want every qubit to touch both X-stabilizer and Z-stabilizer.

In case it was not cutted , there would be qubits that will be left without Z stabilizer (Red circle) and some without X stabilizers (Purple circle):

enter image description here

https://www.slideshare.net/RonCohen53/xxxx-es-treepptx-252234967 look ar this presentation that demonstrates the development of the strings around the surface code

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