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The theory of QEC with stabilizer codes defines an alternative way to represent a quantum state in terms of operators. To understand better what I am concerning about, let's consider the 7-qubit Steane code:

$$ \left|0\right\rangle_L \equiv \frac{1}{\sqrt{8}}(\left|0000000\right\rangle + \left|1010101\right\rangle + \left|0110011\right\rangle + \left|1100110\right\rangle + \left|0001111\right\rangle + \left|1011010\right\rangle + \left|0111100\right\rangle + \left|1101001\right\rangle) $$ $$ \left|1\right\rangle_L \equiv \frac{1}{\sqrt{8}}(\left|1111111\right\rangle + \left|0101010\right\rangle + \left|1001100\right\rangle + \left|0011001\right\rangle + \left|1110000\right\rangle + \left|0100101\right\rangle + \left|1000011\right\rangle + \left|0010110\right\rangle) $$

Papers and books directly define the relative generators as:

$$ K^1 = IIIXXXX $$ $$ K^2 = XIXIXIX $$ $$ K^3 = IXXIIXX $$ $$ K^4 = IIIZZZZ $$ $$ K^5 = ZIZIZIZ $$ $$ K^6 = IZZIIZZ $$ where I is identity matrix and X,Z Pauli's matrices.

Therefore, my questions are: how these generators have been calculated? How can I calculate the generators for a generic superposition of n qubits?

Thank you

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I think, we usually guess the generators by(specially in the CSS case) by their classical dual, and then find the logical qubits and check whether they are useful or not. –  Ali Oct 14 '13 at 11:47
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An algorithm would respect the rules : a) $|0_L\rangle$ and $|1_L\rangle$ are eigenvectors of all the generators $K_i$, with eigenvalue $1$ b) For each error operator $E$ (in the $X_i, Y_i, Z_i$), there is at least one generator $K_j$ which anticommutes with it (such as $E|0\rangle$ and $E|1\rangle$ are eigenvectors of $K_j$, with eigenvalue $-1$). c) For each error operator $E$ (in the $X_i, Y_i, Z_i$), the set of generators ${K_j}$ which anticommutes with $E$ is unique (there is not an other error operator with the same set of generators) –  Trimok Oct 14 '13 at 17:38
    
Now, practically, one may, for instance, try to restrict to generators build with $I,Z$ or $I,X$. –  Trimok Oct 14 '13 at 17:44
    
This might be a good opportunity to encourage posters to spell out abbreviations (at least once). Abbreviations may not be obvious to readers with a different background. –  Qmechanic Nov 1 '13 at 11:19

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