I am going through the idea of stabilizer formalism.

Defined what is a Pauli group $P_n$ and its properties, we describe a stabilizer set $S$ as: $$S\subset P_n$$

The stabilizer set establishes valid codewords for a state if the equation $$s\left|\psi\right\rangle=\left|\psi\right\rangle,\;\;\;\forall s \in S \;\;\;\;\; (1)$$ is satisfied. That means $\left|\psi\right\rangle$ is a +1 eigenstate of $s$.

Each valid codeword belongs to $V$, that is a set of qubits stabilized by $S$. Therefore, if $(1)$ is satisfied, then $\left|\psi\right\rangle \in V$.

Let's consider the Steane code of 7 qubits. The followings are the stabilizer codes for such encode: $$ K^1 = IIIXXXX $$ $$ K^2 = XIXIXIX $$ $$ K^3 = IXXIIXX $$ $$ K^4 = IIIZZZZ $$ $$ K^5 = ZIZIZIZ $$ $$ K^6 = IZZIIZZ $$

These reduce the $2^7$ Hilbert space into a two-dimensional subspace. These stabilizers generate valid codewords for the Steane code: $$ \left|0\right\rangle_L \equiv \frac{1}{\sqrt{8}}(\left|0000000\right\rangle + \left|1010101\right\rangle + \left|0110011\right\rangle + \left|1100110\right\rangle + \left|0001111\right\rangle + \left|1011010\right\rangle + \left|0111100\right\rangle + \left|1101001\right\rangle) $$ $$ \left|1\right\rangle_L \equiv \frac{1}{\sqrt{8}}(\left|1111111\right\rangle + \left|0101010\right\rangle + \left|1001100\right\rangle + \left|0011001\right\rangle + \left|1110000\right\rangle + \left|0100101\right\rangle + \left|1000011\right\rangle + \left|0010110\right\rangle) $$

Here my doubt come; Each stabilizer is used as "filters of the input", so if an input, on which are applied one or more of those stabilizers, does not satisfy the equation $(1)$ ($\left|\psi\right\rangle$ -1 eigenvalue of $s$?), then we can say that an error occurred. Through syndrome measurement we can identify where the error occurred and correct it.

Another issue: verifying $(1)$ means, for example, $\;K^1 \left|1010101\right\rangle = \left|1011010\right\rangle$. Since both $\left|1010101\right\rangle$ and $\;\left|1011010\right\rangle$ represent $\left|0\right\rangle_L$, we say that $(1)$ is satisfied?

Finally: $\;K^4 \left|1010101\right\rangle = ?$

Thank you.

Added Last trouble:

Quantum circuit to prepare the [[7,1,3]] logical |0⟩ state

The state of the system is represented by:

$$\left|\psi\right\rangle_F={1\over 2}(\left|\psi\right\rangle_I+U\left|\psi\right\rangle_I)\left|0\right\rangle + {1\over 2}(\left|\psi\right\rangle_I-U\left|\psi\right\rangle_I)\left|1\right\rangle$$

We apply $K^1,K^2,K^3$ to the input and we measure the ancilla qubits to verify the integrity of the input (if $\left|\psi\right\rangle_I$ is +1 eigenstate of $K^1,K^2,K^3$). If the equation $(1)$ is not satisfied, then the corrupted qubit is corrected with a $Z$ gate addressed by syndrome measurement of ancilla qubits. This is how does the system work?


1) If there is an error $E_j$, the new states $E_j|0\rangle_L$ and $E_j|1\rangle_L$ are eigenvectors, with eigenvalue $-1$, of all the stabilizers $s_j$ belonging to some set subset $S_j$ of $S$. (the elements of $S_j$ anticommute with $E_j$).This subset $S_j$ identifies uniquely the error $E_j$.

2) $|0\rangle_L$ and $|1\rangle_L$ are eigenvectors, with eigenvalue $1$, of all the stabilizers $s$ belonging to $S$ (this is not true for the "components" of $|0\rangle_L$ and $|1\rangle_L$ like, for instance, $|1010101\rangle$).For a stabilizer $s$, you just calculate $s|0\rangle_L$ and $s|1\rangle_L$, and you check that the result is $|0\rangle_L$ or $|1\rangle_L$.

For instance :

$K^1\left|0\right\rangle_L = (IIIXXXX) \\\frac{1}{\sqrt{8}}(\left|0000000\right\rangle + \left|1010101\right\rangle + \left|0110011\right\rangle + \left|1100110\right\rangle + \left|0001111\right\rangle + \left|1011010\right\rangle + \left|0111100\right\rangle + \left|1101001\right\rangle)= \\ \frac{1}{\sqrt{8}}(\left|0001111\right\rangle + \left|1011010\right\rangle + \left|0111100\right\rangle + \left|1101001\right\rangle + \left|0000000\right\rangle + \left|1010101\right\rangle + \left|0110011\right\rangle + \left|1100110\right\rangle)\\ =\left|0\right\rangle_L$

3) $K^4 \left|1010101\right\rangle = IIIZZZZ |1010101\rangle$. With $Z |0\rangle = |0\rangle$, and $Z |1\rangle = -|1\rangle$, you get :

$K^4 \left|1010101\right\rangle = |1010101\rangle$

  • $\begingroup$ it is clear. Thank you. I edited my question with an example, perhaps you can tell me if I am wrong or not $\endgroup$ – camillo_benso Oct 25 '13 at 13:00
  • $\begingroup$ @AntonioVerlotta : Please, make a new question for this. $\endgroup$ – Trimok Oct 25 '13 at 17:38

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