Syndrome extraction operator as matrix?

Question

I am trying to understand how to achieve the syndrome extraction operator matrix in quantum repetition code (if it even exists). It is given that the syndrome is defined here (page 4) as:

[perform] controlled-nots from the first and second received qubits to the first ancilla qubit, then from the first and third received qubits to the second ancilla bit.

First of all, are the ancilla qubits entangled with $\vert \psi \rangle$?

Here is what I have understood so far. Given $\vert \psi \rangle = \alpha \vert x_1, x_2, x_3\rangle + \beta \vert \overline{x_1}, \overline{x_2}, \overline{x_3}\rangle$ (where $\overline{x_{i}} = x_i + 1 \quad (\text{mod }2)$), we can define an operator $S$ as:

$$S: \alpha \vert x_1, x_2, x_3\rangle + \beta \vert \overline{x_1}, \overline{x_2}, \overline{x_3}\rangle \rightarrow \underbrace{\vert x_1 \oplus x_2\rangle}_{\text{1$^{\text{st}}$ ancilla qubit}} \underbrace{\vert x_1 \oplus x_3\rangle}_{\, \text{ 2$^{\text{nd}}$ ancilla qubit}},$$

or equivalently as:

$$S: \alpha \vert x_1, x_2, x_3\rangle + \beta \vert \overline{x_1}, \overline{x_2}, \overline{x_3}\rangle \rightarrow \underbrace{\vert \overline{x_1} \oplus \overline{x_2}\rangle}_{\text{1$^{\text{st}}$ ancilla qubit}} \underbrace{\vert \overline{x_1} \oplus \overline{x_3}\rangle}_{\, \text{ 2$^{\text{nd}}$ ancilla qubit}}.$$

This will yield the ancilla values corresponding to the following states: $$ \begin{align*} \text{State$\qquad~~$} & \text{$~~~~$Ancilla qubits}\\ \alpha \vert 000\rangle + \beta \vert 111\rangle &\qquad\qquad \vert 00\rangle\\ \alpha \vert 100\rangle + \beta \vert 011\rangle &\qquad\qquad \vert 11\rangle\\ \alpha \vert 010\rangle + \beta \vert 101\rangle &\qquad\qquad \vert 10\rangle\\ \alpha \vert 001\rangle + \beta \vert 110\rangle &\qquad\qquad \vert 01\rangle\\ \alpha \vert 110\rangle + \beta \vert 001\rangle &\qquad\qquad \vert 01\rangle\\ \alpha \vert 101\rangle + \beta \vert 010\rangle &\qquad\qquad \vert 10\rangle\\ \alpha \vert 011\rangle + \beta \vert 100\rangle &\qquad\qquad \vert 11\rangle\\ \alpha \vert 111\rangle + \beta \vert 000\rangle &\qquad\qquad \vert 00\rangle\\\\ \end{align*} $$

How can I perform such operation using a matrix?

Answer 1

First of all, are the ancilla qubits entangled with $|\psi\rangle$?

Yes, depending on what the state is. Let's say you started with $$ \alpha|000\rangle+\beta|111\rangle, $$ but it has experienced an error of $(\cos\theta I+i\sin\theta X)$ on the first qubit. So, your state becomes $$ \cos\theta(\alpha|000\rangle+\beta|111\rangle)+i\sin\theta(\alpha|100\rangle+\beta|011\rangle). $$ When you perform syndrome extraction, you'll get $$ \cos\theta(\alpha|000\rangle+\beta|111\rangle)|00\rangle+i\sin\theta(\alpha|100\rangle+\beta|011\rangle)|11\rangle. $$ This state has entanglement between the computational qubits register and the ancilla register.

Of course, in the next step, you measure the ancilla qubits, and that projects the ancilla qubits into a basis state, removing all entanglement between system and ancillas.

How can I perform such operation using a matrix?

As for what the matrix is, I strongly recommend against performing such calculations. Once you're up to 5 qubits, you're talking a $32\times 32$ matrix, which is fairly awful. It is far better to just talk about the action on (relevant) basis states, as you did. However, since you have explicitly asked:

I calculated this in Mathematica using

cNOT[i_, j_] := 
 IdentityMatrix[32] + 
  KroneckerProduct[IdentityMatrix[2^(i - 1)], {{0, 0}, {0, 1}}, 
   IdentityMatrix[2^(j - i - 1)], {{-1, 1}, {1, -1}}, 
   IdentityMatrix[2^(5 - j)]]
TeXForm[cNOT[1, 4].cNOT[1, 5].cNOT[2, 4].cNOT[3, 5]]

(the first 3 qubits are the system qubits, and qubits 4&5 are the ancillas) giving the answer $$ \left( \begin{array}{cccccccccccccccccccccccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) $$

We can then verify the calculation that I showed before:

initial = {\[Alpha], 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \[Beta], 0, 0, 0};
witherror = 
 KroneckerProduct[{{Cos[\[Theta]], I Sin[\[Theta]]}, {I Sin[\[Theta]],
      Cos[\[Theta]]}}, IdentityMatrix[2^4]].initial
U.witherror