In the case of the simple three-qubit repetition code, the encoding consists of the mappings $|0\rangle \rightarrow\left|0_{\mathrm{L}}\right\rangle \equiv|000\rangle$ and $|1\rangle \rightarrow\left|1_{\mathrm{L}}\right\rangle \equiv|111\rangle$. For the bit flip channel there are four error syndromes, corresponding to the four projection operators:$$ \begin{aligned} P_{0} &=|000\rangle\langle 000|+| 111\rangle\langle 111|, \\ P_{1} &=|100\rangle\langle 100|+| 011\rangle\langle 011|, \\ P_{2} &=|010\rangle\langle 010|+| 101\rangle\langle 101|, \\ P_{3} &=|001\rangle\langle 001|+| 110\rangle\langle 110| . \end{aligned} $$ In Nilsen and Chuang's book, they stated the syndrome analysis is a measurement, e.g. $\langle\psi\mid P_1\mid\psi\rangle.$ But we only have one $\mid\psi\rangle$ at the beginning while four projection operators. I think the measurement should be like: enter image description here i.e., the way we would not disturb the original state. But it seems in the textbook, they didn't mention this kind of measurement. Did I understand right?
The reason I have this doubt is that I saw a picture in a paper:enter image description here where the $R$ in the picture stands for error recovery while there seems no measurement in $R$.

  • $\begingroup$ One of your concerns seem to be related to the fact you only have one state $|\psi\rangle$, but 4 projectors. You're comparing the wrong numbers here. You only need that the dimension of the Hilbert space that $|\psi\rangle$ is defined on be greater than or equal to the number of projectors. $\endgroup$
    – DaftWullie
    May 17, 2021 at 6:30
  • $\begingroup$ Could you link the paper in which you saw this circuit ? And I agree with the previous comment, can you explain why exactly having four projector operator but one state disturbs you ? $\endgroup$ May 17, 2021 at 7:54
  • $\begingroup$ The link you want for the paper is here. $\endgroup$
    – narip
    May 17, 2021 at 8:28

1 Answer 1


What you have is somewhat equivalent to the book in the following sense.

Your ancillae are here measuring the parity between your qubits. If you measure the ancillae in the computational basis you will find:

  • $|00\rangle$: all qubit have the same parity. The state of the data qubits is thus projected in $Im(P_0)$
  • $|01\rangle$: two first qubit have same parity but not the two last. The state of the data qubits is thus projected in $Im(P_3)$
  • $|10\rangle$: two first qubit don't have same parity, but two last do. The state of the data qubits is projected in $Im(P_1)$
  • $|11\rangle$: two first and two last qubits don't have the same parity. The state of the data qubits is in $Im(P_2)$

In conclusion:

Performing the circuit you drawed, and measuring the ancillae qubit will project your system in the space associated to one of your projector depending on the outcome you find on those ancillae. Then you can apply the appropriate recovery to correct.

[edit] To answer some of the issues about the projector.

In Q.M postulate we do not "choose" a projector when we measure. A measurement can be defined by a collection of projectors $\{P_k\}$ verifying $\sum_k P_k = I$. If you measure (and read the outcome), the density matrix will be transformed as:

$$ \rho \to P_k \rho P_k/ Tr(\rho P_k)$$ with probability $p_k = Tr(\rho P_k)$

In practice all the projectors play a role in the measurement. Indeed, as you see, you must compute a probability of outcome for each of the projector. In your example, in your comment you would find $Tr(\rho P_2)=0$ for $\rho=|\psi_e \rangle \langle \psi_e |$ that you defined. Thus such projection has a probability $0$ to occur.

In the context of this question, in the non autonomous scenario, if ancillae are found in $|00\rangle$, the density matrix of data qubits will become $P_0 \rho P_0/Tr(\rho P_0)$, if the ancillae are found in $|01\rangle$, the density matrix of data qubits will becomes $P_3 \rho P_3/Tr(\rho P_3)$ and so on.

  • $\begingroup$ In fact, I don't need to measure my ancilla. I only need to add some more control gates while the ancilla controls the first three lines(sorry that I didn't draw it directly in the picture of my question). But what I really don't understand is how can measurement in Nielsen's book be realized while I only have one initial state while having four projectors? $\endgroup$
    – narip
    May 17, 2021 at 6:18
  • $\begingroup$ I've added some content to my question. $\endgroup$
    – narip
    May 17, 2021 at 6:24
  • $\begingroup$ @narip Indeed you are not forced to measure them. If you apply gate between ancillary qubits and data qubits to correct for the errors it is called autonomous error correction. What I presented here is non autonomous case. About your question, can you explain in more details what disturbs you if we have more projector than state for a measurement ? There is no problem with that. $\endgroup$ May 17, 2021 at 7:55
  • $\begingroup$ I think I might have some misunderstanding in the reasoning next. If we have the initial state: $a\mid000\rangle+b\mid111\rangle$. After the error, it becomes $\mid\psi_e\rangle\equiv a\mid100\rangle+b\mid011\rangle$. If unluckily I choose the projector $P_2$ instead of $P_1$(the notation is the same as my question). The state after the measurement is $\frac{P_2\mid\psi_e\rangle}{\sqrt{\langle\psi_e\mid P_2\mid\psi_e\rangle}}$, which is exactly 0. Hence we lose the information of the initial state. $\endgroup$
    – narip
    May 17, 2021 at 8:32
  • 1
    $\begingroup$ @narip I added further info. Is it more clear ? $\endgroup$ May 17, 2021 at 11:19

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