Editing Quantum error correction (section)

==Bit-flip code==

The repetition code works in a [[Classical information channel|classical channel]], because classical bits are easy to measure and to repeat. This approach does not work for a quantum channel in which, due to the [[no-cloning theorem]], it is not possible to repeat a single qubit three times. To overcome this, a different method has to be used, such as the ''three-qubit bit-flip code'' first proposed by Asher Peres in 1985.<ref>{{cite journal
 | last = Peres
 | first = Asher
 | title = Reversible Logic and Quantum Computers
 | journal =  Physical Review A
 | volume = 32
 | issue = 6
 | pages = 3266–3276
 | year = 1985 
 | doi = 10.1103/PhysRevA.32.3266
 | pmid = 9896493
 | bibcode = 1985PhRvA..32.3266P
 }}</ref> This technique uses [[Quantum entanglement|entanglement]] and syndrome measurements and is comparable in performance with the repetition code.

[[File:Quantum error correction of bit flip using three qubits.svg|upright=1.35|thumb|right|[[Quantum circuit]] of the bit flip code]]
Consider the situation in which we want to transmit the state of a single qubit <math>\vert\psi\rangle</math> through a noisy [[Quantum channel|channel]] <math>\mathcal E</math>. Let us moreover assume that this channel either flips the state of the qubit, with probability <math>p</math>, or leaves it unchanged. The action of <math>\mathcal E</math> on a general input <math>\rho</math> can therefore be written as <math>\mathcal E(\rho) = (1-p) \rho + p\cdot\rho </math>.

Let <math>|\psi\rangle = \alpha_0|0\rangle + \alpha_1|1\rangle</math> be the quantum state to be transmitted. With no error-correcting protocol in place, the transmitted state will be correctly transmitted with probability <math>1-p</math>. We can however improve on this number by ''encoding'' the state into a greater number of qubits, in such a way that errors in the corresponding logical qubits can be detected and corrected. In the case of the simple three-qubit repetition code, the encoding consists in the mappings <math>\vert0\rangle\rightarrow\vert0_{\rm L}\rangle\equiv\vert000\rangle</math> and <math>\vert1\rangle\rightarrow\vert1_{\rm L}\rangle\equiv\vert111\rangle</math>. The input state <math>\vert\psi\rangle</math> is encoded into the state <math>\vert\psi'\rangle = \alpha_0 \vert000\rangle + \alpha_1 \vert111\rangle</math>. This mapping can be realized for example using two CNOT gates, entangling the system with two [[Ancilla bit#Ancilla qubits|ancillary qubits]] initialized in the state <math>\vert0\rangle</math>.<ref>{{cite book |last1=Nielsen |first1=Michael A. |last2=Chuang |first2=Isaac L. |author-link1=Michael A. Nielsen |author-link2=Isaac L. Chuang |year=2000 |title=Quantum Computation and Quantum Information |publisher=Cambridge University Press}}</ref> The encoded state <math>\vert\psi'\rangle</math> is what is now passed through the noisy channel.

The channel acts on <math>\vert\psi'\rangle</math> by flipping some subset (possibly empty) of its qubits. No qubit is flipped with probability <math>(1-p)^3</math>, a single qubit is flipped with probability <math>3p(1-p)^2</math>, two qubits are flipped with probability <math>3p^2(1-p)</math>, and all three qubits are flipped with probability <math>p^3</math>. Note that a further assumption about the channel is made here: we assume that <math>\mathcal E</math> acts equally and independently on each of the three qubits in which the state is now encoded. The problem is now how to detect and correct such errors, while not corrupting the transmitted state''.''

[[File:Fidelity Error Correction Bit Flips.svg|thumb|upright=1.7|Comparison of output ''minimum'' fidelities, with (red) and without (blue) error correcting via the three qubit bit flip code. Notice how, for <math>p\le 1/2</math>, the error correction scheme improves the fidelity.]]

Let us assume for simplicity that <math>p</math> is small enough that the probability of more than a single qubit being flipped is negligible. One can then detect whether a qubit was flipped, without also querying for the values being transmitted, by asking whether one of the qubits differs from the others. This amounts to performing a measurement with four different outcomes, corresponding to the following four projective measurements:<math display="block">\begin{align}
    P_0 &=|000\rangle\langle000|+|111\rangle\langle111|, \\
    P_1 &=|100\rangle\langle100|+|011\rangle\langle011|, \\
    P_2 &=|010\rangle\langle010|+|101\rangle\langle101|, \\
    P_3 &=|001\rangle\langle001|+|110\rangle\langle110|.
\end{align}</math>This reveals which qubits are different from the others, without at the same time giving information about the state of the qubits themselves. If the outcome corresponding to <math>P_0</math> is obtained, no correction is applied, while if the outcome corresponding to <math>P_i</math> is observed, then the Pauli ''X'' gate is applied to the <math>i</math>-th qubit. Formally, this correcting procedure corresponds to the application of the following map to the output of the channel:
<math display="block">\mathcal E_{\operatorname{corr}}(\rho)=P_0\rho P_0 + \sum_{i=1}^3 X_i P_i \rho\, P_i X_i.</math>

Note that, while this procedure perfectly corrects the output when zero or one flips are introduced by the channel, if more than one qubit is flipped then the output is not properly corrected. For example, if the first and second qubits are flipped, then the syndrome measurement gives the outcome <math>P_3</math>, and the third qubit is flipped, instead of the first two. To assess the performance of this error-correcting scheme for a general input we can study the [[Fidelity of quantum states|fidelity]] <math>F(\psi')</math> between the input <math>\vert\psi'\rangle</math> and the output <math>\rho_{\operatorname{out}}\equiv\mathcal E_{\operatorname{corr}}(\mathcal E(\vert\psi'\rangle\langle\psi'\vert))</math>. Being the output state <math>\rho_{\operatorname{out}}</math> correct when no more than one qubit is flipped, which happens with probability <math>(1-p)^3 + 3p(1-p)^2</math>, we can write it as <math>[(1-p)^3+3p(1-p)^2]\,\vert\psi'\rangle\langle\psi'\vert + (...)</math>, where the dots denote components of <math>\rho_{\operatorname{out}}</math> resulting from errors not properly corrected by the protocol. It follows that <math display="block">F(\psi')=\langle\psi'\vert\rho_{\operatorname{out}}\vert\psi'\rangle\ge (1-p)^3 + 3p(1-p)^2=1-3p^2+2p^3.</math>This [[Fidelity of quantum states|fidelity]] is to be compared with the corresponding fidelity obtained when no error-correcting protocol is used, which was shown before to equal <math>{1-p}</math>. A little algebra then shows that the fidelity ''after'' error correction is greater than the one without for <math>p<1/2</math>. Note that this is consistent with the working assumption that was made while deriving the protocol (of <math>p</math> being small enough).