11.6 Some errors can be corrected on some states

Alice prepares a quantum object in some state |\psi\rangle and sends it to Bob. The object is intercepted by a malicious Eve who changes its state by applying one of the prescribed unitary operations U_1,\ldots, U_n, with probabilities p_1,\ldots, p_n, respectively. Alice and Bob know the set of possible unitaries (errors), and the associated probabilities, but they do not know which particular unitary operation was chosen by Eve. Can Bob reconstruct the state |\psi\rangle? The answer is affirmative, at least for some states |\psi\rangle.

Let \mathcal{H} be the Hilbert space pertaining to the object, and let \mathcal{C} be a subspace of \mathcal{H}. Suppose |\psi\rangle\in\mathcal{C}, and that, for each vector in \mathcal{C}, we have \langle\psi|U^\dagger_i U_j|\psi\rangle = \delta_{ij} Any error U_k transforms the subspace \mathcal{C} into the subspace \mathcal{C}_k, which is orthogonal to \mathcal{C} and also to any other subspace \mathcal{C}_j for j\neq k. All Bob has to do is - perform a measurement, defined by projectors on the subspaces \mathcal{C}_j for j=1,\ldots n, - identify k, and - apply U_k^\dagger.

As an example, consider an object composed of three qubits and the subspace \mathcal{C} spanned by the two basis vectors |000\rangle and |111\rangle. Suppose Eve applies one of the following four unitary operations: U_0=\mathbf{1}\otimes\mathbf{1}\otimes \mathbf{1}, U_1 =X\otimes\mathbf{1}\otimes \mathbf{1}, U_2 =\mathbf{1}\otimes X\otimes \mathbf{1}, and U_3=\mathbf{1}\otimes\mathbf{1}\otimes X. That is, the identity, or bit-flip on the first, second, or third qubit. Each operation is chosen randomly with the same probability of 1/4. It is easy to see that the four operations generate four subspaces: \begin{aligned} \mathcal{C} = \Big\langle|000\rangle,|111\rangle\Big\rangle &\qquad \mathcal{C}_1 = \Big\langle|100\rangle,|011\rangle\Big\rangle \\\mathcal{C}_2 = \Big\langle|010\rangle,|101\rangle\Big\rangle& \qquad \mathcal{C}_3 = \Big\langle|001\rangle,|110\rangle\Big\rangle. \end{aligned} The eight dimensional Hilbert space of the three qubits is then decomposed into the sum of orthogonal subspaces \mathcal{C} \oplus \mathcal{C}_1 \oplus\mathcal{C}_2 \oplus \mathcal{C}_3 So suppose Alice prepares |\psi\rangle=\alpha|000\rangle+\beta|111\rangle and Eve applies the bit-flip to the third qubit. This generates the state \mathbf{1}\otimes\mathbf{1}\otimes X|\psi\rangle=\alpha|001\rangle+\beta|110\rangle\in \mathcal{C}_3. The projective measurement on these subspaces tells Bob that the new state is in the subspace \mathcal{C}_3, and hence the original state can be recovered by the operation \mathbf{1}\otimes\mathbf{1}\otimes X.