12.4 Some errors can be corrected on some states

In another one of their escapades, Alice prepares a quantum object in some state |\psi\rangle and sends it to Bob. But en route, the object is intercepted by a malicious third party, Eve, who changes its state by applying one of a prescribed list of unitary operations U_1,\ldots,U_n, chosen with probability p_1,\ldots,p_n, respectively. Let’s assume that Alice and Bob know the set of possible unitaries (which we can think of as errors), along with the associated probabilities, but they do not know which particular unitary operation was chosen by Eve in this run of their experiment. Can Bob reconstruct the state |\psi\rangle?

The answer is “yes, but only sometimes” — it depends on the original state |\psi\rangle.

Let \mathcal{H} be the Hilbert space pertaining to the object, and let \mathcal{C} be a subspace of \mathcal{H} that contains |\psi\rangle. Write \mathcal{C}_i to denote the subspace of \mathcal{H} given by the image of U_i applied to \mathcal{C}, i.e. \mathcal{C}_i \coloneqq \{U_i|\phi\rangle \mid |\phi\rangle\in\mathcal{C}\}. Suppose that these subspaces are all orthogonal to one another: for each |\phi\rangle\in\mathcal{C}, we have \langle\phi|U^\dagger_i U_j|\phi\rangle = \delta_{ij} (and if none of the U_i are equal to the identity, then we also ask for each \mathcal{C}_i to be orthogonal to \mathcal{C}). Assume that the unitary applied by Eve was U_k. Then all Bob has to do to reconstruct |\psi\rangle is

• perform a measurement, defined by projectors on the subspaces \mathcal{C}_i for i=1,\ldots n
• identify k
• apply U_k^\dagger.

Things hopefully become clearer with an example. Consider an object composed of three qubits, and define the subspace \mathcal{C} to be the span of the two basis vectors |000\rangle and |111\rangle. Suppose Eve applies one of the following four unitary operations, each with equal probability: \begin{aligned} 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} \\U_3 &= \mathbf{1}\otimes\mathbf{1}\otimes X \end{aligned} That is, the identity, or the bit-flip on the first, second, or third qubit. These four operations generate the four subspaces \begin{aligned} \mathcal{C} = \mathcal{C}_0 &= \Big\langle|000\rangle,|111\rangle\Big\rangle \\\mathcal{C}_1 &= \Big\langle|100\rangle,|011\rangle\Big\rangle \\\mathcal{C}_2 &= \Big\langle|010\rangle,|101\rangle\Big\rangle \\\mathcal{C}_3 &= \Big\langle|001\rangle,|110\rangle\Big\rangle. \end{aligned} which actually span the eight-dimensional Hilbert space \mathcal{H} of the three qubits, giving us a decomposition into orthogonal subspaces: \mathcal{H} \cong \mathcal{C}_0 \oplus \mathcal{C}_1 \oplus\mathcal{C}_2 \oplus \mathcal{C}_3

So suppose Alice prepares |\psi\rangle=\alpha|000\rangle+\beta|111\rangle (which is indeed a vector in \mathcal{C}), and Eve ends up applying U_3, 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 U_3^\dagger=\mathbf{1}\otimes\mathbf{1}\otimes X.