## 7.2 Random isometries

There is another invertible operation in quantum theory: an isometry, which is a combination of adding another quantum system and then applying a unitary transformation to the resulting composite system. So let us take a quick look at a simple generalisation of random unitaries, namely random isometries V_i, which give \rho\longmapsto \rho' = \sum_{i=1} p_i V_i \rho V^\dagger_i. An isometry V is similar to a unitary operator except that it maps states in the Hilbert space \mathcal{H} to states in a larger Hilbert space \mathcal{H}' (cf. {the appendix on isometries}(#isometries)). Here \mathcal{H} is associated with the input and \mathcal{H}'=\mathcal{H}_\mathcal{A}\otimes\mathcal{H} with the dilated system (i.e. the ancilla \mathcal{A} plus our system of interest). Isometries satisfy V^\dagger V=\mathbf{1}, and they are usually implemented by adding an ancilla in a fixed state and then applying a unitary operation to the resulting composed system. They can be then reversed by applying the inverse of that unitary and discarding the ancilla.

Now, if \mathcal{H}' is sufficiently larger than \mathcal{H}, and if the images \mathcal{H}'_i of \mathcal{H} in \mathcal{H}' under the different isometries V_i do not overlap102, then we can reverse the action of the channel: we can, at least in principle, perform a measurement on \mathcal{H}', defined by the partition \mathcal{H}'=\mathcal{H}'_1\oplus\mathcal{H}'_2\oplus\ldots, and find out which subspace contains the output state; once we know which subspace the input was sent to, we know which particular isometry V_k was applied by the channel; then we simply apply V^\dagger_k.

In order to see this consider the following simple, but important, example, which we will revisit several times in different disguises.

Alice constructs a quantum channel which is a mixture of four isometries. The input is a single qubit, and the output is a dilated system composed of three qubits. She prepares the input qubit in a state103 |\psi\rangle and then combines it with the two ancillary qubits which are in a fixed state |0\rangle|0\rangle. Then she applies one of the four, randomly chosen, unitary operations to the three qubits, to generate the following four isometries: \begin{aligned} V_1 &= |000\rangle\langle 0| + |111\rangle\langle 1| \\V_2 &= |001\rangle\langle 0| + |110\rangle\langle 1| \\V_3 &= |010\rangle\langle 0| + |101\rangle\langle 1| \\V_4 &= |100\rangle\langle 0| + |011\rangle\langle 1|. \end{aligned}

The three qubits, which form the output of the channel, are given to Bob, whose task is to recover the original state |\psi\rangle of the input qubit. In this scenario, Bob, who knows the four isometries, can find out which particular isometry was applied. He knows that104

• V_1 maps \mathcal{H} to \mathcal{H}'_1, which is a subspace of \mathcal{H}' spanned by |000\rangle and |111\rangle;
• V_2 maps \mathcal{H} to \mathcal{H}'_2, which is a subspace of \mathcal{H}' spanned by |001\rangle and |110\rangle;
• V_3 maps \mathcal{H} to \mathcal{H}'_3, which is a subspace of \mathcal{H}' spanned by |010\rangle and |101\rangle;
• V_4 maps \mathcal{H} to \mathcal{H}'_4, which is a subspace of \mathcal{H}' spanned by |100\rangle and |011\rangle.

Given that these subspaces are mutually orthogonal, and \mathcal{H}'=\mathcal{H}'_1\oplus\mathcal{H}'_2\oplus\mathcal{H}'_3\oplus\mathcal{H}'_4, Bob can perform a measurement defined by the projectors on these subspaces. For example, if Alice randomly picked V_2, then the input state |\psi\rangle=\alpha_0|0\rangle+\alpha_1|1\rangle will be mapped to the output state \alpha_0|001\rangle+\alpha_1|110\rangle in the \mathcal{H}'_2 subspace. Bob’s measurement will then detect \mathcal{H}'_2 as the subspace where the output state resides, but the measurement (i.e. the corresponding projection) will not affect any state in that subspace. Bob can now simply apply V_2^\dagger and obtain |\psi\rangle.

Just in case you are curious (as you should be!), below is a diagram of how the four isometries are implemented. How would you reverse these operations? Single unitaries or isometries apart, it turns out that the only reversible, or correctable, channels (i.e. channels in which the input state can be recovered) are exactly the mixtures of mutually orthogonal isometries V^\dagger_i V_j=\delta_{ij}\mathbf{1}. We shall return to these channels later on.

1. That is, if the subspaces \mathcal{H}'_i are mutually orthogonal.↩︎

2. Our arguments here can be easily extended to any mixed state \rho, but for simplicity we consider the case of a pure state.↩︎

3. !!!to-do!!! picture↩︎