Logical qubit initialization for the surface code

Question

I am reading Fowler et al's paper on the surface code.. I do not understand how to initialize a logical qubit in an arbitrary state with the surface code. I do understand how to initialize the qubit in logical $|{0}\rangle$ and $|{1}\rangle$, but not how to initialize in an arbitrary superposition.

In Appendix B he shows that after the measurements in the circuit below,

the middle two qubits are in the one of the following four states:

Can they also be in a superposition of one of these states?

Answer 1

For $|0\rangle$, $|1\rangle$, $|+\rangle$, and $|-\rangle$ you do transversal initialization (initialize all physical qubits to the desired state, then turn on the stabilizers).

For $|i\rangle$ and $|-i\rangle$ you do topological initialization using twists.

For $T|+\rangle$ states and other states with nice state distillation protocols, you do noisy low code distance initialization followed by distillation at high code distances. Basically, just prepare a physical qubit into the desired state, pretend it's a surface code qubit with code distance one, and increase its code distance while hoping nothing went wrong. Then use a distillation procedure to ensure that nothing went wrong. Read A magic state's fidelity can be superior to the operations that created it and Efficient magic state factories with a catalyzed |CCZ> to 2|T> transformation for the state of the art.

For other single qubit states, you approximate them using a series of operations. Read Efficient synthesis of universal Repeat-Until-Success circuits for the state of the art.

Answer 2

I've not read the cited paper, so I don't know how this corresponds to anything that they say, but one way that I would think about it is, if I have an unknown qubit state stored on a single qubit, how do I copy this onto a surface code already initialised in logical 0?

Now, if it weren't logical qubits, we can easily write down a circuit that would accomplish this (left-hand circuit, which has output $|0\rangle|\psi\rangle$). If we express this in terms of Pauli operators (middle circuit), we can replace these with logical operators (right-hand circuit). It is this right-hand circuit that I would implement.

Answer 3

You can initialize a qubit to any arbitrary state by gate $U3$ (abbreviation used on IBM Q):

$$ U3(\theta,\phi,\lambda)= \begin{pmatrix} \cos(\theta/2) & -\mathrm{e}^{i\lambda} \sin(\theta/2) \\ \mathrm{e}^{i\phi}\sin(\theta/2) & \mathrm{e}^{i(\lambda+\phi)} \cos(\theta/2) \end{pmatrix} $$

It is also possible to prepare any multi-qubit quantum state by a method described in this paper: Transformation of Quantum States Using Uniformly Controlled Rotations.

In your case, the arbitrary states is probably that generated by Hadamard gate $H$.

Regarding measurement in your circuit: If the measurement is done in z-basis, the only four possibilites are those you listed in your question. Yes, qubits can be in superposition but after measurement, they end up in one of four combination of $|0\rangle$ and $|1\rangle$.