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1

You can simply control $X$ gates with qubits $q3$ and $q4$. You DO NOT have to measure them firstly and then use classical bits for controlling. The reason is that in quantum computing, controling some qubit with other qubits or with their measured results in classical register is the same. Hence, you can implement the algorithm on real quantum computer.

2

I don't think that this will be possible on real current quantum hardware. An alternative would be to run it on a simulator with a realistic noise model. This means that the circuit will be run in a non-ideal environment, and so should incur errors similar to how it would if it was executed on a real device. This tutorial teaches you how to build a noise ...

1

AFAIK, this is impossible on IBM's current hardware. See this github issue: https://github.com/Qiskit/qiskit-textbook/issues/119

2

I will attempt to provide some insight regarding your first question. For starters, both quantum surface codes and quantum block codes are stabilizer codes, which means that although they are significantly different in terms of their construction and utility, they still share some common ground. With regard to which code family is more promising, I believe ...

0

I have added the layout from the paper. A Z error on Db will fire Xb and Xa. A Z error on Dc will fire Xa. Thus these two are distinguishable. If a X error occurs on Dc this will fire Zb. This can be corrected by applying X on Dc. If a X error occurs on Db this will also fire Zb. It is also corrected by applying X on Dc. At the end Db and Dc have been ...

6

The errors that are described by the Master equation are continuous errors. The action of error correction is to discretize those errors. For example, dephasing noise can be described by the Master equation. The net effect is that an initial state $\rho$ is transformed into $$\rho\mapsto (1-p)\rho+pZ\rho Z,$$ where $p$ is a function of time. However, ...

1

Distinguishing $X$ and $Z$ errors is easy. $X$ errors anti-commute with the $Z$-type stabilizers, and so when you perform a measurement of those parity checks, you get and answer '1'. Similarly, $Z$ errors give you a '1' answer only on the $X$-type parity checks. Also, note that, in the bulk (i.e. not on the edges), you never get a '1' on only one weight-4 ...

3

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 ...

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 ...

3

Each vertex has a physical data qubit. But what exactly do the operators (green circles) represent? Is there an X and Z operator on every vertex site, or only some of them (Figure 1b)? The circles on, e.g. the bottom-left of Fig. 1, show you how to describe each of the stabilizers. So, for every mustard yellow square in the top-left diagram, there is a ...

2

First of all, are the ancilla qubits entangled with $|\psi\rangle$? Yes, depending on what the state is. Let's say you started with $$\alpha|000\rangle+\beta|111\rangle,$$ but it has experienced an error of $(\cos\theta I+i\sin\theta X)$ on the first qubit. So, your state becomes $$\cos\theta(\alpha|000\rangle+\beta|111\rangle)+i\sin\theta(\alpha|100\... 1 You can initialize a qubit to any arbitrary state by gate U3 (abbreviation used on IBM Q):$$ 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} $$It is also possible to prepare any multi-qubit quantum ... -5 It was my understanding that error-correction does not really apply to quantum computing. The elements in the set are not numbers, they are probabilities. Probability elements are never in one state or another. They are always in all states. That is the beauty of quantum computing. 5 Quantum error correction concerns errors that happen on qubits; it does not provide any protection against errors in operations on those qubits. Note however, that an error on an operation can be seen as the perfect operation plus some error 'on the qubit'. It is, however, the case that, without any precaution, the added operations introduced by error ... 3 AFAIK, what you want can't be run on the hardware right now. See this github issue. However, you can do this in the simulators. If, for example, c[0] and c[1] make up a two-bit classical register c, you can do this: qc.x(q[0]).c_if(c,3) qc.x(q[1]).c_if(c,3) 0 You will need to know how long it takes for each gate of the circuit to be performed. Then the decoherence error rate is simply$$e^{t_{gate}/t_{decoherence}}

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