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3

It is not an error. Under the subsection IBM's modified definition, they define the quantum volume as: $\log_2 V_Q = \arg\max_{n \le N} \{\min [n, d(n)]\}$ That is, if you take the minimum of the number of qubits and the circuit depth over those qubits (i.e. $10$, because the circuit may well be deeper than the number of qubits, so the number of qubits is ...

1

The circuit in the diagram is for fault-tolerantly measuring the error syndrome of the logical qubit $|\psi_L\rangle$. I.e., $|\psi_L\rangle$ is the logical qubit (a qubit encoded into 7 qubits using the Steane code) whose errors you want to correct. So removing it wouldn't make sense.

1

I'm not sure how possible what you described is. I think the best you can possibly do is to project into the $n$ qubit symmetric space by applying $$\text{SymmetricProjector}(n) = \sum_{k=0}^{n} \left|{n \atop k}\right\rangle\left\langle{n \atop k}\right|$$ where the "n choose k state" is the superposition of all n-bit numbers with k bits set: \...

1

I believe what you try to do is impossible, because it violates the no-cloning theorem. Note that what you want to do is not the standard way to do quantum error correction, because you have $n-1$ copies of a state, not an encoded state. If you have some more knowledge about the state, you might want to look into stuff like magic state distillation, e.g. ...

0

Well, You should know Logical qubits are having topological feature. those are not spanning in a single state like you said. Logical qubit has to be written in superposition states. like this. this is arbitrary state for shor code with distance 3. and if you are wondering about CNOT operation between two Logical qubits read this paper. https://arxiv.org/abs/...

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I guess you are confusing Stabilizer types. Brown plaquettes(X Stabilizer) & Yellow plaquettes(Z Stabilizer) So if there is Error Z_topleft or Error Z_beneath, it gives same Syndrome. if this is your question, Imagine if there are only Z Error occurred on the Rotated Surface Code ( Figure c). Z_topleft Z_beneath is one of stabilizer. So we can choose ...

2

You can use stim for this, although you do have to write the stabilizer projection procedure for yourself. Write some methods to project a system into the +1 eigenstate of several stabilizers: from typing import List import stim def find_compatible_tableau(stabilizers: List[stim.PauliString]) -> stim.Tableau: num_qubits = max(len(e) for e in ...

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