On page 26 of arXiv:quant-ph/0504218, it is written that in Knill's method of fault-tolerant error correction (FTEC), the output of the transversal bell measurement becomes $(P_m \otimes I) | \Phi_0 \rangle ^{\otimes n}$, and $P_m$, which is the necessary Pauli operator for teleportation, completes the teleportation by acting on the second ancilla block. However, I don't understand why $P_m$ completes the teleportation. I can understand why this would be the case in normal teleportation using three physical qubits, but I don't understand why it can be written in the same form in this case where we are performing transversal bell measurement using three logical qubits. Also, it seems that Knill's method can be applied not only to CSS codes, but also to others. So, does this mean that the transversal CNOT is not applying a logical CNOT?
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$\begingroup$ Your link to the paper doesn't work. $\endgroup$– Marco Fellous-AsianiAug 4 at 17:36
1 Answer
I can understand why this would be the case in normal teleportation using three physical qubits, but I don't understand why it can be written in the same form in this case where we are performing transversal bell measurement using three logical qubits.
Not quite sure how to convince you that this works. The whole point is creating a $[[n,1,d]]$ code is that the logical states, composed of $n$-qubits, operate in the same way as acting on a single (k=1) physical qubit. Of course, as long as you only apply logical operations on them. If you have multiple of those logical qubits, and you apply logical operations to them, they will still function as expected.
Perhaps you should write a simulation of this and see that it indeed works.
Also, it seems that Knill's method can be applied not only to CSS codes, but also to others. So, does this mean that the transversal CNOT is not applying a logical CNOT?
I haven't worked closely with Knill's method, but it seems that if you want it to be fault-tolerant, you will need the logical CNOT to be transversal. While CSS codes always have a transversal CNOT, non-CSS codes can have it as well with some caveats (see answers here).
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$\begingroup$ If a transversal CNOT is a logical CNOT, then I think it should be performing a logical Bell measurement, so the output should be $(P_m \otimes I) | \bar{\Phi_0} \rangle $. Why is it $(P_m \otimes I) | \Phi_0 \rangle ^{\otimes n}$? I'm confused because I think the tensor product of $n$ Bell states is not the same as a logical Bell state. $\endgroup$– kongJul 2 at 20:20
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$\begingroup$ @kong We are not creating the tensor product of $n$ Bell states. We are creating a $2n$ qubit state of the form $(|\bar{0}\rangle_n |\bar{0}\rangle_n + |\bar{1}\rangle_n |\bar{1}\rangle_n)/\sqrt{2}$, where $|\bar{i}\rangle_n$ for $i=0,1$ are the logical basis states of the code, each composed to $n$ qubits. $\endgroup$ Jul 2 at 20:33