0

In the phase error correction code for three qubits, why do we need to use the hadamard gate? Can someone explain this with math. Quantum circuit for phase error correction code.

LittleBlue
  • 61
  • 4

1 Answers1

0

The maths is, effectively, $HZH=X$. In other words, a system with phase ($Z$) noise, and conjugated by Hadamards, is exactly the same as a system with $X$ noise on it. So if you understand how the bit flip encoding (circuit before the first Hadamards) and correction (circuit after the last Hadamards) detects & corrects bit flip errors, then you understand how this circuit detects & corrects phase errors.

DaftWullie
  • 57,689
  • 3
  • 46
  • 124
  • I know that part. But I wanted to abstract myself from bit flip code and understand the phase code independently. – LittleBlue May 12 '23 at 11:08
  • Then the important feature is that if you take the basis $|\pm\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$, the $Z$ operator maps between the two: $Z|+\rangle=|-\rangle$. This means that if you have an encoded state $\alpha|+++\rangle+|---\rangle$, a single $Z$ error will be detectable and correctable, effectively by doing a majority vote of elements in the $|\pm\rangle$ basis. – DaftWullie Feb 06 '24 at 13:32