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I already got part of the answer in thread. Which says that if I want to perform a state tomography on a known state, the estimation can be simplified.

In the case of "GHZ-class", citing from the thread above: if the state is $|\psi_{a,b}\rangle= a|000\rangle+b|111\rangle$, then $$\langle XXX\rangle = 2ab, \qquad \langle YYY\rangle = 0, \qquad \langle ZZZ\rangle = a^2-b^2,$$

Now the point is that I am actually trying to optimise QPT for logical computing, for the case of operation between two logical qubits, both encoded in GHZ-class states.

The computation is also logical, so I wonder whether the above equations still hold after the gates.

Daniele Cuomo
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  • I don't quite understand the question. What does it mean "this equation still hold after the gates"? What gates are you referring to? And what do you mean with "the computation is also logical"? – glS May 30 '22 at 09:15
  • I intentionally left the gates undefined with the only restriction to be logical. This means that after computation, the state is still logical -- i.e. it belongs the codeword space --. – Daniele Cuomo May 30 '22 at 11:48
  • What do you mean by 'both encoded in GHZ-states', i.e. encoding a qubit into a single state? – JSdJ May 30 '22 at 15:23
  • I got confused, I fixed it. I meant the classical encoding from 1 qubit to 3. – Daniele Cuomo May 30 '22 at 19:48

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