I want to get some information about, the state below:


in which $$a^2 + b^2 =1.$$

I want to know its name. Is there any article or information about it?


2 Answers 2


For Bell states, we have the even parity states $|00\rangle \pm |11\rangle$ and the odd parity states $|01\rangle \pm |10\rangle$.

These are more generally GHZ states, which what you wrote is a 3-qubit GHZ state. Following the Bell-state convention, I would imagine the most appropriate descriptor for $|101\rangle \pm |010\rangle$ to be an odd parity GHZ state. But it's not a state with any significant importance so it's not generally given a name.


There's not a common name for this specific state.

It's a GHZ state $|000\rangle + |111\rangle$ with the middle qubit flipped.

  • 1
    $\begingroup$ This answer is true for some $a$ and $b$. $\endgroup$
    – Mauricio
    Commented Feb 6, 2023 at 12:25
  • $\begingroup$ The question in the title asks about |000> + |111> (where a=b=1). $\endgroup$
    – hft
    Commented Feb 7, 2023 at 1:57

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