What are the quantum relations of four-particle GHZ states?

Question

How to represent a four particle GHZ states in quantum cryptography and what is the relation between them?

Answer 1

$\newcommand\ket[1]{\left|#1\right>}$ Since you ask about 4-particle GHZ state, I’ll assume you’re familiar with the more usual 3-particle GHZ state.

The 4-particle GHZ state is $(\ket{0000}+\ket{1111})/\sqrt2$. Like the 3-particle GHZ state, loosing a single particle transforms this state to a mere classical mixture of $000$ and $111$. The relation between the particles can be expressed by the stabilizer group generated by : $$\begin{gather} XXXX\\ZZII\\IZZI \\ IIZZ. \end{gather}$$ This group contains

1. the 6 permutations of $ZZII$, meaning, that if you measure any two particle in the $Z$ direction, you will obtain the same result
2. $XXXX$ : meaning that if you measure all particle in the $X$ direction, you will have an even number of them in each direction (i.e. all in the same direction, or two in one direction, 2 in the other, but never 3 in one direction and 1 in the other)
3. $YYYY$ : meaning that if you measure all particle in the $Y$ direction, you will have an even number of them in each direction.
4. $ZZZZ$ : meaning that if you measure all particle in the $Z$ direction, you will have an even number of them in each direction. (This is consistent with 1.)
5. the 6 permutations of $-XXYY$ : if you measure two particle in the $X$ direction, and two in the $Y$ direction, you will have an odd number of $-1$ results.

I thought combining 2,3, and 5 leads to a contradiction similar to the GHZ paradox, but I didn’t manage to find the correct combination, except that measuring any of the qubit the in $X$ or $Y$ direction project the 3 others in a 3-particle GHZ state.