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

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

• You really should get an intro to qubits and calculation. This would take hours to discuss to be complete because you need to start from the wave functions to not do any hand waving. – b degnan Feb 20 '17 at 20:42
• Didn't some geezer say "If you think you understand quantum mechanics, you don't understand quantum mechanics?" – Paul Uszak Feb 20 '17 at 22:15
• A set of quadruplets is usually denoted using Xi ( ξ ) as seen here: arxiv.org/pdf/quant-ph/0509033.pdf. . If you want a start on how the particles relate to each other I recommend reading up on mixed states: en.wikipedia.org/wiki/Quantum_state#Mixed_states – J.A.K. Feb 20 '17 at 23:25

$\newcommand\ket[1]{\left|#1\right>}$ 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.