Quantum Graph states hand computation

Question

I am reading an article on Quantum Graph states. I wanted to ask a few questions. The Graph state is $$|G\rangle=\prod_{e\in G}CZ |+\rangle^{\otimes n}$$ Now my first question is if I apply the Controlled Z gate say between vertices $a$ and $b$ then do I also apply the Controlled Z between $b$ and $a$. I have done the calculation on 3 qubits with edges between vertices $(1,2)$ and $(1,3)$ with applying CZ between $(1,2)$ and $(1,3)$ and not $(2,1)$ and $(3,1)$. My calculations are $$|G\rangle=\prod_{e\in G}CZ |+\rangle^{\otimes 3}=|000\rangle+|010\rangle+|100\rangle-|110\rangle+|001\rangle+|011\rangle-|101\rangle-|111\rangle$$ Is this correct?

Further, the paper that I am reading said that if I apply the $S$ gate on the first qubit the state is changed to $$|0++\rangle+i|1--\rangle$$ but I am getting $$|0++\rangle-i|1--\rangle$$

The paper that I am reading is https://journals.aps.org/pra/abstract/10.1103/PhysRevA.78.042309.

Answer 1

The calculation in the paper is correct. One of the easiest ways to think about this is that you start all qubits in the $|+\rangle$ state, so $|+++\rangle=(|0\rangle+|1\rangle)|++\rangle)/\sqrt{2}$. Now you've got to co controlled-phase between 1 and 2, and 1 and 3. Let's take qubit 1 to be the control both times (controlled-phase is symmetric, so it doesn't matter which is the control, and which the target). So if qubit 1 is in $|0\rangle$, do nothing. If qubit 1 is in $|1\rangle$, apply $Z$ to the other two qubits. Now, $Z|+\rangle=|-\rangle$, so you get $$ |0++\rangle+|1--\rangle. $$

Now if you apply $S=|0\rangle\langle 0|+i|1\rangle\langle 1|$ to the first qubit, you get $$ |0++\rangle+i|1--\rangle. $$

Also note that in your state $|G\rangle$, the $|111\rangle$ term has the wrong sign - you get two -1 phases because of the (1,2) pair and the (1,3) pair.