# Quantum Graph states hand computation

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.

• controlled-Z is symmetric. Applying it between $a$ and $b$ is the same as applying it between $b$ and $a$. Jun 22, 2021 at 13:50

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.
• The gate written in the paper is $S=|0><0| -i|1><1|$ Jun 22, 2021 at 14:22
• About the sign on $1111$ yes that is a typo Jun 22, 2021 at 14:24
• Then yes, there's a typo somewhere. I've used the standard definition of $S$. Jun 22, 2021 at 15:24