Continuing from my last question that I posted about a paper on Graph States. I have another doubt regarding a possible typo error in the paper. Here it goes. The authors define a 'labeled state' as $$|G_{\vec{l}}\rangle=\bigotimes_{i}X_i^{l_{i1}}Z_i^{l_{i2}}|G'\rangle$$where G is a graph state. Then they define an 'encoded graph state' as $$ |G_{\vec{l}_{*2}}\rangle=\bigotimes_{i}Z_i^{l_{i2}}|G'\rangle $$ and say that the encoded graph state is the labeled graph state with $l_{1i}=0, \forall i$.

Here $\vec{l}_{i*}=(l_{i1},l_{i2})$ for the i-th vertex, $\vec{l}_{*j}=(l_{1j},l_{2j},....l_{nj})$ for the $j$ th bit over all the $n$ vertices, and $\vec{l}=(\vec{l}_{1*}, \vec{l}_{2*},........\vec{l}_{n*}$, each $l_{ij}\in \{0,1\}$.

My question is shouldn't the condition be $l_{i1}=0, \forall i$, because only then the X gate is removed. Can somebody check?

The link for the paper is https://journals.aps.org/pra/abstract/10.1103/PhysRevA.78.042309

  • $\begingroup$ Can you specify what is $\vec{l}$? And what are the superscripts $l_{i1}$ and $l_{i2}$? Is $\vec{l}$ a $n \times 2$ 'vector'? Is it some $n \times n$ matrix? If so, is it Hermitian/Symmetric? Because then clearly $l_{1i} = l_{i1}$. Also, is there an open-access version of the paper? $\endgroup$
    – JSdJ
    Jun 23, 2021 at 10:52
  • $\begingroup$ Here is the open access version arxiv.org/abs/0808.1532 $\endgroup$
    – Upstart
    Jun 23, 2021 at 11:36
  • 1
    $\begingroup$ Why is there a downvote on this question? $\endgroup$
    – Upstart
    Jun 23, 2021 at 16:20

1 Answer 1


Sure, I would say this is pretty clearly a typo. They very helpfully write out what they mean immediately after that statement so you can compare your understanding!

  • $\begingroup$ Okay, but aren't these typos very intricate to be typos in the first place. Since throughout the paper they are used. Similar was the case with the first typo in ths S gate. Anyways thanks for the clarification. $\endgroup$
    – Upstart
    Jun 24, 2021 at 11:13

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