1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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 at 10:52
  • $\begingroup$ Here is the open access version arxiv.org/abs/0808.1532 $\endgroup$
    – Upstart
    Jun 23 at 11:36
  • 1
    $\begingroup$ Why is there a downvote on this question? $\endgroup$
    – Upstart
    Jun 23 at 16:20
2
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ 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 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.