3
$\begingroup$

The appendix to the paper Graph States as a Resource for Quantum Metrology states that when graph states subjected to finite erasures, $$G\Rightarrow Tr_\vec{y}G.$$ While more explicitly he explains the formula as $$G_\vec{y} = \mathrm{const}\times\sum_{\vec{y}}Z_\vec{j}\lvert G\rangle\langle G\rvert Z_\vec{j}.$$ I was wondering, how can these two statements the same, i.e., why the erasure error can be stated as the operation of $Z$? I know $I + Z$ can be the erasure error, but how can $Z$ alone stand for the erasure error?

$\endgroup$
5
  • $\begingroup$ Have you checked section $6$ (specifically proposition $8$ on page $46$) of the review article "entanglement in graph states"? The proof is on the next page. $\endgroup$
    – JSdJ
    Apr 11, 2021 at 10:01
  • $\begingroup$ I've checked, and the question is amended, avoiding passing the wrong meaning. $\endgroup$
    – narip
    Apr 11, 2021 at 10:26
  • $\begingroup$ I just realized you changed the question - so that makes my answer moot. Have you tried factoring out the $\frac{1}{2}(I + Z)$? If you delete for instance $2$ qubits, you'll get a summation over $II, ZI, IZ$ and $ZZ$ - exactly a sum over $j \in \{00,01,10,11\}$, or in other words every combination of $0$'s and $1$'s for the traced away qubits. $\endgroup$
    – JSdJ
    Apr 11, 2021 at 10:43
  • $\begingroup$ @narip Hi Narip is the reasoning of the final paragraph of the accepted answer below clear to you. In particular how "This is $2^{|L_y|}\sigma_x$ if $x'$ is 0 on all the sites $y$ and its neighbours. It's 0 otherwise" is obtained? $\endgroup$
    – John Doe
    Apr 4 at 7:19
  • $\begingroup$ @JohnDoe Hi. I didn't thoroughly reread the paper. However, I hope the following will be helpful to you. If $x^\prime$ is zero, then all the summands in term $\sum_{z\in \{0,1\}^n}^{\prime}(-1)^{x^\prime\cdot z}$ will be $1$. Therefore, the result of this summation will be how many terms are in $\sum^\prime$. Otherwise, if there's one $1$ in the summation, there will always be one $-1$. $\endgroup$
    – narip
    Apr 6 at 14:28

1 Answer 1

4
$\begingroup$

The idea of erasure being projection onto $|0\rangle$ is perhaps misleading in this context (my fault for mentioning it in a comment without having looked at the full details of what this specific paper did). This paper does not project the set of qubits $y$ onto the $|0\rangle$ state. Instead, they trace out those qubits. Perhaps the best way of writing this, to be more consistent with the notation is $$ G\rightarrow \text{Tr}_yG\otimes \frac{I_y}{2^{|y|}}. $$ If you think of $G$ as some sum of terms in the Pauli basis, $$ G=\sum_{x\in\{0,1,2,3\}^n}\alpha_x\sigma_x, $$ then what the partial trace is doing is selecting all the $x$ for which the $y$ components of $\sigma_x$ are all $I$.

So, let us consider terms $$ \sum_{z\in\{0,1\}^{n}}'Z_z\sigma_xZ_z. $$ I'm using the notation $\sum'$ to denote the fact that while $z\in\{0,1\}^n$, I'm only summing over terms where $z$ is 0 on every site not specified by $y$ or the neighbours of $y$ (i.e. those not in the set $L_y$ in the paper's notation). There are two possibilities for the fixed $x$ and a particular $z$ - either $Z_z$ commutes or anti-commutes with $\sigma_x$. $$ Z_z\sigma_xZ_z=(-1)^{x'\cdot z}\sigma_x. $$ Here I'm using $x'$ to denote a binary string derived from $x$ which is 1 for a given site if $x$ yielded an $X$ or $Y$. Thus, $$ \sum_{z\in\{0,1\}^{n}}'Z_z\sigma_xZ_z=\sigma_x\sum_{z\in\{0,1\}^{n}}'(-1)^{x'\cdot z}. $$ This is $2^{|L_y|}\sigma_x$ if $x'$ is 0 on all the sites $y$ and its neighbours. It's 0 otherwise. This immediately excludes any stabilizer (product of generators) that contains a generator (of the form $XZZZ\ldots Z$) where the $X$ acts on one of the sites to be traced out, or one of its neighbours. Since all generators act on a single vertex and its neighbours, this means that the only terms remaining do not act at all on the sites being traced out, i.e. they are $I$ on those sites. Exactly the set of stabilizers you were trying to select.

$\endgroup$
1
  • $\begingroup$ Could you please elaborate on how you get "This is $2^{|L_y|}\sigma_x$ if $x'$ is 0 on all the sites $y$ and its neighbours. It's 0 otherwise." $\endgroup$
    – John Doe
    Apr 4 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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