# Graph States subjected to finite erasures

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?

• 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.
– JSdJ
Apr 11, 2021 at 10:01
• I've checked, and the question is amended, avoiding passing the wrong meaning. Apr 11, 2021 at 10:26
• 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.
– JSdJ
Apr 11, 2021 at 10:43
• @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? Apr 4 at 7:19
• @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 $-1$. Apr 6 at 14:28

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.
• 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." Apr 4 at 6:21