dg(tr t ) under the action of a CQCA T, because b = 1 . Starting from an arbitrary translation-invariant pure sta- bilizer state, the entanglement might decrease in the first k steps, e.g. if t k = b − $^{1}$, but then starts to increase lin- early with dg(tr t ). This behavior is shown in Figure 7. 8 sion relation for c ( k ) and d ( k ). c ( k + 1) = (tr t ) 2 $^{$^{k}$+1}$− 1 = (tr t ) 2 $^{k}$− $^{1}$(tr t ) 2 k = c ( k )(tr t ) 2 $^{k}$, d ( k + 1) = k +1 ∑ i =1 (tr t ) 2 $^{$^{k}$+1}$− 2 i = k i =1 (tr t ) 2 $^{k}$+2$^{k}$− 2 i + (tr t ) 0 = d ( k )(tr t ) 2 k + 1 . lim k →∞ 1 2 $_{k}$dg( t 2 k b ( 0 1 ) = lim k →∞ 1 2 $_{k}$dg( a · c k ) + b · d ( k )) = lim k →∞ 1 2 $_{k}$dg( a · c ( k − 1)(tr t ) 2 $^{k$_{−}$}$1 + b · d ( k − 1)(tr t ) 2 $^{k$_{−}$}$1 + b ) = lim k →∞ 1 2 $_{k}$dg((tr t ) 2 $^{$^{k$_{−}$}$1}$r ( k − 1)) = lim k →∞ 1 2 k ( dg(tr t ) 2 k 2 + dg( r ( k − 1)) ) = lim k →∞ ( k − k$_{0}$ ∑ i =1 1 2 $_{i}$dg(tr t ) + lim k →∞ 1 2 $_{k}$dg( r ( k$_{0}$ )) ︸ ︷︷ ︸ =0 = dg(tr t ) . Now we are able to calculate the limit: In the tripartite setting, again starting from the “all- spins-up” state, the entanglement between the finite re- gion and the rest of the chain grows under the action of a CQCA T with 2 · dg(tr t ) until it reaches L . Then it remains constant. If we start with a general translation- invariant stabilizer state, again the entanglement might decrease at first. After some time it starts increasing and reaches L , where it remains if the CQCA is not periodic. The entanglement generation is twice as fast as for the bipartite setting, because we have two cuts. Results are shown in Figure 8. In both cases, the asymptotic entanglement generation only depends on the degree of the trace of the automa- ton. A CQCA with gliders generates entanglement just as fast as a fractal automaton if the degree of the trace is the same. Of course, periodic automata destroy all en- tanglement they just generated in the next steps. This corresponds to the fact that their trace is a constant. The entanglement generation rate is however not directly gov- erned by the neighborhood of the CQCA. Of course, the size of the neighborhood bounds the possible rate of en- tanglement generation from above, but even automata with a huge neighborhood can be periodic and thus gen- erate no entanglement at all. It is worth mentioning that CQCAs saturate the bound on the entanglement generation rate for translation- invariant operations acting on translation-invariant states derived in [2]. This means that there is no This means that starting from the “all-spins-up” stabi- lizer product state, the entanglement grows linearly with Proof. By Theorem III.3, every translation-invariant sta- bilizer state with stabilizer generator length 2 n + 1 en- tangles n qubit pairs maximally with respect to any bi- partite cut. Lemma IV.2 shows that under the action of a CQCA T the length of the stabilizer generators grows asymptotically with dg(tr t ) qubit pairs per step. This implies the proposition. Theorem IV.3. The asymptotic bipartite entanglement generation rate (maximally entangled qubit pairs per step) of a general centered CQCA T is the degree of its trace polynomial, dg(tr t ) . Now we only have to put Theorem III.3 together with Lemma IV.2 to prove the following theorem: In the third step we used that dg( b ) < dg( t 2 $^{k}$b ( 0 1 ) , ∀ k ≥ k$_{0}$ . In the second case we have dg( t 2 $^{k}$b ( 0 1 ) ≤ dg( b ) , ∀ k . Thus, dg( t 2 $^{k}$b ( 0 1 ) ) is bounded by dg( b ). Therefore, n ( t, ξ ) is bounded and Δ n Δ t = 0. But n ( t, ξ ) bounded also implies t periodic and therefore dg(tr t ) = 0. Thus, we have Δ n Δ t = 0 = dg(tr t ). This completes the proof. Figure 7: Entanglement generation for the stabilizer state with w ( ξ ) = Y ⊗ X ⊗ Y in a bipartite setting with differ- ent CQCAs. One can see that entanglement can also be de- stroyed, but grows asymptotically linearly with the number of time steps t . The rate is given by the degree of the trace of the CQCA matrix. per iod i c C Q C A, tr = 0 T$_{F}$ , t r = u $_{−}$1 + 1 + u 2- s te p gl i der C Q C A , t r = u $_{−}$2 + u 2 T$_{G}$ , tr = u $_{−}$1 + u t E ( t ) 20 15 10 5 0 40 30 20 10 0