3 Figure 2: The existence of zero energy ground states for a qudit chain with d -dimensional qudits and r projectors per pair. We highlight two notable cases: d = 2 , r = 1 and d = 4 , r = 4 . This means that in case 1, it is still not possible to construct solutions for a long chain. In cases 2 and 3, we can construct sets of states with s$_{n}$ growing more slowly than D$_{n}$ . valid for all n . The question now is how to choose s$_{n}$ as we go along the chain. The only constraint on Γ [1] ,i$_{1}$ α$_{1}$ is linear independence, which requires s$_{1}$ ≤ d ( s$_{0}$ = 1 as the first pair have r constraints). If we choose the equality sign in the recur- sion relationship above at each step we obtain D$_{n}$ linearly independent zero energy states, where with dimensions $^{rs$_{n}$}$− $_{1}$× ds$_{n}$ . If ds$_{n}$ ≥ $^{rs$_{n}$}$− 1 and the ma- trix C has rank $_{1}$, the conditions (9) are independent and we can construct ds$_{n}$ − $^{rs$_{n}$}$− 1 linearly independent Γ i n $_{+1}$, [ n +1] α n ,α $^{n}$+1 , corresponding to solutions for the n +1 qudit chain (see the appendix for further discussion of the rank of C ). The freedom we have now is to use only a subset of them for constructing solutions. Thus, we obtain the formula s$_{n}$$_{+1}$ ≤ ds$_{n}$ − $^{rs$_{n}$}$− $_{1}$, (10) D$_{n}$ = $^{dD$_{n}$}$− 1 − $^{rD$_{n}$}$− $_{2}$, (11) for all n with D$_{0}$ = 1 and D$_{1}$ = d . The solution of this recursion relation is • For example when d ≤ r ≤ d 2 4 , the recursion rela- tion (10) can be satisfied also by s$_{n}$ = h n provided that h 2 − dh + r ≤ 0 . This requires g ≤ h ≤ f , so the lower bound on h is the least integer ≥ g . • On the other hand, for r < d the simple choice s$_{n}$ = 1 also satisfies the recursion (10). This means one can just solve the system from left to right as a linear system of equations. This results in a prod- uct state solution in the form which we can construct by starting with any ψ [1] and finding every ψ [ n +1] from the previous ones. ψ i$_{1}$i$_{2}$...i n = ψ [1] $^{,i$_{1}$}$ψ [2] ,i$_{2}$ · · · ψ [ n ] ,i n (12) There are three interesting regimes for r and d which yield different behaviors of D$_{n}$ (Figure 2): 2. r = d 2 4 results in D$_{n}$ = ( d 2 ) $^{n}$( n +1) , an exponential growth in n (except when d = 2 , which gives linear growth). 1. r > d 2 4 gives D$_{n}$ = r n 2 sin( n +1) θ sin θ with cos θ = d $_{2}$√$_{r}$ . D$_{n}$ becomes negative when n + 1 > π θ and thus no zero energy states can be constructed for a long chain if r > d 2 4 . 3. r < d 2 4 implies f > d 2 and f > g so for large n , D$_{n}$ ∼ f n ( 1 − g f ) − 1 and the number of zero energy states grows exponentially. III. NUMERICAL INVESTIGATION USING MATRIX PRODUCT STATES In this Section we numerically search for the ground states of our class of random projector Hamiltonians (4). We probe the relations obtained in the previous Section, and see how well the energy coming from our small- χ MPS imaginary time evolution converges to zero. The numerical technique we use is similar to Vidal’s [7, 15]. We use imaginary time evolution to bring the system from a known state to its ground state: | Ψ$_{grd}$ 〉 = $^{lim$_{τ}$}$→∞ e $_{−}$Hτ | Ψ$_{0}$ 〉 || e $_{−}$Hτ Ψ$_{0}$ 〉|| . In our numerical work we normal- ize the state after every time step [5]. We start from a uniform superposition of all the states and Trotterize by evolving first the odd pairs of sites and then the even pairs. Our experimentation with the parameters for a lin- ear chain of length N = 20 is shown in Figures 3 , 4 and Any set of s$_{n}$ that satisfies the inequality (10) must have s$_{n}$ ⩽ D$_{n}$ . To show this, we rewrite (10) as s$_{0}$ = 1 s$_{1}$ − ds$_{0}$ = − u$_{1}$ s$_{n}$ − $^{ds$_{n}$}$− $_{1}$+ $^{rs$_{n}$}$− 2 = − u$_{n}$, n ⩾ 2 with u$_{n}$ ⩾ 0 , n ⩾ 1 . s$_{n}$ = D$_{n}$ − ∑ n l $_{=1}$u$_{l}$ $^{D$_{n}$}$− $_{l}$, n ⩾ 1 These relations can be inverted to give from which s$_{n}$ ⩽ D$_{n}$ follows at once. D$_{n}$ = f n +1 − g n +1 f − g f = d 2 + √ d 2 4 − r, g = d 2 − √ d 2 4 − r. with f + g = d and fg = r . Hence,