The graph g 0

The graph g₀ shown in Figure 9.1is constructed using the method of Chapter 8, with the single qubit circuit corresponding to a sequence of H and HT gates. The intuition behind this graph is to force the ground state to correspond to the history state of these simple computations, thus allowing us to reference the computational value of the corresponding qubit in several disparate

locations. We will eventually combine several of these g₀ graphs using projectors, which will remove certain states from the ground space depending on the value of the corresponding qubits.

As such, let k = 4 + 2b^dmax₂ c, and then let us look at the single-qubit circuit C0 with k gates U_j, for j∈ [k], where

U₁ = HT U₂ = (HT )^† (9.4)

and the rest of the Uj = H. We will use the second, third, and fourth of these time steps as computations in the eventual gadgets, while the remaining time steps act as padding to ensure that the computational time steps used in the eventual graph occur at a distance at least d_max from each other. As the circuit C0 implements an identity operation, we can easily concatenate several of these circuits and examine the graph corresponding to the circuit using circular time, as in the construction used in Chapter 8. For our purposes, we will want to use 8 copies ofC0 in series, as the eventual gadgets used in our proof have 8 possible locations for interactions with other copies of g₀.

In particular, we will have that the 0-1 Hamiltonian corresponding to the eventual adjacency matrix of g0 acts on the Hilbert spaceH(g0) =C²⊗ C^8k⊗ C⁸. If we then remember that B(U ) is the operator that takes ω7→ S, where ω = e^iπ/4 and S is the shift operator acting on C⁸, we have that the component of the Hamiltonian corresponding to the circuit is

Hprop =−√ 2

8k−1X

t=0

B(Vt)13⊗ |t + 1iht| + B(Vt^†)13⊗ |tiht + 1|, (9.5)

where the B(U )₁₃act on the Hilbert spacesC²⊗ C⁸ while the clock acts on theC^8k Hilbert space, and where

V_t=







HT t = 0 mod 8 (HT )^† t = 1 mod 8 H otherwise.

(9.6)

This term, along a penalty to theC⁸ Hilbert space given by Hpen=IC² ⊗ I_C^8k ⊗ S³+ S⁴+ S³

, (9.7)

which forces the third register into a particularly useful state, allows us to guarantee that the ground state is a history state. Altogether, we then have that the adjacency matrix of g(0) is given by

A(g₀) = H_prop+ H_pen, (9.8)

where each vertex is labeled by a computational basis state in the Hilbert space, namely (z, t, j) with z ∈ F2, t ∈ [8k], and j ∈ [8]. The graph g0 in the special case that dmax = 0 is shown in Figure 9.1.

We can then use the results ofChapter 8to calculate the smallest eigenvalue of A(g₀), the corre-sponding eigenvectors, and the eigenvalue gap. In particular, we have that the smallest eigenvalue is

e₁ =−1 − 3√

2 =−5.24 . . . , , (9.9)

t = 0 HT

t = 1 T^†H

t = 2 H t = 3

H t = 4

Figure 9.1: The graph g₀ for the case d_max≤ 1. Vertices are arranged with each ray corresponding to a specific time t proceeding clockwise, with the outer 8 vertices corresponding to logical 0 and the inner 8 corresponding to logical 1, with the further breakdown into 8 vertices corresponding to the ancillary register. The difference in color for some edges is an attempt to highlight those edges corresponding to the penalty term (bottom of the figure) and the circuit (top left of the figure).

corresponding to a four dimensional ground space spanned by the states

Additionally, we have that the energy gap is at least λ²₁(g₀)≥√

which is constant for all interactions with a given d_max.

Note that the amplitudes of |ψz,0i in the above basis contain the result of computing either the identity, Hadamard, or HT gate acting on the “input” state |zi, while the amplitudes |ψz,1i corresponds to the result of the identity, Hadamard or HT gate acting on the “input” state.

Additionally, this information is repeated at least 8 times, once for each copy of C0.

With these bounds on the single particle eigenvalues and their corresponding eigenstates, we can now show that the graph g₀has no two-particle frustration-free states. ByLemma 14, it follows that g₀ has no N -particle frustration-free states for N ≥ 2. While we would like this to be true for all interactions, in the case of only onsite interactions (dmax = 0) no anti-symmetric state is penalized and thus there is a 4-particle frustration-free state. However, we show that this is the only case for which this is true, and further that if we restrict ourselves to the symmetric subspace, we again have that there are no frustration-free two-particle states.

Lemma 19. If dmax > 0, then λ¹₂(g₀) > 0 for all states. If dmax = 0, then when restricted to symmetric states, λ¹₂(g₀) > 0.

Proof. Suppose (for a contradiction) that |Qi ∈ H(g0)^⊗2 is a nonzero vector in the nullspace of H(g₀, 2), so since A(g₀) has smallest eigenvalue e₁ and the interaction term is positive semidefinite. We can therefore write

for all vertices u, v∈ g0.

We then have by assumption that U_d^(1,1)_max > 0 if d_max> 0 or U0² > 0 if d_max= 0, and thus for all vertices u, v of distance d_max,

hu, v |Qi = 0. (9.18)

We will use this equation to show a contradiction, so that |Qi cannot exist.

Note that vertices of the form (x, T, j) and (z, T +t, k) are at least a distance t apart for all times t < 4k, since edges only connect vertices with corresponding times that differ by at most 1. Further, note that the portion of the Hamiltonian that connect adjacent times only arise from the terms corresponding to the circuit. Since each unitary for k > t≥ 2 corresponds to a Hadamard, and the corresponding term in the Hamiltonian only connects vertices with the same j or j’s that differ by 4, we have that only vertices of the form (x, 2, j) and (z, 2 + t, j) or (x, 2, j) and (z, 2 + t, j + 4) can be a distance t apart; all other pairs of vertices with these two times must be at a distance of at least t + 1.

With all of this in mind, let us assume that d_max is an even integer greater than zero. We then have that the vertices (0, 2, j) and (0, 1+d_max, j +4) are a distance d_max−1 apart. Further, we have that (0, 1 + dmax, j + 4) is also connected to the vertices (0, 1 + dmax, j + 1) and (0, 1 + dmax, j− 1) (from the penalty term of the Hamiltonian), and thus we have that the vertices u = (0, 2, j) and v = (0, 1 + d_max, j + `) are a distance d<sub>max</sub> apart for all j∈ [8] and for ` = ±1. Using (9.18)with these pairs of vertices we then have that

hu, v |Qi = X

x,a,z,b∈F2

Q_xa,zbh0, 2, j |ψx,aih0, 1 + dmax, j + `|ψzbi (9.19)

= 1 64k

X

x,a,z,b∈F2

Q_xa,zbh0|xih0|H|z iω^{(−1)aj+(−1)b(j+`)} (9.20)

= 1

64√

2k (Q_00,00+ Q_00,10)ω^{2j+`</sup>+ (Q<sub>00,01</sub>+ Q<sub>00,11</sub>)ω<sup>−`}

+ (Q_01,00+ Q_01,10)ω^{`</sup>+ (Q<sub>01,01</sub>+ Q<sub>01,11</sub>)ω<sup>−2j−`}

, (9.21)

and thus we have that Q_0a,0b =−Q0a,1b for all a, b ∈ F2. Using the same reasoning with vertices u = (0, 2, j) and v = (1, 1 + d_max, j + `) with ` =±1 then gives us

hu, v |Qi = X

x,a,z,b∈F2

Q_xa,zbh0, 2, j |ψx,aih1, dmax, j + `|ψzbi (9.22)

= 1 64k

X

x,a,z,b∈F2

Q_xa,zbh0|xih1|H|z iω^{(−1)aj+(−1)b(j+`)} (9.23)

= 1

64√

2k (Q_00,00− Q00,10)ω^{2j+`</sup>+ (Q<sub>00,01</sub>− Q00,11)ω<sup>−`}

+ (Q_01,00− Q01,10)ω^{`</sup>+ (Q<sub>01,01</sub>− Q01,11)ω<sup>−2j−`}

, (9.24)

which combined with our previous results show that Q_0a,zb = 0 for all a, b, z∈ F2. Again using the

same reasoning with u = (1, 2, j) and v = (1, 1 + d_max, j + `) for ` = 3 or ` = 5 gives us

which combined with our previous results implies that Q_1a,zb = 0 for all a, b, z ∈ F2. Putting this together, we then have each Qxa,zb= 0, and thus |Qi does not exist; in other words, if dmax> 0 is even, then the nullspace of H(g₀, 2) is empty.

Now let us assume that d_max is a positive odd integer. For all such d_max, we can then use equation(9.18)with vertices u = (y, 2, j) and v = (y, 1 + d_max, j + `) for y ∈ F2, j∈ [8], and ` = ±1

where in the last line we used the fact that Q_za,zb = 0. A similar result with u = (0, 2, j) and v = (1, 2 + d_max, j + 4) then gives us that Q_0a,zb = 0. Finally, repeating this same procedure with u = (1, 2, j) and v = (1, 2 + d_max, j) and with u = (1, 2, j) and v = (0, 2 + d_max, j + 4) gives us that Q_1a,zb = 0. Putting this all together, we have that each Q_xa,zb = 0 and thus |Qi does not exist if d_maxis an odd integer.

Finally, let us assume that d_max = 0, and that the state |Qi is symmetric (so that Qxa,zb = Q_zb,xa). With these assumptions, let us examine equation(9.18) with u = v = (y, 0, j) for y ∈ F2

and j∈ [8]:

hu, v |Qi = X

x,a,z,b∈F2

Q_xa,zbhy, 0, j |ψx,aihy, 0, j |ψzbi (9.38)

= 1 64k

X

x,a,z,b∈F2

Q_xa,zbhy |xihy |z iω^{(−1)aj+(−1)bj} (9.39)

= 1

64k Q_y0,y0ω^2j + Q_y1,y1ω^−2j + 2Q_y1,y0)

. (9.40)

Evaluating these equations together then gives us that Qxa,xb = 0 for all a, b, x∈ F2. If we now use equation (9.18)with u = v = (0, 3, j) for all j ∈ [8], we find that

hu, v |Qi = X

x,a,z,b∈F2

Q_xa,zbh0, 3, j |ψx,aih0, 3, j |ψzbi (9.41)

= 1 64k

X

x,a,z,b∈F2

Q_xa,zbh0|H|xih0|H|z iω^{(−1)aj+(−1)bj} (9.42)

= 1

128k 2Q_00,10ω^2j+ 2Q_01,11ω^−2j+ (2Q_01,10+ 2Q_00,11)

. (9.43)

and thus Q_00,10 = Q_01,11 = 0 and Q_01,10=−Q00,11. If we now use(9.18) with the only remaining vertices leading to novel restrictions, namely u = v = (0, 1, 0), we find

hu, v |Qi = X

x,a,z,b∈F2

Q_xa,zbh0, 1, j |ψx,aih0, 1, j |ψzbi (9.44)

= 1

64k 2Q01,10h0|HT |0ih0|HT |1i + 2Q00,11h0|HT |0ih0|HT |1i

(9.45)

= 1

64k Q_01,10ω + Q_00,11ω⁻¹) (9.46)

= Q01,10

64k (ω− ω⁻¹) (9.47)

must be zero, and thus each Q_xa,zb = 0. Hence, if d_max = 0 no symmetric state |Qi is in the nullspace of H(g₀, 2).

Combining all of this together, we have that if dmax> 0, then the nullspace of H(g0, 2) is empty and λ¹₂(g₀) > 0, while if d_max= 0 then no symmetric state is in the nullspace of H(g₀, 2), and thus when restricted to symmetric states, λ¹₂(g₀) > 0.

In document The computational power of many-body systems (Page 150-156)