Diagram elements

We use several different graphs closely related to the graph g₀, with some depicted inFigure 9.2. We call these figures diagram elements, which are also the simplest examples of gate diagram, which we will define shortly. The idea behind these graphs is to encode a single qubit computation, complete with inputs and outputs.

(0, 2)

Figure 9.2: Diagram elements from which a gate diagram is constructed. Each diagram element is a schematic representation of the graph g0 shown in Figure 9.1.

Each diagram element corresponds to two copies of the graph g₀, along with self-loops and edges between the two copies. The idea of these elements is to ensure that the ground state of the diagram element is closely related to that of the g0 graph, but where almost all of the vertices of the diagram element has a self-loop. The only vertices without such a self-loop are those corresponding to input and outputs of the diagram element, which will have a self-loop added to them in the final gate diagram.

Along these lines, each diagram element will be labeled by the unitary it computes, along with four numbers between zero and two, corresponding to the number of inputs “nodes” and output

“nodes” for each logical state of the diagram. Each such node will correspond to 16 vertices of the underlying graph representing one logical state and time of the two g₀ graphs. These nodes are placed so that the minimal distance between two vertices in separate nodes will be at least d_max.

Explicitly, each diagram element will be labeled by a unitary U ∈ {I, H, HT }, along with four integers n_0,in, n_0,out, n_1,in, and n_1,out, each between 0 and 2. These numbers correspond to the number of nodes for each particular input or output. We shall label such a diagram element a U_(n^{(n0,in,n1,in)}

0,out,n1,out) element. The vertex set for the corresponding diagram element corresponds to two copies of g₀(namely, 2× 8k × 8 × 2 vertices, labeled as (z, t, j, d) for z, d ∈ F2, t∈ [8k], and j ∈ [8]).

For each node of the gate diagram, we will associate a time for which the underlying history state has computed the correct unitary. Further, we will have these times each be a distance of at least d_max apart, to ensure that each node is at least a distance d_max apart. Namely, for each logical input and output, we will associate two times:

• 0-input: tⁱⁿ0,1 = 2 and tⁱⁿ_0,2= k + 2,

• 1-input: tⁱⁿ1,1 = 2k + 2 and tⁱⁿ_1,2= 3k + 2,

• 0-output: t^out0,1 = 4k + ` and t<sup>out</sup><sub>0,2</sub> = 5k + `,

• 1-output: t^out1,1 = 4k + ` and t<sup>out</sup><sub>1,2</sub> = 5k + `,

where ` is 1, 2, or 3, depending on whether the labeled unitary is HT , I, or H, respectively.

For a given diagram element U_(n^{(n0,in,n1,in)}

0,out,n1,out), it will be useful to have defined the set of logical states and corresponding times explicitly used as input and output in the diagram element. As such, let T ⊂ F2× [8k] be defined as

T = [

z∈F2

{(z, tⁱⁿz,j) : j≤ nz,in} ∪ {(z, t^outz,j) : j≤ nz,out}, (9.48)

and note that T contains those nodes (i.e., the sets of times and logical states) used as input and output for the given gate diagram, and that|T | corresponds to the number of nodes in the diagram element.

With T defined, the adjacency matrix for the corresponding diagram element U_(c,d)^(a,b) will be A

G(a,b),(c,d) U



= A(g₀)⊗ I2+ X

(z,t) /∈T,j∈[8]

|z, t, j ihz, t, j | ⊗ X

a,b∈F2

|aihb| (9.49)

= A(g₀)⊗ I2+ 2Π_¬T ⊗ I8⊗ |+ih+| (9.50) In particular, the graph for G(a,b),(c,d)

U will simply correspond to two copies of g₀, along with a projector onto the equal superposition between the two graphs for each vertex not used in a node of the diagram.

Because of the very similar form between G(a,b),(c,d)

U and g₀, their ground spaces and ground energies are closely related. As the second term in (9.50) is positive semi-definite, we have that the ground energy of A(G(a,b),(c,d)

U ) is at least that of A(g₀). With more exact results, we have the following lemma:

Lemma 20. Let G(a,b),(c,d)

U be the graph corresponding to a diagram element. The ground space of A(G(a,b),(c,d)

U ) is

S = span{|ψz,a,−i : z, a ∈ F2}. (9.51)

Proof. Note that A(G(a,b),(c,d)

U ) commutes with I2⊗ I8k ⊗ I8 ⊗ |+ih+|, and thus there exists an eigenbasis for the adjacency matrix in which each vector is of the form |φi|+i or |φi|−i. For states of this latter form, the second term in (9.50) vanishes, so |ψ, −i is in the ground space of A(g(a,b),(c,d)

U ) if and only if |ψ i is in the ground space of A(g0), and thus we have that S is a subspace of the nullspace.

Now let us examine |α, +i for any state |αi. Since the second term of (9.50)is positive semi-definite, we have that the ground energy of A(G(a,b),(c,d)

U ) is at least e₁. Hence, if |α, +i is in the ground space, then

α, +

A G(a,b),(c,d)

U
α,+ = e₁ =hα|A(g0)|αi (9.52) and thus

hα|Π¬S ⊗ I8|αi = 0, (9.53)

with|αi in the ground space of A(g0).

However, note that for all diagram elements (and all d_max), (z, 0) is not in T . We then have that

Π_¬S ≥ I2⊗ |0ih0| ⊗ I8. (9.54)

As this operator is strictly positive when restricted to the ground space of A(g₀), hψx,γ|I2⊗ |0ih0| ⊗ I8|ψz,δi = 1

8kδ_γ,δδ_x,z, (9.55)

we also have that Π_¬S is strictly positive when restricted to the ground space of A(g0), and thus

|α, +i is not in the ground space of A(G(a,b),(c,d)

U ).

Putting this together, we have that the ground space of A(G(a,b),(c,d)

U ) is S, as claimed

With this bound on the form of the ground space of A(G(a,b),(c,d)

U ), we can then use our knowledge of the two-particle interaction Hamiltonian on g0to relate this to the two-particle interaction Hamil-tonian on G(a,b),(c,d)

U . Namely, we show that since there does not exist a two-particle frustration-free state on g₀, there also does not exists a two-particle frustration-free state on G(a,b),(c,d)

U .

Lemma 21. If d_max > 0, then λ¹₂(G(a,b),(c,d)

U ) > 0 for all states. If d_max= 0, then when restricted to symmetric states, λ¹₂(G(a,b),(c,d)

U ) > 0.

Proof. Note that usingLemma 20, the ground space of A(G(a,b),(c,d)

U ) is in one-to-one correspondence with the ground space of A(g₀), by the transformation

|φx,a,−i ↔ |φx,ai. (9.56)

Namely, by attaching (or removing) a second register in the |−i state, corresponding to having equal and opposite amplitudes between the two copies of g0 present in G(a,b),(c,d)

U , we can transform between these two single-particle ground spaces.

We will use this relation, along with the fact that λ¹₂(g₀) > 0 from Lemma 19, to show that λ¹₂(G(a,b),(c,d)

U ) > 0 with the same assumptions.

Let us then look at any two-particle state that minimizes the movement term. In particular, it takes the form

|φi = X

α,β,x,z∈F2

Q^x,z_α,β|ψx,α,−i|ψz,β,−i. (9.57) Additionally, let us define the related two-particle state on g0 as

|φi = X

α,β,x,z∈F2

Q^x,z_α,β|ψx,αi|ψz,βi. (9.58) We can then see what the expectation of the interaction term of the Hamiltonian is under the state|φi:

hφ|Hint|φi = X

u,v∈V (G(a,b),(c,d)

U )

hφ|Ud(u,v)(ˆnu, ˆnv)|φi (9.59)

= X

u,v∈V (g0),d1,d2∈F2

hφ|Ud((u,d1),(v,d2))(ˆn_(u,d1), ˆn_(v,d2))|φi (9.60)

≥ X

u,v∈V (g0),d1∈F2

hφ|Ud(u,v)(ˆn_(u,d1), ˆn_(v,d1))|φi (9.61) where in the third line we only count the contributions to the interaction when both particles are in the same copy of g₀. As the interaction is positive-semidefinite, this can only decrease the expectation.

Now, from the form of|φi, we have that for any two u, v ∈ V (g0) and either copy of g₀, hφ|Ud(u,v)(ˆn_(u,d1), ˆn_(v,d1))|φi

= X

x1,x2,z1,z2∈F2

α1,α2,β1,β2∈F2

(Q^x_α^1,z1

1,β1)^∗Q^x_α^2,z2

2,β2hψx1,α1|hψz1,β1|Ud(u,v)(ˆn_(u,d1), ˆn_(v,d1))|ψx2,α2i|ψz2,β2i (9.62)

≥

hd1|−i⁴ X

x1,x2,z1,z2∈F2

α1,α2,β1,β2∈F2

(Q^x_α¹₁^,z_,β¹₁)^∗Q^x_α²₂^,z_,β²₂hψx1,α1|hψz1,β2|Ud(u,v)(ˆn_u, ˆn_v)|ψx2,α2i|ψz2,β2i (9.63)

= 1

4hφ|Ud(u,v)(ˆnu, ˆnv)|φi. (9.64)

HT

1

1

2

Figure 9.3: A gate diagram with two diagram elements labeled q = 1 (left) and q = 2 (right).

Hence, we have that

hφ|Hint|φi ≥ X

u,v∈V (g0),d1∈F2

hφ|Ud(u,v)(ˆn_(u,d1), ˆn_(v,d1))|φi (9.65)

≥ 1 4

X

u,v∈V (g0),d1∈F2

hφ|Ud(u,v)(ˆn_u, ˆn_v)|φi (9.66)

= 1 4

X

d1∈F2

hφ|Hint|φi (9.67)

= 1

2hφ|Hint|φi. (9.68)

UsingLemma 19, we have that (9.68)is larger than zero for all states and interactions that satisfy the conditions ofLemma 19, and thushφ|Hint|φi > 0. As such, there does not exist a two-particle frustration-free state on the graph G(a,b),(c,d)

U under the same assumptions as for g0.

In document The computational power of many-body systems (Page 156-160)