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Published byTerence Simmons Modified over 8 years ago
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On the spectrum of Hamiltonians in finite dimensions Roberto Oliveira Paraty, August 14 th 2007. Joint with David DiVincenzo and Barbara Terhal @ IBM Watson.
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In one slide: Ground state energy is hard. Bulk of the spectrum is Gaussian and universal.
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The setup H = Hamiltonian on a set V of N spin ½ particles. Spec(H) = { 0 · 1 · 2 …}. We will assume Tr(H)=0. 2 (H) = “variance” = Tr(H 2 )/2 N Write H = X ½ V H X, where H X acts on the spins in X. Can assume Tr(H X H Y )=0 if X Y.
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site term H {k} = h z [k] E.g.: Ising with transverse field H = (bond terms) + (site terms) bond term H {i,j} = J x x [i,j] ij k H= k H {k} + i~j H {i,j}
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site term H {k} = h z [k] E.g.: Ising with transverse field H = (bond terms) + (site terms) bond term H {i,j} = J x x [i,j] ij k H= k H {k} + i~j H {i,j} + F H F face term H F
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Dimensionality assumption 9 metric and d, C, c, such that: Radius R center x B(x,R) = {v 2 V : (v,x) · R} cR d · X ½ B(x,R) 2 (H X ) · CR d X ½ B(x,R) ||H X || · CR d X ½ B(x,R) 2 (H X ) · CR d-
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E.g. nearest neighbor in Z d L 1 norm ¼ R d terms inside ¼ R d-1 at the boundary Radius R
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Gaussian spectrum Plot a histogram of Spec(H/ (H)). with small but fixed bin size b>0. That will approach a Gaussian as N adiverges, for fixed parameters.
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A bit more formally There is a probability measure on the line given by H: H = 2 -N 2 Spec(H) We show that this measure is approximately normal in the sense that for all a<b, as N grows: H (a (H),b (H)) ! (2 ) -1/2 s a b exp(-t 2 /2)dt
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Even more formally Recall: strength inside ball ¼ R d, strength across boundary ¼ R d- , with C,c extra parameteres. Then for all a<b, | H [a (H),b (H)]-(2 ) -1/2 s a b exp(-t 2 /2)dt| · D(C,c,d, ) (Diam( )) - d/8
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A bit less formally again Inside ¼ R d, boundary ¼ R d- Radius R center x B(x,R) = {v 2 V : (v,x) · R} H/ (H) [x,y] ¼ (2 ) -1/2 s x y exp(-t 2 /2)dt
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A simple case We will now explain a special case of the Theorem. Nearest neighbor interactions on a n x n patch of the planar square lattice (N=n 2 spins). Also assume that all terms in the Hamiltonian have norm of constant order.
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site term H {k} = h z [k] Back to that old slide H = (bond terms) + (site terms) bond term H {i,j} = J x x [i,j] ij k H= k H {k} + i~j H {i,j}
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What do we do? Main idea: ignore terms acting on red lines (i.e. treat non-interacting systems). Then put them back in via a perturbation argument.
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Omitting the interactions G = s G s, subsystems might have high dimension. How does one compute the global spectrum? Answer: convolution of the individual spectral distributions. By the usual Central Limit Theorem, G G) is approximately Gaussian as long as some conditions are satisfied.
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Which conditions? Many terms in the sum. The influence of any given term is small. m ¿ n 1/2 suffices.
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Putting red lines back in
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Next step Recall that we have. H/ (H) = 2 -N 2 Spec(H) We know that G/ (G) is approx. gaussian. G/ (G) = 2 -N 2 Spec(G) G We will show that (H-G) is small. By a perturbation theory argument, this implies that H/ (H) ¼ G/ (G).
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H-G) is small Variance ¼ # of qubits Total # of qubits ¼ n 2 Qubits on red lines ¼ m(n/m) 2 = n 2 /m ) (H-G) 2 · (H) 2 /m, small if m À 1
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To conclude Take some 1 ¿ m ¿ n 1/2 (e.g. m=n 1/3 ). Then G/ (G) is approx. Gaussian. Also H/ (H) ¼ G/ (G). So we are done. The key step: m x m boxes have m 2 vertices but only ¼ m vertices on their boundaries.
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General case Inside ¼ R d, boundary ¼ R d- Radius R center x B(x,R) = {v 2 V : (v,x) · R} H/ (H) [a,b] ¼ (2 ) -1/2 s a b exp(-t 2 /2)dt
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General result: proof sketch Main idea: break the system into subparts with small total boundary strength. Treat isolated systems via standard CLT. Put the boundary back in via perturbation theory.
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Conclusions Spectral distribution is approximately Gaussian with std. deviation ¼ N 1/2. This is universal for quantum spin systems in finite dimensional structures when long-range interactions decay fast enough.
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Further work Bounds are actually weak for many problems. Are there better bounds for specific systems? Fermions? Bosons? Applications?
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