On the spectrum of Hamiltonians
in finite dimensions
Roberto
Oliveira
Joint with David
DiVincenzo and
Barbara Terhal
@ IBM Watson.
Paraty, August 14th 2007.
In one slide:
Ground state energy is hard.
Bulk of the spectrum is Gaussian
and universal.
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(H2)/2N
• Write H = X½ V HX ,
where HX acts on the spins in X. Can
assume Tr(HXHY)=0 if X Y.
E.g.: Ising with transverse field
H = (bond terms)
+ (site terms)
i
k
site term
H{k} = h z[k]
j
bond term
H{i,j} = J xx[i,j]
H=k H{k} + i~jH{i,j}
E.g.: Ising with transverse field
face term HF
H = (bond terms)
+ (site terms)
i
k
site term
H{k} = h z[k]
j
bond term
H{i,j} = J xx[i,j]
H=k H{k} + i~jH{i,j}
+ FHF
Dimensionality assumption
• 9 metric  and d, C, c, >0 such that:
Radius R
cRd· X½B(x,R)2(HX)·CRd
X½ B(x,R)||HX||·CRd
center x
B(x,R) = {v2 V : (v,x)· R}
X ½  B(x,R)2(HX)·CRd-
E.g. nearest neighbor in Zd
Radius R
L1 norm
¼ Rd terms
inside
¼ Rd-1 at the
boundary
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.
A bit more formally
• There is a probability measure on the
line given by H:
m=mH= 2-N 2 Spec(H) 
• We show that this measure is
approximately normal in the sense that
for all a<b, as N grows:
mH (a(H),b(H)) ! (2)-1/2sab exp(-t2/2)dt
Even more formally
• Recall: strength inside ball ¼Rd, strength
across boundary ¼ Rd-, with C,c extra
parameteres.
• Then for all a<b,
|mH [a(H),b(H)]-(2)-1/2sab exp(-t2/2)dt|
· D(C,c,d,) (Diam())-d/8
A bit less formally again
• Inside ¼ Rd, boundary¼ Rd-.
Radius R
mH/(H)[x,y]
center x
B(x,R) = {v2 V : (v,x)· R}
¼ (2)-1/2sxy exp(-t2/2)dt
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=n2
spins).
• Also assume that all terms in the
Hamiltonian have norm of constant
order.
Back to that old slide
H = (bond terms)
+ (site terms)
i
k
site term
H{k} = h z[k]
j
bond term
H{i,j} = J xx[i,j]
H=k H{k} + i~jH{i,j}
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.
Omitting the interactions
• G = s Gs, 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,
mG/(G) is approximately Gaussian as
long as some conditions are satisfied.
Which conditions?
• Many terms in the sum.
• The influence of any given term is small.
• m¿ n1/2 suffices.
Putting red lines back in
Next step
• Recall that we have.
mH/(H) = 2-N 2 Spec(H) /(H)
• We know that mG/(G) is approx.
gaussian.
mG/(G) = 2-N 2 Spec(G) /(G)
• We will show that (H-G) is small.
• By a perturbation theory argument, this
implies that mH/(H)¼ mG/(G).
(H-G) is small
Variance ¼ # of qubits
Total # of qubits ¼ n2
Qubits on red lines ¼ m(n/m)2 = n2/m
) (H-G)2· (H)2/m, small if mÀ 1
To conclude
• Take some 1¿ m¿ n1/2 (e.g. m=n1/3).
• Then mG/(G) is approx. Gaussian.
• Also mH/(H)¼ mG/(G). So we are done.
• The key step: m x m boxes have m2
vertices but only ¼ m vertices on their
boundaries.
General case
• Inside ¼ Rd, boundary¼ Rd-.
Radius R
mH/(H)[a,b]
center x
B(x,R) = {v2 V : (v,x)· R}
¼ (2)-1/2sab exp(-t2/2)dt
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.
Conclusions
• Spectral distribution is approximately
Gaussian with std. deviation ¼ N1/2.
• This is universal for quantum spin
systems in finite dimensional structures
when long-range interactions decay fast
enough.
Further work
• Bounds are actually weak for many
problems. Are there better bounds for
specific systems?
• Fermions? Bosons?
• Applications?