Applications of Reflection Positivity:
Graphene and Other Examples
Elliott Lieb
Princeton University
Joint work with R. Frank
Outline of the talk
1. The Coulomb potential: An ancient example (except that the ancients didn’t com-pletely recognize it). This is an introduction for people who are not familiar with the subject, but it has surprising generalizations (with Frank)
2. Periodicity in systems with competing ferro- and antiferromagnetic interactions (The problem of stripes. (with Giuliani and Lebowitz)
3. Spin-space reflection positivity
4. Flux-Phase problem in all dimensions in the Hubbard model (with Loss) 5. Peierls instability in 1 dimension (with Kennedy and Nachtergaele)
6. Possible lattice distortions of graphene (in the Hubbard model context). (with Frank) 7. Long range order and infrared bounds in systems with continuous symmetry.
Coulomb Energy
The Coulomb interaction between two charge distributions f and g is
E(f, g) =
Z
Rn Z
Rn
f(x)|x − y|−λg(y)dx dy
with λ = n − 2 (for n ≥ 3). The kernel |x − y|−λ is positive definite for all 0 < λ < n, so
E(f, f) > 0. Now let P be a plane in Rn. let f be supported in the right half-space and
g in the left. what can we say about E(f, g)?.
Answer: If f is the reflection through P of g, i.e., f = θg, then E(θg, g) ≥ 0. This implies that
|E(f, g)|2 ≤ E(θg, g)E(f, θf)
Here are some more recent, even startling results:
1. While |x − y|−λ is positive definite for all 0 < λ < n, it is RP if and only if
n − 2 ≤ λ < n (Lopes-Maris, Frank-L) (Includes the pseudo-relativistic Laplacian.) |p|.
2. Take a sphere S ⊂ Rn, let F be supported outside S, and let θSF now denote an (appropriately normalized) reflection of F in S. Then E(θSF, F) ≥ 0.
Y.Y Li and Zhu proved that the only continuous functions F such that for every z ∈ Rn
there is a sphere centered at z such that F = θSF are
F(x) = c ¡a2 + |x − w|2¢−p for some a, c, and w ∈ Rn.
(Frank & L extended this to measures.)
Stripes and competing interactions
Stripes are sometimes seen as ground states in condensed matter. In an attempt to understand this as an effect of competing interactions, Giuliani, Lebowitz, and L considered an Ising spin Hamiltonian in 1D
H = X
x6=y
|x − y|−pσx σy − J
X
x
σx σx+1 .
I.e., long range antiferro- and short range ferro-magnetic interactions.
Question: Is the ground state periodic? When J = 0 the ground state is the alternating
+ − + − + − +− Ne´el state because |x − y|−p is RP. If J À 1 it is an Ising + + +...
state. In between it is periodic, with period 2k, of the form
+ + + + − − − − + + + + − − − − + + + + − − − − .
Spin-space reflection positivity
Consider the Hubbard model on a graph with U ≤ 0 and real hopping matrix tx,y:
H = X
x,y
X
σ=±1
tx,yc†x,σ cy,σ + h.c. + U X x
(nx,+1 − 1
2)(nx,−1 − 1 2)
with nx,σ = c†x,σ cx,σ. Suppose we want to find the ground state with a given particle
number 2N = P
x,σ nx,σ
By SU(2) invariance we can always look for this in the subspace where N particles have
σ = +1 and N have σ = −1. Let Ψ(X, Y ) be the g.s. as a function of the coordinates of the +1 particles X = (x1, . . . , xN) and the -1 particles Y = (y1, . . . , yN).
Think of Ψ(X, Y ) as a matrix indexed by N−tuples X and Y . The energy E(Φ, Ψ) :=
hΦ|H|Ψi is then a quadratic form. It is not too hard to see, using U ≤ 0, that this form satisfies:
E(Ψ, Ψ) ≥ 12E(|Ψ|, |Ψ|) + 12E(|Ψ∗|, |Ψ∗|)
This implies that it is best to substitute either |Ψ| or |Ψ∗| in place of Ψ, which implies that the g.s. is a positive definite matrix. The up and the down spins are, therefore, positively correlated.
One consequence is that Ψ(X, X) > 0 for some configuration X, i.e., the spin-up and the spin-down particles can be on top of each other. Hence the g.s. has total spin = 0. I do not know any other way to prove this (obvious ??) fact.
By the well known “hole-particle” transformation, this result translates into the following
unexpected fact about the g.s. of the U > 0 Hubbard model on a bipartite graph with
the number of particles equal to the number of sites (half-filling) and with a A-sites and
Flux-Phase Problem
Again we consider the Hubbard model, but this time the graph is the cubic lattice on a torus (i.e., periodic boundary conditions). We are at half-filling again, but the sign of the on-site interaction U does not matter. We introduce a magnetic field, possibly non-homogeneous, that acts on the orbital motion, not the spin magnetic moment. This is done by multiplying the real tx,y = t > 0 by a phase exp(iθx,y), where θx,y = Rxy A(z)·dz. The flux through each square plaquette is arg{tx,yty,ztz,wtw,x}. It is easy to see that the
g.s. energy depends only on the fluxes.
Question: What choice of fluxes minimizes the g.s. energy?
Answer: Flux π through every square face of the lattice, i.e., the sign of the t’s around a loop are + + +−.
To prove this we cut the lattice in 2 pieces through bonds and then make a hole-particle transformation on the right half: cx,σ → c†x,σ. So c†c across a bond becomes c†c†, etc.
Flux-Phase Problem (continued)
This now looks reflection symmetric. But then c†c on the right is changed to c c†, which equals −c†c. To get rid of the unwanted minus sign on the right we have to change c to
−c everywhere on the right, but this means that the flux across the cut bonds has become
π.
With these changes we see that if we define πL to be the set of fluxes on the left, and πR
those on the right, we see that the g.s. energy satisfies
E(πL, πR) > 12E(πL, πL) + 12E(πR, πR) ,
and thus the lowest energy occurs when πL = πR.
The flux π through every square that we cut has not changed. By repeating this argument through every symmetry plane we conclude that we need flux π everywhere.
Peierls Instability
Hubbard model in 1D, in which the (real) hopping ti,i+1 is allowed to vary:
H =
N X
i
ti,i+1 hc†ici+1 + c†i+1ci i
+ U N X
i
n↑i n↓i + g N X
i
(ti − 1)2 .
The last term is the energy penalty for changing ti from its preferred value 1.
For a half-filled band and N = ∞, Peierls observed that letting the ti dimerize
(long-short-long-short ) would open a gap and lower the total energy. Actually, he considered only U = 0 (H¨uckel model). Two Questions:
a.) If the chain is finitely long, or U 6= 0, will dimerization still be optimal? E.g., N = 6? b.) What if (ti − 1)2 is replaced by some other F(ti − 1)? What about “spin-Peierls” for
Answers:
1.) For periodic boundary conditions (i.e., the chain is wrapped in a circle) and
N = 2 mod 4, the energy minimizing configuration has period 2. Nothing more compli-cated can happen.
2.) If N = 0 mod 4, more complicated things can happen (e.g., for N = 4 we can get a trapezoid instead of a rectangle), but as N → ∞, we get period 2.
3. For finite N we can get period 1, which is a special case of period 2. That is, the minimizing configuration has ti = constant. This happens for Benzene (N=6). The question of dimerization or not depends on the parameters of the problem.
4. The distinction between N = 0 mod 4 and N = 2 mod 4 is connected with the flux through the large N− loop.
Graphene model
We now consider the Hubbard model on the honeycomb (hexagonal) lattice. As in the 1D case, just discussed, we ask for possible energy - minimizing lattice distortions.
We cannot generalize the previous 1D result to the usual square lattice because of the flux-phase problem. That is, we get RP only if we have flux π in every square plaquette, but such a flux is ridiculously beyond reach, physically. But the optimum flux though a hexagon is π + π = 0 since a hexagon is two squares with a side missing (think of a brick wall). Thus, we get RP through the 3 possible reflection lines of the hexagonal lattice.
A
A
A
A
A
A
B
B
B
C
C
C
The left picture shows that good hopping matrix elements (or physical bond lengths) have period 3. That is, there can be 3 lengths distributed periodically, as shown. The well known Kekul´e configuration is the special choice in which 2 of the 3 bond-lengths are equal.
