A diamagnetic inequality for

semigroup differences

(Irvine, November 10–11, 2001)

Barry Simon and 100DM

The integrated density of states (IDS)

Schrödinger operator:

H := H(V) := −\frac{1}{2}Δ + V_{ω} =: H(0, V),

or, with a magnetic vector potential A,

H := H(A, V) := \frac{1}{2}(−i∇ − A)^{2} + V_{ω}

on L^{2}(ℝ^{d}).

• To model disordered systems, the potential V is often taken to be a random potential, e.g.,

V (x) = V_{ω}(x) = _{n∈ℕ}^{∑} f(x − x_{n}(ω))

where x_{n} are randomly distributed points in ℝ^{d}, or

V (x) = V_{ω}(x) = _{n∈ℤ^{d}}^{∑} λ_{n}(ω)f(x − x_{n})

where the (λ_{n}) are i.i.d. random variables. We will assume that V ∈ L_{loc}^{1} (ℝ^{d}) and v ≥ 0, for simplicity.

• The magnetic vector potential A gives rise to a magnetic field B := dA. Again, B can be thought of as given by a random process or is fixed.

Let Λ ⊂ ℝ^{d} be an open set. H_{Λ}^{#} (A, V_{ω}) is the restriction of H(A, V_{ω}) to Λ with Dirichlet (# = D), respectively Neumann (# = N), boundary conditions.

Definition (IDS) The finite volume integrated density of states for Dirichlet, respectively Neumann, boundary conditions is given by

ρ_{Λ,ω}^{#} (s) := \frac{1}{|Λ|} #\{eigenvalues λ_{j}(H_{Λ}^{#} (A, V_{ω})) ≤ s\}

ρ_{ω}^{#} := lim_{Λ→ℝ^{d}} ρ_{Λ,ω}^{#}

Natural questions

Question 1: Do the limits ρ_{ω}^{#} exist?

Question 2: If so, how are they related? In particular, are they the same (= independence of the boundary conditions)?

Fact:

• Λ → |Λ|ρ_{Λ,ω}^{D} (resp. |Λ|ρ_{Λ,ω}^{N} ) is a sub (resp. super) additive ergodic process.

This implies that the macroscopic limits

ρ_{ω}^{#} = lim_{Λ→ℝ^{d}} ρ_{Λ,ω}^{#}

exist almost surely and are non-random, i.e.,

ρ_{ω}^{#} = 𝔼[ρ_{ω}^{#} ] almost all ω

(= self-averaging property of the IDS).

Independence of the Boundary conditions

We will fix some potential V and magnetic vector potential A and have the the finite volume IDS ρ_{Λ}^{#} for these fixed potentials. It will turn out that the independence of the boundary conditions of the macroscopic limits of ρ_{Λ}^{#} is independent of their existence!

Let f : ℝ → ℝ be a nice function, then

∫ f(E) dρ_{Λ}^{#} (E) = \frac{1}{|Λ|} tr_{L^{2}(Λ)} [f(H_{Λ}^{#} (A, V))].

for # =N (Neumann), resp. =D (Dirichlet) boundary conditions.

• We will often write tr[f(H_{Λ}^{#} (A, V))] instead of tr_{L^{2}(Λ)} [f(H_{Λ}^{#} (A, V))] as long as there can be no confusion.

• Choosing f(E) = e^{−tE} we get the Laplace transforms of the measures dρ_{Λ}^{#} , i.e., the Laplace transform is the trace of the corresponding semigroup.

Theorem 1 (S. Nakamura, S.-i. Doi et al). Take Λ = [−L, L]^{d}, V, B = dA uniformly bounded, and f ∈ C_{0}^{1}(ℝ). Then

|tr [f(H_{Λ}^{N}(A, V)) − f(H_{Λ}^{D}(A, V))]| ≤ C \frac{|∂Λ|}{|Λ|} = \frac{C}{L}.

Sketch (of Nakamura’s proof): Recall the Krein spectral shift:

tr[f(A_{1}) − f(A_{2})] = ∫ f′(E) ξ_{A_{1},A_{2}}(E) dE

with

∥ξ_{A_{1},A_{2}} ∥_{L^{1}} ≤ ∥A_{1} − A_{2}∥_{1} .

Take A_{1} := (H_{Λ}^{N} + M)^{−p}, A_{2} := (H_{Λ}^{D} + M)^{−p}, then

f(H_{Λ}^{N}) = g(A_{1}) with f(E) = g((E + m)^{−p}).

So using Krein, it is enough to show that

∥(H_{Λ}^{N} + M)^{−p} − (H_{Λ}^{D} + M)^{−p}∥ _{1} ≤ C|∂Λ|.

However, this is rather tricky and requires a good knowledge of the domains of the restricted operators, which is complicated.

A completely different approach:

Theorem 2 (Barry Simon, 100DM). Let Λ ⊂ ℝ^{d} be any open set, A ∈ L_{loc}^{2} , V ≥ 0, V ∈ L_{loc}^{1} . Then a) |(e^{−tH_{Λ}^{N}(A,V)}f)(x)| ≤ (e^{−tH_{Λ}^{N}(0,V)} |f|)(x) for x ∈ Λ b) |((e^{−tH_{Λ}^{N}(A,V)} − e^{−tH_{Λ}^{D}(A,V)})f)(x)|

≤ ((e^{−tH_{Λ}^{N}(0,V)} − e^{−tH_{Λ}^{D}(0,V)} )|f|)(x)

≤_{V ≥0} ((e^{−tH_{Λ}^{N}(0,0)} − e^{−tH_{Λ}^{D}(0,0)} )|f|)(x).

In particular, tr(e^{−tH_{Λ}^{N}(A,V)} − e^{−tH_{Λ}^{D}(A,V)} )

≤ tr(e^{−tH_{Λ}^{N}(0,0)} − e^{−tH_{Λ}^{D}(0,0)} ) = O(|∂Λ|).

(Weyl asymptotic for the free case!)

Motivation: The Feynman-Kac-Itô formula

(e^{−tH_{Λ}^{D}(A,V)}f)(x) = 𝔼^{x}[e^{−iS^{t}(A)(b)−∫_{0}^{t} V (b_{s})ds} χ_{Λ_{t}} (b)f(b_{t})],

where t → b_{t} is a Brownian motion process,

S^{t}(A) := ∫_{0}^{t} A(b_{s}) db_{s} + \frac{1}{2} ∫_{0}^{t} divA(b_{s}) ds

is the “line integral” of A along a Brownian path, and we integrate only over the region

Λ_{t} := {b| b_{s} ∈ Λ for all 0 ≤ s ≤ t}.

With Neumann boundary conditions:

(e^{−tH_{Λ}^{N}(A,V)}f)(x) = 𝔼̃^{x}[e^{−iS^{t}(A)(b̃)−∫_{0}^{t} V (b̃_{s})ds} f(b̃_{t})]

where t → b̃_{t} is the so-called reflected Brownian motion (in Λ).

Note that, at least morally, b̃ = b for paths b ∈ Λ_{t} (if Brownian motion did not hit the boundary up to time t it could not have been reflected, yet.)

Assuming this, we immediately get

|(e^{−tH_{Λ}^{N}(A,V)} − e^{−tH_{Λ}^{D}(A,V)})f| =

= |𝔼̃^{x}[e^{−iS^{t}(A)(b̃)−∫_{0}^{t} V (b̃_{s})ds} \underbrace{(1 − χ_{Λ_{t}} (b̃))}_{≥0}f(b̃_{t})]|

≤ 𝔼̃^{x}[e^{− ∫_{0}^{t} V (b̃_{s})ds}(1 − χ_{Λ_{t}} (b̃))|f(b̃_{t})|]

= (e^{−tH_{Λ}^{N}(0,V)} − e^{−tH_{Λ}^{D}(0,V)} )|f|

= 𝔼̃^{x}[ \underbrace{e^{− ∫_{0}^{t} V (b̃_{s})ds}}_{≤1}(1 − χ_{Λ_{t}} (b̃))|f(b̃_{t})|]

≤ 𝔼̃^{x}[(1 − χ_{Λ_{t}} (b̃))|f(b̃_{t})|]

= (e^{−tH_{Λ}^{N}(0,0)} − e^{−tH_{Λ}^{D}(0,0)} )|f| Sketch of the proof of Theorem 2:

a) ⇒ b): Take a potential W ≥ 0. We have duHamel’s formula, for A, A + B ≥ 0

e^{−tA} − e^{−t(A+B)} =

= ∫_{0}^{t} \underbrace{\frac{d}{ds}(e^{−sA}e^{−(t−s)(A+B)} )}_{=e^{−sA}(−A+A+B)e^{−(t−s)(A+B)}} ds

= ∫_{0}^{t} e^{−sA}Be^{−(t−s)(A+B)} ds.

Choose A = H_{Λ}^{N}(A, 0), B = W, i.e., A + B = H_{Λ}^{N}(A, W). Then

|(e^{−tH_{Λ}^{N}(A,0)} − e^{−tH_{Λ}^{N}(A,W)})f|

≤ ∫_{0}^{t} \underbrace{|e^{−sH_{Λ}^{N}(A,0)} W e^{−(t−s)(H_{Λ}^{N}(A,W))} f|}_{≤ e^{−sHN Λ(0,0)} |W |e^{−(t−s)(HN Λ(0,W))} |f|} ds

=_{W ≥0}(e^{−tH_{Λ}^{N}(0,0)} − e^{−tH_{Λ}^{N}(0,W)} )|f|

Now reconstruct Dirichlet b.c.: Set W (x) := W_{n}(x) := n1_{Λ^{c}} (x) and note that (morally)

s − lim_{n→∞} e^{−tH_{Λ}^{N}(A,W_{n})} = e^{−tH_{Λ}^{D}(A,0)}

Proof of a): Let D = ∇ − iA, u_{ε} := √|u|^{2} + ε^{2} , and s_{ε} := \frac{u}{u_{ε}}. Then the quadratic form domain of the operator with magnetic field and Neumann b. c. is the domain of D.

Lemma 3 (Quadratic form version of Kato’s inequality). u ∈ D(D) ⇒ |u| ∈ D(∇) and for φ ≥ 0, φ ∈ D(∇), u ∈ D(D) we have

Re(\overline{D(s_{ε}φ)} · Du) ≥ ∇φ∇u_{ε}
= |s_{ε}|∇φ∇|u|

Remark: ∇u_{ε} = \frac{1}{u_{ε}}u_{ε}∇u_{ε} = \frac{1}{u_{ε}}∇u_{ε}^{2} = \frac{1}{u_{ε}}∇|u|^{2} =

\frac{|u|}{u_{ε}}∇u_{ε} = |s_{ε}|∇|u|.

How to use this Lemma: Note that \overline{s_{ε}}u = |s_{ε}||u|, hence we have

⟨s_{ε}φ, u⟩ = ∫ |s_{ε}|φ|u| dx ≥ 0,

and, using the above bound, we see

∫ |s_{ε}|(∇φ∇|u| + λφ|u|) dx
≤ Re (⟨D(s_{ε}φ), Du⟩ + λ⟨s_{ε}φ, u⟩)
= Re⟨s_{ε}φ, v⟩ ≤ ⟨|s_{ε}|φ, |v|⟩ ≤ ⟨φ, |v|⟩

for all λ > 0 and u = (H_{Λ}^{N}(A, 0) + λ)^{−1}v.

Taking ε → 0, we get

⟨(H_{Λ}^{N}(0, 0) + λ)φ, |u|⟩
= ⟨∇φ, ∇|u|⟩ + λ⟨φ, |u|⟩
≤ ⟨φ, |v|⟩.

Now choose φ = (H_{Λ}^{N}(0, 0) + λ)^{−1}ψ, ψ ≥ 0. Then

⟨ψ, |(H_{Λ}^{N}(A, 0) + λ)^{−1}v|⟩
≤ ⟨(H_{Λ}^{N}(0, 0) + λ)^{−1}ψ, |v|⟩
= ⟨ψ, (H_{Λ}^{N}(0, 0) + λ)^{−1}|v|⟩

for all ψ ≥ 0 and v ∈ L^{2}(Λ). I.e.,

|(H_{Λ}^{N}(A, 0) + λ)^{−1}v| ≤ H_{Λ}^{N}(0, 0) + λ)^{−1}|v| and |(H_{Λ}^{N}(A, 0)+λ)^{−n}v| ≤ H_{Λ}^{N}(0, 0)+λ)^{−n}|v| for all n ∈ ℕ.

The result for the Neumann semigroup follows, since

e^{−tH_{Λ}^{N}} = s − lim_{n→∞} (\frac{n}{t})^{n}(H^{N} + \frac{n}{t})^{−n}.