Schrödinger’s equation with random potentials
Marius Beceanu Jürg Fröhlich Avy Soffer Pasadena February 2015
- M. Beceanu J. Fröhlich A. Soffer
Random Potentials 1/15
Schrdingers equation with random potentials Marius Beceanu Jrg - - PowerPoint PPT Presentation
Schrdingers equation with random potentials Marius Beceanu Jrg Frhlich Avy Soffer Pasadena February 2015 M. Beceanu J. Frhlich A. Soffer Random Potentials 1/15 Introduction Consider the linear Schrdinger equation in R d with
Marius Beceanu Jürg Fröhlich Avy Soffer Pasadena February 2015
Random Potentials 1/15
Consider the linear Schrödinger equation in Rd with random time-dependent potential i∂tψ(x, t) − ∆ψ(x, t) + Vω(x)ψ(x, t) = 0, ψ(0) = ψ0. Here Vω := V (x, Xt). Conserved quantities for constant V : M[ψ] :=
E[ψ](t) :=
Dispersive inequalities (same with more derivatives): ψL∞
t L2 x + PcψL2 t L2d/(d−2) x
ψ2 (Strichartz) D1/2PcψL2
t L2 x(Q) |Q|1/2dψ2 (local smoothing).
Constant V : E and M conserved; Pcψ disperses. Time-dependent V : M is conserved.
Random Potentials 2/15
Consider the equation on R3 (or d ≥ 3) iψt − ∆ψ + V (x, Xt)ψ + ǫ(χ ∗ |ψ|2)ψ = 0, ψ(0) = ψ0 given. Xt standard Brownian motion on bounded subset of Riemannian manifold; nontrivial V (x, y) ∈ Cy(L∞
x ∩ L1 x); Hartree-type potential with
small coupling constant. For any ψ0 ∈ L2 and |ǫ| < ǫ(ψ02), there is a.s. a global solution ψ s.t. Eψ2
L2
t L6,2 x
ψ02
2.
Moreover, if AyV (x, y) ∈ L∞
y (L1 x ∩ L∞ x ), then the energy remains
bounded on average: E(∇ψω(t)2
L2
x) ≤ ∇ψ02
L2
x + Cψ02
L2
x.
Model problem: −∆ + V0(x) + g(Wt)V1(x), V0 ∈ C ∞ large fixed potential, Wt standard Brownian motion on T1, V1 perturbation, coupling g(y) = 0.1 sin(y). Other simple perturbations: e.g. −∆ + V0(x) + V1(x − Wt), rotations,
Random Potentials 3/15
1974 Ovchinnikov. 1985, 1986 Pillet: started the study, Feynman-Kac formula, linear L2 wave operators. 1989, 1990 Cheremshantsev: Brownian motion over the whole space. 2009 Kang, Schenker: discrete Schrödinger, translation-invariant potential. 2010 De Roeck, Fröhlich, Pazzo: For a quantum particle interacting with infinitely many thermal reservoirs, they proved a central limit theorem, diffusive scaling for the second momentum, with no corrections, and a distribution law for finite energy states. 2013 Beceanu, Soffer: Strichartz estimates, other properties; Brownian motion on the whole space.
Random Potentials 4/15
Consider the inhomogenous linear equation with random potential Vω := V (x, Xt) i∂tψω − ∆ψω + Vωψω = Ψω, ψω(0) := ψ0. Define g(x, y, t) := E(ψ(x, t) | Xt = y), f (x1, x2, y, t) := E(ψ(x1, t)ψ(x2, t) | Xt = y). Conditional expectations of ψ, respectively of the density matrix ψ ⊗ ψ, at time t and under the condition that Xt = y.
Random Potentials 5/15
As shown by Pillet, i∂tg − ∆xg − iAyg + V (x, y)g = G, g(0) := ψ0(x)µ0(y), i∂tf − ∆x1f + ∆x2f − iAyf + (V ⊗ 1 − 1 ⊗ V )f = F, f (0) := ψ0(x1)ψ0(x2)µ0(y), where µ0(y) is the initial distribution of Xt, i.e. of X0, and G(x, y, t) := E(Ψ(x, t) | Xt = y), F(x1, x2, y, t) := E(Ψ(x1, t)ψ(x2, t) − ψ(x1, t)Ψ(x2, t) | Xt = y).
Random Potentials 6/15
Pillet also proved the following Feinman-Kac-type formula: E
We compute the right-hand side by using the equation of f . The left-hand side controls Strichartz estimates: Eψω2
L2
t L6,2 x
ψ02
2 + E
Random Potentials 7/15
Consider the solution g to i∂tg − ∆xg + iAyg + V (x1, y)g = 0. Enough to prove that g ∈ L2
ωL2 t,yL6,2 x .
Non-triviality assumption on V : For almost every x in some open set O there exist y1, y2 such that V (x, y1) = V (x, y2). Then −∆x + iAy + V has no bound states for ℑλ ≤ 0. Proof: bound state φ(x, y) = ⇒ independent of y = ⇒ for each y solves (−∆ + V (x, y))φ(x) = λφ(x). V (x, y) are distinct = ⇒ φ = 0 on
⇒ φ ≡ 0.
Random Potentials 8/15
Let WX := t−3/2L∞
t B(X). Suppose T ∈ WX is s.t.
limδ→0 T(ρ) − T(ρ − δ)WX = 0. If I + ˆ T(λ) is invertible in B(X) for every λ ∈ R, then 1 + T possesses an inverse in WX of the form 1 + S.
Consider the equation, for V ∈ Cy(L∞
x ∩ L1 x) and nontrivial,
i∂tf − ∆xf + iAyf + V (x, y)f = 0, f (0) = f0 ∈ L1
yL2 x.
Then Strichartz: f L2
y L2 t L6,2 x
f0L2
y L2 x.
Moreover, f Lp
y (L2 x+L∞ x ) t−3/2f0Lp y (L1 x∩L2 x).
Random Potentials 9/15
Rewrite Duhamel’s identity symetrically I − χt>0i|V |1/2(x, y)eit(−∆x+V (x,y)+iAy )|V |1/2 sgn V (x, y) = =
−1. Apply Wiener’s theorem. Leads to I − χt>0i|V |1/2(x, y)eit(−∆x+iAy +V (x,y))|V |1/2 sgn V (x, y) being L1
t,yL2 x-bounded. Conclusion follows by Duhamel.
Random Potentials 10/15
Repeat for density matrix. Let V ⊗ 1 + 1 ⊗ V = V1V2, where V2 :=
1 ⊗ |V |1/2 sgn V
1 ⊗ |V |1/2 . The free resolvent is RA(λ) := (−∆x1 + ∆x2 − iAy − λ)−1. Need to prove that V2RA(λ)V1 is L2-bounded and, after taking the Fourier transform, I + V2RA(λ)V1 is invertible for every λ. Matrix form I + T = I + T11 T12 T21 I + T22
Random Potentials 11/15
I+i T =
T11 I + i T22 I i(I + i T11)−1 T12 i(I + i T22)−1 T21 I
Diagonal terms are invertible by reduction to previous case: (I+iT11)−1 = I−iχt>0eit∆x2 |V |1/2(x1)eit(−∆x1+V (x1,y)+iAy )|V |1/2 sgn V (x1). Now we need to study I +
T11)−1 T12 i(I + i T22)−1 T21
S1 i S2
S. Then S2 = S1S2 S2S1
After squaring, these off-diagonal terms are compact. One can use Fredholm’s alternative for I + S2 = (I − iS)(I + iS) (same for both). Just as importantly, Wiener’s theorem applies to I + S2.
Random Potentials 12/15
Write (I + iT)−1 in terms of (I + S2)−1 and the diagonal terms (I + iTkk)−1. Get almost all integrable components, convolved with something explicit. Use this to evaluate f . In fact much more complicated. Conclusion:
L2.
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Proof: Contraction scheme. By L2 norm conservation, a priori ψ ∈ L∞
ω L∞ t L2 x.
The inhomogenous source term is small in L1
ωL1 t J1x1,x2 because
(χ ∗ |ψ|2)ψ ⊗ ψL1
ωL1 t J1x1,x2 ψ2
L2
ωL2 t L6,2 x ψ2
L∞
ω L∞ t L2 x.
and ψ belongs to L2
ωL2 t L6,2 x
by Strichartz.
Random Potentials 14/15
Random Potentials 15/15