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Unique continuation for the Hohenberg-Kohn theorem Louis Garrigue - - PowerPoint PPT Presentation

Unique continuation for the Hohenberg-Kohn theorem Louis Garrigue Banff, January 28, 2019 Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem Hohenberg-Kohn theorem N N H N ( v ) := i + w ( x i x j )


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Unique continuation for the Hohenberg-Kohn theorem

Louis Garrigue Banff, January 28, 2019

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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Hohenberg-Kohn theorem

HN(v) :=

N

  • i=1

−∆i +

  • 1i<jN

w(xi − xj) +

N

  • i=1

v(xi) ρΨ(x) := N

  • Rd(N−1) |Ψ|2 (x, x2, . . . , xN)dx2 · · · dxN

Theorem (Hohenberg-Kohn, 1964) Let w, v1, v2 ∈ ?. If there are two ground states Ψ1 and Ψ2 of HN(v1) and HN(v2), such that ρΨ1 = ρΨ2, then v1 = v2 + E1−E2

N

Works for bosons and fermions, in any dimension d (1983) Lieb remarked that it relied on a strong unique continuation property (SUCP). He conjectured that ? = L

d 2 (R3) + L∞(Rd)

We can take ? = L

dN 2 (Rd) + L∞(Rd) by Jerison-Kenig

(1985), but this covers Coulomb singularities only for N = 1. We need a power independent of N

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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Proof of the HK theorem (1964)

1 E1

  • Ψ2, HN(v1)Ψ2
  • = E2 +
  • Rd ρ(v1 − v2).

2 Exchanging 1 ↔ 2 gives E1 − E2 =

  • Rd ρ(v1 − v2).

3 The above is an =, hence Ψ2 is a ground state for HN(v1),

so HN(v1)Ψ2 = E1Ψ2.

4 Substracting the two eigenvalue equations for Ψ2 gives

  • E1 − E2 +

N

  • i=1

(v2 − v1)(xi)

  • Ψ2 = 0.

5 By strong unique continuation, |{Ψ2(X) = 0}| = 0, so

E1 − E2 + N

i=1(v2 − v1)(xi) = 0 and integrating on

[0, L]d(N−1), we obtain v1 = v2 + c.

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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Strong UCP

Theorem (Strong UCP for many-body Schr¨

  • dinger operators)

Assume that the potentials satisfy v, w ∈ Lp

loc(Rd)

with p > max 2d 3 , 2

  • .

If Ψ ∈ H2

loc(RdN) is a non zero solution to HN(v)Ψ = EΨ, then

|{Ψ(X) = 0}| = 0. In 3D, we can take ? = Lp>2(R3) + L∞(R3) in the HK

  • theorem. Covers the physical case, ie Coulomb-like

singularities Works for excited states Already known that {Ψ(X) = 0} is not an open set (Georgescu 1980)

  • L. Garrigue, Unique continuation for many-body Schr¨
  • dinger operators and the Hohenberg-Kohn theorem.
  • II. The Pauli Hamiltonian, (2019), arXiv:1901.03207.

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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History

Date Weak

  • r Strong

Number of particles Hypothesis

  • n v (loc)

Magnetic ?

Carleman 39 W 1 (and N) L∞ No H¨

  • rmander

63 W 1 L2d/3 No Georgescu 80 W N L2d/3 No Schechter-Simon 80 W N Ld No Jerison-Kenig 85 S 1 Ld/2 No Kurata 97 S 1 Many Yes Koch-Tataru 01 S 1 Ld/2 Yes

Laestadius-Benedicks-Penz

18 S N Many Yes Garrigue 19 S N Lp>2d/3 Yes Other related works : Kinzebulatov-Shartser (2010), Zhou (2012,2019), Lammert (2018)

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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Carleman-type inequality

If |{Ψ(X) = 0}| > 0, then

  • |∆Ψ|2

|X−X0|τ is finite for all τ for some

X0 (we take X0 = 0) Theorem (Carleman-type inequality) Define φ(X) := (− ln |X|)−1/2. We have τ 3

  • B1/2

φ5

  • e(τ+2)φΨ

|X|τ+2

  • 2

+ τ

  • B1/2

φ5

  • e(τ+1)φΨ

|X|τ+1

  • 2

+ τ −1

  • B1/2

φ5

eτφΨ |X|τ

  • 2

c

  • B1/2
  • eτφ∆Ψ

|X|τ

  • 2

. Proved using techniques of H¨

  • rmander, Koch-Tataru, ...

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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Fractional Carleman-type inequality

With Hardy’s inequality |X|−2s (−∆)s, transforms into a fractional Carleman inequality Corollary (Fractional Carleman inequality) For any δ > 0, s ∈ [0, 1], s′ ∈

  • 0, 1

2

  • , τ τ0, u ∈ C ∞

c (B1\ {0}),

τ 3−4s

  • (−∆)(1−δ)s
  • eτφ u

|X|τ

  • 2

L2

+ τ 1−4s′

n

  • i=1
  • (−∆)(1−δ)s′

eτφ ∂iu |X|τ

  • 2

L2 κn

δ5/2

  • eτφ ∆u

|X|τ

  • 2

L2 .

This Carleman inequality pairs very naturally with Sobolev multipliers assumptions |Vmany-body|2 ǫ(−∆)

3 2 −ǫ + c in the

proof of the SUCP

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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Magnetic case, the Pauli Hamiltonian

HN(v, A) :=

N

  • j=1
  • (σj · (−i∇j + A(xj)))2 + v(xj)
  • +
  • 1i<jN

w(xi − xj) Theorem (Strong UCP for the many-body Pauli operator) Assume that the potentials satisfy div A = 0 and A ∈ Lq

loc(Rd)

with q > 2d, curl A, v, w ∈ Lp

loc(Rd)

with p > max 2d 3 , 2

  • .

If Ψ ∈ H2

loc(RdN) is a non zero solution to HN(v, A)Ψ = EΨ, then

|{Ψ(X) = 0}| = 0.

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

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Hohenberg-Kohn for the Maxwell-Schr¨

  • dinger model

Tellgren (2018), Ruggenthaler et al. (2014). Ev,A(Ψ, a) :=

  • Ψ, HN(v, a + A)Ψ
  • +

1 8πα2

  • |curl a|2

=

  • Ψ, HN

0 Ψ

  • +
  • ρΨ
  • v + |a + A|2

+ 2

  • (jΨ + curl mΨ) · (A + a) +

1 8πα2

  • |curl a|2

Internal current j(Ψ,a) := jΨ + curl mΨ + ρΨa Theorem (Hohenberg-Kohn for Maxwell DFT) Let p > 2 and q > 6 and let w, v1, v2 ∈ Lp(R3) + L∞(R3), A1, A2 ∈

  • Lq

loc ∩ H1

(R3) be such that Ev1,A1 and Ev2,A2 are bounded from below and admit ground states (Ψ1, a1) and (Ψ2, a2). If ρΨ1 = ρΨ2 and j(Ψ1,a1) = j(Ψ2,a2), then A1 = A2 and v1 = v2 + (E1 − E2)/N.

Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem