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 ) + v ( x i ) i =1 1 � i < j � N i =1 � R d ( N − 1) | Ψ | 2 ( x , x 2 , . . . , x N ) d x 2 · · · d x N ρ Ψ ( x ) := N Theorem (Hohenberg-Kohn, 1964) Let w , v 1 , v 2 ∈ ? . If there are two ground states Ψ 1 and Ψ 2 of H N ( v 1 ) and H N ( v 2 ) , such that ρ Ψ 1 = ρ Ψ 2 , then v 1 = v 2 + E 1 − E 2 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 d 2 ( R 3 ) + L ∞ ( R d ) ? = L d N 2 ( R d ) + L ∞ ( R d ) by Jerison-Kenig We can take ? = L (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

Proof of the HK theorem (1964) Ψ 2 , H N ( v 1 )Ψ 2 1 E 1 � � � � = E 2 + R d ρ ( v 1 − v 2 ). 2 Exchanging 1 ↔ 2 gives E 1 − E 2 = � R d ρ ( v 1 − v 2 ). 3 The � above is an =, hence Ψ 2 is a ground state for H N ( v 1 ), so H N ( v 1 )Ψ 2 = E 1 Ψ 2 . 4 Substracting the two eigenvalue equations for Ψ 2 gives N � � � E 1 − E 2 + ( v 2 − v 1 )( x i ) Ψ 2 = 0 . i =1 5 By strong unique continuation, |{ Ψ 2 ( X ) = 0 }| = 0, so E 1 − E 2 + � N i =1 ( v 2 − v 1 )( x i ) = 0 and integrating on [0 , L ] d ( N − 1) , we obtain v 1 = v 2 + c . Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

Strong UCP Theorem (Strong UCP for many-body Schr¨ odinger operators) Assume that the potentials satisfy � 2 d � v , w ∈ L p loc ( R d ) with p > max 3 , 2 . If Ψ ∈ H 2 loc ( R dN ) is a non zero solution to H N ( v )Ψ = E Ψ , then |{ Ψ( X ) = 0 }| = 0 . In 3 D , we can take ? = L p > 2 ( R 3 ) + L ∞ ( R 3 ) 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¨ odinger operators and the Hohenberg-Kohn theorem. • II. The Pauli Hamiltonian , (2019), arXiv:1901.03207. Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

History Weak Number of Hypothesis Date Magnetic ? or Strong particles on v (loc) L ∞ Carleman 39 W 1 (and N ) No L 2 d / 3 H¨ ormander 63 W 1 No L 2 d / 3 Georgescu 80 W N No L d Schechter-Simon 80 W No N L d / 2 Jerison-Kenig 85 S 1 No Kurata 97 S 1 Many Yes L d / 2 Koch-Tataru 01 S 1 Yes 18 S Many Yes N Laestadius-Benedicks-Penz L p > 2 d / 3 Garrigue 19 S Yes N Other related works : Kinzebulatov-Shartser (2010), Zhou (2012,2019), Lammert (2018) Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

Carleman-type inequality | ∆Ψ | 2 � If |{ Ψ( X ) = 0 }| > 0, then | X − X 0 | τ is finite for all τ for some X 0 (we take X 0 = 0) Theorem (Carleman-type inequality) Define φ ( X ) := ( − ln | X | ) − 1 / 2 . We have 2 2 � � � �� � e ( τ +2) φ Ψ e ( τ +1) φ Ψ � � � � � � τ 3 φ 5 φ 5 + τ � ∇ � � � � | X | τ +2 | X | τ +1 � � � � B 1 / 2 B 1 / 2 � � � 2 2 � e τφ Ψ e τφ ∆Ψ � �� � � � � + τ − 1 φ 5 � � � � � ∆ . � c � | X | τ � � | X | τ � � � � B 1 / 2 B 1 / 2 Proved using techniques of H¨ ormander, Koch-Tataru, ... Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

Fractional Carleman-type inequality With Hardy’s inequality | X | − 2 s � ( − ∆) s , transforms into a fractional Carleman inequality Corollary (Fractional Carleman inequality) For any δ > 0 , s ∈ [0 , 1] , s ′ ∈ 0 , 1 � � , τ � τ 0 , u ∈ C ∞ c ( B 1 \ { 0 } ) , 2 2 � � � �� � e τφ u τ 3 − 4 s � � ( − ∆) (1 − δ ) s � � � � � | X | τ � � � � � L 2 n 2 2 � � � ( − ∆) (1 − δ ) s ′ � �� � � � � � e τφ ∂ i u L 2 � κ n � e τφ ∆ u � + τ 1 − 4 s ′ � � � � � � � � L 2 . � � | X | τ � � � � | X | τ � � δ 5 / 2 � � � � � � i =1 This Carleman inequality pairs very naturally with Sobolev multipliers assumptions | V many-body | 2 � ǫ ( − ∆) 3 2 − ǫ + c in the proof of the SUCP Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

Magnetic case, the Pauli Hamiltonian N ( σ j · ( − i ∇ j + A ( x j ))) 2 + v ( x j ) � � � � H N ( v , A ) := + w ( x i − x j ) j =1 1 � i < j � N Theorem (Strong UCP for the many-body Pauli operator) Assume that the potentials satisfy div A = 0 and A ∈ L q loc ( R d ) with q > 2 d , � 2 d � curl A , v , w ∈ L p loc ( R d ) with p > max 3 , 2 . If Ψ ∈ H 2 loc ( R dN ) is a non zero solution to H N ( v , A )Ψ = E Ψ , then |{ Ψ( X ) = 0 }| = 0 . Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem

Hohenberg-Kohn for the Maxwell-Schr¨ odinger model Tellgren (2018), Ruggenthaler et al. (2014). 1 � � � | curl a | 2 Ψ , H N ( v , a + A )Ψ E v , A (Ψ , a ) := + 8 πα 2 � � � � v + | a + A | 2 � Ψ , H N = 0 Ψ + ρ Ψ � 1 � | curl a | 2 + 2 ( j Ψ + curl m Ψ ) · ( A + a ) + 8 πα 2 Internal current j (Ψ , a ) := j Ψ + curl m Ψ + ρ Ψ a Theorem (Hohenberg-Kohn for Maxwell DFT) Let p > 2 and q > 6 and let w , v 1 , v 2 ∈ L p ( R 3 ) + L ∞ ( R 3 ) , L q loc ∩ H 1 � ( R 3 ) be such that E v 1 , A 1 and E v 2 , A 2 are � A 1 , A 2 ∈ bounded from below and admit ground states (Ψ 1 , a 1 ) and (Ψ 2 , a 2 ) . If ρ Ψ 1 = ρ Ψ 2 and j (Ψ 1 , a 1 ) = j (Ψ 2 , a 2 ) , then A 1 = A 2 and v 1 = v 2 + ( E 1 − E 2 ) / N. Louis Garrigue Unique continuation for the Hohenberg-Kohn theorem