On the strong Scott conjecture for Chandrasekhar atoms Konstantin - PowerPoint PPT Presentation

On the strong Scott conjecture for Chandrasekhar atoms Konstantin Merz 1 Joint work with Rupert L. Frank 2, 3 , Heinz Siedentop 2 , and Barry Simon 3 1 Technische Universit¨ at Braunschweig, Germany 2 Ludwig–Maximilians–Universit¨ at M¨ unchen, Germany 2 California Institute of Technology Pasadena, CA, U.S.A. Luminy, October 22, 2019 K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 1

Chandrasekhar operator ( � = m = e = 1) Z Z �� � 1 − c 2 ∆ ν + c 4 − c 2 − Z � � � L 2 ( R 3 ) C Z := + | x ν − x µ | in | x ν | ν =1 1 ≤ ν<µ ≤ Z ν =1 where c is the velocity of light. Under x �→ x / c , C Z is unitarily equivalent to   Z � √ c − 1 γ � c 2 � � − ∆ ν + 1 − 1 − +   | x ν | | x ν − x µ | ν =1 1 ≤ ν<µ ≤ Z with γ := Z / c . ◮ C Z is bounded from below, if and only if γ ≤ 2 /π (Kato’s inequality) ◮ Study of properties of ground states as Z , c → ∞ keeping γ ≤ 2 /π fixed. K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 2

Ground state energy as Z , c → ∞ with γ fixed � 1 � Z 2 + O ( Z 47 / 24 ) . inf spec( C Z ) = E TF ( Z ) + 4 − s C ( γ ) Here, � � ρ ( x ) ρ ( y ) E TF ( Z ) := inf {E TF [ ρ ] : 0 ≤ ρ ∈ L 5 / 3 ( R 3 ) , | x − y | dx dy < ∞} = e TF Z 7 / 3 Z ( x ) = Z 2 ρ TF ( Z 1 / 3 x ), and is the non-relativistic Thomas–Fermi energy, ρ TF 1 ��� � � p 2 p 2 + 1 − 1 − γ � 2 − γ � s C ( γ ) := γ − 2 Tr − > 0 | x | | x | − − The leading order was proven by Sørensen (’05), the Scott correction ( Z 2 ) by Frank–Siedentop–Warzel (’08) and Solovej–Sørensen–Spitzer (’10). What can be said about the density on these length scales? � � 2 dx ′ � ψ ( x , x ′ ) � � ρ ( x ) := N R 3( N − 1) for a ground state ψ of C Z K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 3

Hydrogen Thomas–Fermi One-particle density as Z , c → ∞ , Z / c = γ (1) Z − 2 ρ ( Z − 1 3 x ) → ρ TF ( x ) weakly and in Coulomb norm where 1 1 ρ TF Z ( x ) = Z 2 ρ TF 3 x ) (with H. Siedentop ) ( Z 1 (2) Strong Scott conjecture: ground state density on the length scale Z − 1 (with R. L. Frank , H. Siedentop , and B. Simon ) ρ ( r ) Z 3 3 on scale Z − 1 ≪ r ≪ Z − 1 2 3 ( Z/r ) Z 2 Z − 1 Z − 1 / 3 1 r K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 4

Ground state density on the length scale Z − 1 R 3( N − 1) | ψ ( x , x ′ ) | 2 dx ′ , and � Let ψ be a ground state of C Z , ρ ( x ) := N 2 � � ρ ℓ ( r ) := Nr 2 (2 ℓ + 1) − 1 � ℓ S 2 Y ℓ, m ( ω ) ψ ( r ω, x ′ ) d ω dx ′ . � � � � R 3( N − 1) m = − ℓ � � The normalized eigenfunctions of the hydrogenic operator � − d 2 dr 2 + ℓ ( ℓ + 1) + 1 − 1 − γ r in L 2 ( R + , dr ) . C H ℓ,γ = r 2 n ,ℓ ( r ) | 2 and ℓ ( r ) := � ∞ are denoted by ψ H n ,ℓ . Then we define ρ H n =0 | ψ H 4 π r 2 ρ H ( r ) := � ∞ ℓ =0 (2 ℓ + 1) ρ H ℓ ( r ). Theorem Let γ ∈ (0 , 2 /π ) , ℓ ∈ N 0 , and U = U 1 + U 2 with U 1 ∈ r − 1 L ∞ comp and U 2 ∈ D γ . Then, for Z , c → ∞ with Z / c = γ fixed, � ∞ � ∞ c − 3 ρ ℓ ( r / c ) U ( r ) dr = ρ H lim ℓ ( r ) U ( r ) dr Z →∞ 0 0 � � R 3 c − 3 ρ ( x / c ) U ( | x | ) dx = R 3 ρ H ( | x | ) U ( | x | ) dx . lim Z →∞ For instance, U ( r ) = r − 1 1 { r < 1 } + r − 3 / 2 − ε 1 { r ≥ 1 } . K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 5

Hydrogenic density Does the limit even exist? (And in which sense?) To formulate the result precisely, we introduce [0 , 1] → [0 , 2 /π ] σ �→ Φ( σ ) := (1 − σ ) tan πσ 2 . Note that Φ is strictly monotone on [0 , 1] with Φ(0) = 0 and lim σ → 1 Φ( σ ) = 2 /π . Thus, there is a unique σ γ ∈ [0 , 1] such that Φ( σ γ ) = γ . Theorem Let γ ∈ (0 , 2 /π ) and 1 / 2 < s < min { 3 / 2 − σ γ , 3 / 4 } . Then ρ H ( r ) � s ,γ r 2 s − 3 1 { r ≤ 1 } + r − 3 / 2 1 { r ≥ 1 } . Recalling ρ H Z , c ( x ) = c 3 ρ H ( cx ) and ρ TF Z ( x ) = Z 2 ρ TF ( Z 1 / 3 x ) ∼ ( Z / | x | ) 3 / 2 , this 1 shows that there is a transition between the length scales Z − 1 and Z − 1 / 3 . K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 6

Remarks ◮ The result is well-known in the non-relativistic case, see Iantchenko–Lieb–Siedentop (’96). They even proved pointwise convergence which is not expected in the relativistic case since the hydrogenic eigenfunctions are not necessarily bounded. ◮ Heilmann–Lieb (’95) studied ρ H in the non-relativistic case very well. They showed that ρ H decreases monotonically and proved an asymptotic expansion as r → ∞ . ◮ Although we show ρ H ( r ) � r − 3 / 2 , the constant is implicit (and presumably far from optimal) and we are lacking a corresponding lower bound. Moreover, the singularities and monotonicity of ρ H are still unknown. K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 7

Strategy of the proof We follow the lines of Iantchenko–Lieb–Siedentop (’96) by employing a linear-response argument. Let C Z ,λ := C Z − λ � Z ν =1 c 2 U ( c | x ν | )Π ℓ,ν , w.l.o.g. U ≥ 0, and λ > 0. By the Scott correction, � ∞ 1 � | ψ �� ψ | ( C Z − C Z ,λ ) � c − 3 ρ ℓ ( r / c ) U ( r ) dr = lim 2 ℓ + 1 lim Z →∞ Tr lim λ c 2 Z →∞ λ ց 0 0 47 Tr( C H ℓ, Z − λ U c ( r )) − − Tr( C H ℓ, Z ) − + aZ 24 ≤ lim lim λ c 2 Z →∞ λ ց 0 with � � − d 2 � dr 2 + ℓ ( ℓ + 1) + c 4 − c 2 − Z r in L 2 ( R + , dr ) . C H ℓ, Z := c 2 r 2 Goal: Interchange lim inf λ ց 0 and Tr to apply standard perturbation theory, i.e., the Feynman–Hellmann theorem or compute the derivative! To prove the convergence of c − 3 ρ ( x / c ), we also need a bound like Tr( C ℓ − V − λ U ( r )) − − Tr( C ℓ − V ) − � λ ( ℓ + 1 / 2) − 2 − ε where 0 ≤ V ≤ γ/ r . K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 8

Strategy of the proof We follow the lines of Iantchenko–Lieb–Siedentop (’96) by employing a linear-response argument. Let C Z ,λ := C Z − λ � Z ν =1 c 2 U ( c | x ν | )Π ℓ,ν , w.l.o.g. U ≥ 0, and λ > 0. By the Scott correction, � ∞ 1 � | ψ �� ψ | ( C Z − C Z ,λ ) � c − 3 ρ ℓ ( r / c ) U ( r ) dr = lim 2 ℓ + 1 lim Z →∞ Tr lim λ c 2 Z →∞ λ ց 0 0 Tr( C H ℓ,γ − λ U ( r )) − − Tr( C H ℓ,γ ) − ≤ lim λ λ ց 0 with � − d 2 dr 2 + ℓ ( ℓ + 1) + 1 − 1 − γ r in L 2 ( R + , dr ) . C H ℓ,γ := r 2 Goal: Interchange lim inf λ ց 0 and Tr to apply standard perturbation theory, i.e., the Feynman–Hellmann theorem or compute the derivative! To prove the convergence of c − 3 ρ ( x / c ), we also need a bound like Tr( C ℓ − V − λ U ( r )) − − Tr( C ℓ − V ) − � λ ( ℓ + 1 / 2) − 2 − ε where 0 ≤ V ≤ γ/ r . K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 8

Generalized Feynman–Hellmann theorems Assume that A is self-adjoint with A − trace class. Assume that B is non-negative, relatively form bounded with respect to A and that that there is 1 / 2 ≤ s ≤ 1 such that for some M > | inf spec( A ) | , ( A + M ) − s B ( A + M ) − s is trace class. Assume that at least one of the following conditions hold. � < ∞ . ◮ lim sup λ → 0 � � ( A + M ) s ( A − λ B + M ) − s � ◮ There is a s ′ such that max { s ′ , 1 / 2 } < s < 1 and an a > 0 such that B 2 s ≤ a ( A + M ) 2 s ′ . Then the one-sided derivatives of λ �→ S ( λ ) := Tr( A − λ B ) − satisfy Tr B χ ( −∞ , 0) ( A ) = D − S (0) ≤ D + S (0) = Tr B χ ( −∞ , 0] ( A ) . In particular, S ( λ ) is differentiable at λ = 0 , if and only if B | ker A = 0 . In our case, A = C H ℓ,γ (which has no zero eigenvalue) and B = U in L 2 ( R + ). K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 9

Remarks (1) If inf σ ess ( A ) > 0, the result follows from the classic Feynman–Hellmann theorem. (2) S ( λ ) is convex by the variational principle, i.e., D ± S exist. Moreover, Tr B χ ( −∞ , 0] ( A ) = Tr(( A + M ) 2 s χ ( −∞ , 0] ( A ))(( A + M ) − s B ( A + M ) − s ) < ∞ by the form trace class condition. (3) The assumption B 2 s ≤ a ( A + M ) 2 s ′ for max { 1 / 2 , s ′ } < s < 1 implies � < ∞ (see also Neidhardt–Zagrebnov � � ( A + M ) s ( A − λ B + M ) − s � lim sup λ → 0 (’99)). (4) We will work with s > 1 / 2 since � ∞ � ∞ krJ ℓ +1 / 2 ( kr ) 2 � U 1 / 2 ( C H ℓ,γ + M ) − 1 / 2 � 2 2 ∼ dr U ( r ) √ dk k 2 + 1 − 1 + M 0 0 diverges logarithmically. (5) Decisive inequality (Frank, M., Siedentop): for − γ/ | x | ≤ V ≤ − ˜ γ/ | x | with 2 /π ≥ γ ≥ ˜ γ and s < 3 / 2 − σ γ , we have | p | 2 s � ( | p | + V ) 2 s . K. Merz – Strong Scott conjecture – CIRM, October 22, 2019 Nr. 10

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