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Critical BCS-temperature in a constant magnetic field Christian HAINZL (Universit at T ubingen) Venice, 21.8.2017 Christian HAINZL (T ubingen) Critical temperature August 18, 2017 1 / 15 We study the bottom of the spectrum of the


  1. Critical BCS-temperature in a constant magnetic field Christian HAINZL (Universit¨ at T¨ ubingen) Venice, 21.8.2017 Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 1 / 15

  2. We study the bottom of the spectrum of the following two-particle operator: This is the second variation of the corresponding BCS-functional. M T , B + V : L 2 ( R 3 × R 3 ) → L 2 ( R 3 × R 3 ) h x + h y M T , B = , V = V ( x − y ) tanh h x 2 T + tanh h y 2 T � 2 � − i ∇ x + B Π x = p x + B − µ = Π 2 h x = 2 ∧ x x − µ, 2 ∧ x with µ chemical potential, T temperature, and B = (0 , 0 , B ) Goal: Obtain T as a function of B under the condition that inf σ ( M T , B + V ) = 0 . In this way we want to obtain the critical temperature as a function of B , i.e., T c ( B ), for small B . Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 2 / 15

  3. Idea The ground-state eigenfunction separates into relative and center-of-mass motion � x + y � α ( x , y ) ≃ α ∗ ( x − y ) ψ = α ∗ ( r ) ψ ( X ) , 2 p r = p x − p y x = r 2 + X p x = p r + p X r = x − y , , 2 , 2 2 X = x + y y = − r 2 + X 2 , p y = − p r + p X , p X = p x + p y , 2 2 Π x = p x + B 2 ∧ x = p r + p X 2 + B 4 ∧ r + B 2 ∧ X = Π r + Π X 2 Π y = p y + B 2 ∧ y = − p r + p X 2 − B 4 ∧ r + B 2 ∧ X = − Π r + Π X 2 � 2 − µ + � 2 − µ � Π r + Π X Π r − Π X � � � 2 2 � α, M T , B α � = α, α � 2 − µ � 2 − µ � Π X � Π X Π r + Π r − 2 2 tanh + tanh 2 T 2 T Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 3 / 15

  4. Idea The ground-state eigenfunction separates into relative and center-of-mass motion � x + y � α ( x , y ) ≃ α ∗ ( x − y ) ψ = α ∗ ( r ) ψ ( X ) , 2 p r = p x − p y x = r 2 + X p x = p r + p X r = x − y , , 2 , 2 2 X = x + y y = − r 2 + X 2 , p y = − p r + p X , p X = p x + p y , 2 2 Π x = p x + B � p r + B � � p X 2 + B � = Π r + Π X 2 ∧ x = 4 ∧ r + 2 ∧ X 2 Π y = p y + B � p r + B � � p X 2 + B � = − Π r + Π X 2 ∧ y = − 4 ∧ r + 2 ∧ X 2 � 2 − µ + � 2 − µ � � Π r + Π X Π r − Π X � � 2 2 � α, M T , B α � = α, α � 2 − µ � 2 − µ � Π X � Π X Π r + Π r − 2 2 tanh + tanh 2 T 2 T Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 4 / 15

  5. Idea α ≃ α ∗ ( r ) ψ ( X ) Π 2 r − µ α ∗ �� ψ � 2 2 + Λ 0 � ψ, Π 2 � α, M T , B α � ≃ � α ∗ , X ψ � + o ( B ) tanh Π 2 r − µ 2 T 0 = � α � =1 � α, ( M T , B + V ) α � inf � � p 2 r − µ � ψ � =1 � ψ, Π 2 ≃ � α ∗ � =1 � α ∗ , inf + V ( r ) α ∗ � + Λ 0 inf X ψ � + o ( B ) tanh p 2 r − µ 2 T T c (0) − T ≃ − Λ 2 + Λ 0 2 B + o ( B ) , (1) T c (0) hence � 1 − Λ 0 � T ( B ) = T c ( B ) = T c (0) 2 B + o ( B ) . Λ 2 Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 5 / 15

  6. Difficulties There are severe difficulties: M T , B is an ugly symbol. B ∧ x is an unbounded perturbation of an unbounded operator the components of ( − i ∇ + B 2 ∧ x ) do not commute We need an operator inequality of the form p 2 r − µ + V ( r ) + c Π 2 M T , B + V ≥ X + o ( B ) , c > 0 tanh p 2 r − µ 2 T As a way out, we will deal with the Birman-Schwinger version: 1 ⇒ V 1 / 2 V 1 / 2 ϕ = ϕ, ϕ = V 1 / 2 α V ≥ 0 : ( M T , B − V ) α = 0 ⇐ M T , B hence we study the equation � 1 � 1 − V 1 / 2 V 1 / 2 0 = inf σ M T , B Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 6 / 15

  7. B = 0 and T c (0) In order to determine T c (0) one has to solve for T : 0 = inf σ ( M T , 0 + V ) We abbreviate M T = M T , 0 which can be represented as multiplication operator. p 2 − µ + q 2 − µ � M T α ( p , q ) = α ( p , q ) ˆ tanh p 2 − µ + tanh q 2 − µ 2 T 2 T One has the algebraic inequality � p 2 − µ q 2 − µ � M T ( p , q ) ≥ 1 + ≥ 2 T , tanh p 2 − µ tanh q 2 − µ 2 2 T 2 T since x ≥ 2 T . x tanh 2 T The task to solve for the critical temperature T c is non-trivial, even in the B = 0 case. Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 7 / 15

  8. M T + V for p X = 0 [HHSS] At p X = 0 K T ( p r ) M T ( p r + p X 2 , − p r + p X 2 ) = M T ( p r , − p r ) p r p 2 r − µ = K T ( p r ) = tanh(( p 2 r − µ ) / 2 T ) p 2 r − µ K T ( − i ∇ ) + V ( r ) : L 2 ( R 3 ) → L 2 ( R 3 ) . Critical temperature: Since the operator K T + V is monotone in T , there exists unique 0 ≤ T c < ∞ such that inf σ ( K T c + V ) = 0 , respectively 0 is the lowest eigenvalue of K T c + V . T c is the critical temperature for the effective one particle system ( p X = 0). [HHSS] C. Hainzl, E. Hamza, R. Seiringer, J.P. Solovej, Commun. Math. Phys. 281 , 349–367 (2008) . Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 8 / 15

  9. Known results about K T + V p 2 − µ = | p 2 − µ | , hence lim T → 0 tanh p 2 − µ 2 T T c > 0 iff inf σ ( | p 2 − µ | + V ) < 0 | p 2 − µ | has same type of singularity as 1 / p 2 in 2 D [S]. 1 � In [FHNS, HS08, HS16] we classify V ’s such that T c > 0. (E.g. V < 0 is enough) In [LSW] shown that | p 2 − µ | + V has ∞ many eigenvalues if V ≤ 0. the operator appeared in terms of scattering theory [BY93] [FHNS] R. Frank, C. Hainzl, S. Naboko, R. Seiringer, Journal of Geometric Analysis, 17 , No 4, 549-567 (2007) [HS08] C. Hainzl, R. Seiringer, Phys. Rev. B, 77 , 184517 (2008) [HS16] C. Hainzl, R. Seiringer, J. Math. Phys. 57 (2016), no. 2, 021101 [BY93] Birman, Yafaev, St. Petersburg Math. J. 4, 1055-1079 (1993) [LSW] A. Laptev, O. Safronov, T. Weidl, Nonlinear problems in mathematical physics and related topics I, pp. 233-246, Int. Math. Ser. (N.Y.), Kluwer/Plenum, New York (2002) [S] B. Simon, Ann. Phys. 97, 279-288 (1976) Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 9 / 15

  10. Lemma (FHSS12) Let the 0 eigenvector of K T c + V be non-degenerate. Then (a) M T c + V � p 2 X (b) inf σ ( M T c + V ) = 0 meaning T c (0) for the two-particle system is determined by T c the critical temperature of the one-particle operator K T + V , which satisfies 0 = inf σ ( K T c + V ) . The proof of is non-trivial, because M T ( p r + p X 2 , − p r + p X 2 ) �≥ M T ( p r , − p r ) = K T ( p r ) . (a) only holds for V = V ( x − y ) , not for general V ( x , y ). In the presence of B the proof is significantly harder and the main difficulty of our work. [FHSS12] R.L. Frank, C. Hainzl, R. Seiringer, J.P. Solovej, J. Amer. Math. Soc. 25 , 667–713 (2012). Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 10 / 15

  11. Main Theorem � 1 � 1 − V 1 / 2 V 1 / 2 0 = inf σ (2) M T , B Theorem (FHL17) Let V ≤ 0 , V and | r | V ( r ) in L ∞ , and the 0 -eigenvector of K T c + V be non-degenerate, then there are parameters Λ 0 , Λ 2 , depending on V , µ , such that there exists a solution T c ( B ) of equation (2) such that for small B � � 1 − Λ 0 T c ( B ) = T c 2 B + o ( B ) , Λ 2 This proves and generalizes a famous result of Helfand, Hohenberg and Werthamer [HHW]. [FHL17] R. L. Frank, C. Hainzl, E. Langmann, (2017) [HHW] E. Helfand, Hohenberg, N.R. Werthamer, Phys. Rev. 147, 288 (1966) Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 11 / 15

  12. Ingredients of the proof Advantage: M − 1 T , B can be expressed in terms of resolvents. � z 1 � 1 1 1 1 � M − 1 � T , B = tanh dz = T 2 i π 2 T z − h x z + h y h x − i ω n h y + i ω n C n ∈ Z with ω n = π (2 n + 1) T . ( M − 1 T , B V 1 / 2 α )( x , y ) = � � 1 1 ( y , y ′ ) V 1 / 2 ( x ′ − y ′ ) α ( x ′ , y ′ ) dx ′ dy ′ T � ( x , x ′ ) (3) h B − i ω n h B + i ω n n ∈ Z We show 1 1 ( x , y ) ≃ e − i B 2 · x ∧ y ( x − y ) z − h B z − h 0 and introduce center-of-mass and relative coordinates, and use e − iZ · ( B ∧ X ) ψ ( X − Z ) = e − iZ · ( B ∧ X ) ( e − iZ · p X ψ )( X ) = ( e − iZ · Π X ψ )( X ) . Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 12 / 15

  13. High- T c -superconductors But if V is more general, V = V ( r , X ), then M T + V does not necessarily attain its infimum for p X = 0 [HL]. This suggests a different type of pair formation in situations where the interaction is not translation-invariant. We suggest that this happens in high-T c -superconductors. [HL] C. Hainzl, M. Loss, EPJ B (2017) Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 13 / 15

  14. Previous results on the influence of A , W on critical temperature In previous works we investigated the change of the critical temperature by bounded external fields Felder h 2 W , h A by means of the full non-linear functional. The shift in the critical temperature happens through the lowest eigenvalue of the linearized Ginzburg-Landau operator D c = 1 Λ 2 ( − i ∇ + 2 A ( x )) 2 + Λ 1 W ( x ) � � inf σ . Λ 0 Theorem ([FHSS12, FHSS16]) If A and W are bounded and periodic, the ground state of K T c + V non-degenerate, then there exist constants Λ 0 , Λ 1 , Λ 2 such that T BCS = T c (1 − D c h 2 ) + o ( h 2 ) . c [FHSS16] R. L. Frank, C. Hainzl, R. Seiringer, J P Solovej, Commun. Math. Phys. 342 (2016), no. 1, 189–216 [FHSS12] R.L. Frank, C. Hainzl, R. Seiringer, J.P. Solovej, J. Amer. Math. Soc. 25 , 667–713 (2012). Christian HAINZL (T¨ ubingen) Critical temperature August 18, 2017 14 / 15

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