Bosonic quadratic Hamiltonians. diagonalization

Transcription

1 and their diagonalization P. T. Nam 1 M. Napiórkowski 2 J. P. Solovej 3 1 Masaryk University Brno 2 University of Warsaw 3 University of Copenhagen Macroscopic Limits of Quantum Systems 70th Birthday of Herbert Spohn

2 Introduction Main object of interest: Here: H = ij h ij a i a j + 1 k ij a i a j + 1 k ij a i a j 2 2 h ij is a self-adjoint matrix (h ij = h ij = h# ij); k ij is the complex conjugate of the matrix k ij ; ij a i /a i are the bosonic creation/annihilation operators on the Fock space: [a i, a j] = [a i, a j ] = 0, [a i, a j] = δ ij ; the sum might be over an infinite set of indices. ij

3 Operators of that type are important in physics! 1 they appear as effective theories for quantum many-body systems. Examples: Bogoliubov theory for bosonic systems BCS theory (and BCS-like theories) for fermionic systems 2 quantum field theory (eg. scalar field with position dependent mass).

4 Operators of that type are important in physics! 1 they appear as effective theories for quantum many-body systems. Examples: Bogoliubov theory for bosonic systems BCS theory (and BCS-like theories) for fermionic systems 2 quantum field theory (eg. scalar field with position dependent mass). Bogoliubov theory: (describes low-energy behaviour of bosonic systems) H Bog = p 2 a pa p + ρ ŵ(p) (2a 2 }{{} pa p + a pa p + a p a p ) p 0 p 0 > 0 Question: what are the spectral properties of this operator? DIAGONALIZATION: UH Bog U = H Bog = E + p 0 e(p)b pb p

5 Diagonalization of Bogoliubov Hamiltonian Introduce In particular: a p = u p b p + v p b p, [b p, b p ] = δ pp (CCR). (CCR) u 2 p v 2 p = 1 u p = cosh(α p ), v p = sinh(α p ) Diagonalization condition: b pb p and b p b p terms vanish if ρŵ(p) 2 which yields coth(2α p ) = p2 +ρŵ(p) ρŵ(p) (u 2 p + v p) 2 +(p 2 +ρŵ(p)) u p v p }{{}}{{} cosh(2α p) and HBog = E + p 0 e(p)b pb p 1 sinh 2αp 2 = 0 with e(p) = p 4 + 2ρŵ(p)p 2.

6 Question: When, in general, can H be diagonalized? We saw a 2-dimensional example. Finite dimensional case solved (essentially due to Williamson s Theorem on symplectic transformations 36). What about infinite dimension?

7 Fock space formalism Fock space: F(h) := N=0 sym N h = C h (h s h) Creation and anihilation operators: a (f N+1 ) f σ(1)... f σ(n) 1 = f σ(1)... f σ(n+1), σ S N + 1 N σ S N+1 a(f N+1 ) f σ(1)... f σ(n) = N 1 2 fn+1, f σ(1) fσ(2)... f σ(n σ S N σ S N for all f 1,..., f N+1 in h, and all N = 0, 1, 2,... Canonical commutation relations: [a(f), a(g)] = 0, [a (f), a (g)] = 0, [a(f), a (g)] = f, g, f, g h.

8 Fock space formalism Assume h > 0. Recall: dγ(h) = m,n 1 f m, hf n a (f m )a(f n ) where {f n } n 1 D(h) is an arbitrary orthonormal basis for h. General form of quadratic operator: H = dγ(h) + 1 ( ) J kf m, f n a(f m )a(f n ) + J 2 kf m, f n a (f m )a (f n ) m,n 1 Here: k : h h is an (unbounded) linear operator with D(h) D(k) (called pairing operator), k = J kj ; J : h h is the anti-unitary operator defined by J(f)(g) = f, g, f, g h.

9 H = dγ(h) m,n 1 ( ) J kf m, f n a(f m )a(f n ) + J kf m, f n a (f m )a (f n ) Remark: The above definition is formal! If k is not Hilbert-Schmidt, then it is difficult to show that the domain is dense. More general approach: definition through quadratic forms! One-particle density matrices: γ Ψ : h h and α Ψ : h h f, γ Ψ g = Ψ, a (g)a(f)ψ, Jf, α Ψ g = Ψ, a (g)a (f)ψ, f, g h A formal calculation leads to the expression Ψ, HΨ = Tr(h 1/2 γ Ψ h 1/2 ) + R Tr(k α Ψ ).

10 Unitary implementability Generalized creation and annihilation operators A(f Jg) = a(f)+a (g), A (f Jg) = a (f)+a(g), f, g h. Definition: A bounded operator V on h h is unitarily implemented by a unitary operator U V on Fock space if U V A(F )U V = A(VF ), F h h. General form of the transformation we have seen: Pick F = f 0. Then A(F ) = a(f). But VF = f 1 Jf 2 and thus A(VF ) = a(f 1 ) + a (f 2 ). We get: U V A(F )U V = U V a(f)u V = A(VF ) = a(f }{{} 1 ) + a (f 2 ). =: b(f)

11 Quadratic Hamiltonians as quantizations of block operators Our goal: Find U such that UHU = E + dγ(ξ). Let ( ) h k A := k JhJ and H A := 1 2 F m, AF n A (F m )A(F n ). m,n 1 Then a calculation gives H = H A 1 2 Tr(h). Thus, formally, H can be seen as quantization of A.

12 Diagonalization If U V A(F )U V = A(VF ), then Thus, if V diagonalizes A: U V H A U V = H VAV. VAV = ( ξ 0 0 JξJ for some operator ξ : h h, then ( U V HU V = U V H A 1 ) 2 Tr(h) U V = dγ(ξ) + 1 Tr(ξ h). 2 ) These formal arguments suggest it is enough to consider the diagonalization of block operators.

13 Questions Question 1: what are the conditions on V so that U V A(F )U V = A(VF )? Question 2: what are the conditions on A so that there exists a V that diagonalizes A?

14 Question 1 - symplectic transformations Recall A(f Jg) = a(f) + a (g) and U V A(F )U V = A(VF ). Conjugate and canonical commutation relations: A (F 1 ) = A(J F 1 ), [A(F 1 ), A (F 2 )] = (F 1, SF 2 ), F 1, F 2 h h where ( ) ( J S =, J = 0 1 J 0 ). S = S 1 = S is unitary, J = J 1 = J is anti-unitary. Compatibility (wrt implementability) conditions J VJ = V, V SV = S = VSV. (1) Any bounded operator V on h h satisfying (1) is called a symplectic transformation.

15 Question 1 - implementability Symplecticity of V implies ( U J J VJ = V V = V J V JUJ ) Fundamental result: Shale s theorem ( 62) A symplectic transformation V is unitarily implementable (i.e. U V A(F )U V = A(VF )), if and only if V 2 HS = Tr(V V ) <. U V, a unitary implementer on the Fock space of a symplectic transformation V, is called a Bogoliubov transformation.

16 Question 2 - example: commuting operators in dim h > 0 and k be commuting operators on h = L 2 (Ω, C) ( ) h k A := > 0 on h h. k h if and only if G < 1 with G := k h 1. A is diagonalized by the linear operator ( G 1 V := G 2 1 G 2 G G 2 in the sense that ( ) VAV ξ 0 = 0 ξ with ξ := h 1 G 2 = h 2 k 2 > 0. V satisfies the compatibility conditions and is bounded (and hence a symplectic transformation) iff G = kh 1 < 1 V is unitarily implementable iff kh 1 is Hilbert-Schmidt )

17 Historical remarks For dim h < this follows from Williamson s Theorem ( 36); Friedrichs ( 50s) and Berezin ( 60s): h µ > 0 bounded with gap and k Hilbert-Schmidt; Grech-Seiringer ( 13): h > 0 with compact resolvent, k Hilbert-Schmidt; Lewin-Nam-Serfaty-Solovej ( 15): h µ > 0 unbounded, k Hilbert-Schmidt; Bach-Bru ( 16): h > 0, kh 1 < 1 and kh s is Hilbert-Schmidt for all s [0, 1 + ɛ] for some ɛ > 0. Our result: essentially optimal conditions

18 Theorem [Diagonalization of block operators] (i) (Existence). Let h : h h and k : h h be (unbounded) linear operators satisfying h = h > 0, k = J kj and D(h) D(k). Assume that the operator G := h 1/2 J kh 1/2 is bounded and G < 1. Then we can define the self-adjoint operator ( ) h k A := k JhJ > 0 on h h by Friedrichs extension. This operator can be diagonalized by a symplectic transformation V on h h in the sense that ( ) VAV ξ 0 = 0 JξJ for a self-adjoint operator ξ > 0 on h. Moreover, we have ( ) 1 + G 1/4 V. 1 G

19 Theorem [Diagonalization of block operators] (ii) (Implementability). Assume further that G is Hilbert-Schmidt. Then V is unitarily implementable and V HS 2 1 G G HS. (ii) (Boundedness from below). Assume further that kh 1/2 is Hilbert-Schmidt. Then the quadratic Hamiltonian H, defined before as a quadratic form, is bounded from below and closable, and hence its closure defines a self-adjoint operator which we still denote by H. Moreover, if U V is the unitary operator on Fock space implementing the symplectic transformation V, then U V HU V = dγ(ξ) + inf σ(h) and inf σ(h) 1 2 kh 1/2 2 HS.

20 Sketch of proof Step 1. - fermionic case. If B is a self-adjoint and such that J BJ = B, then there exists a unitary operator U on h h such that J UJ = U and UBU = ( ξ 0 0 JξJ Step 2. Apply Step 1 to B = A 1/2 SA 1/2. Step 3. Explicit construction of the symplectic transformation V: V := U B 1/2 A 1/2. ). Step 4. A detailed study of V V = A 1/2 B A 1/2.

21 Analysis of V V = A 1/2 B A 1/2. Step 4a. If V V 1 is Hilbert-Schmidt, then V is implementable. Step 4b. Using functional calculus V V 1 = 1 π 0 1 t + A Then use Cauchy-Schwarz with This gives where (SAS A) 2 }{{} =: E X := 1 t + A 2 EA1/2 B 1, A 1/2 1 t + B 2 A 1/2 tdt. 1 Y := B t + B 2 A 1/2. ±2(V V 1) ɛ 1 K + ɛa 1/2 B A 1/2 = ɛ 1 K + ɛv V K := 2 π 0 1 t + A 2 ESA 1 SE 1 tdt. t + A 2

22 Step 4c. We rewrite ±2(V V 1) ɛ 1 K + ɛv V ɛ 1 K + Cɛ. Step 4d. Tr K <. Step 4e. Lemma: If L = L boundedand there exists a trace class operator K 0 such that ±2L ɛ 1 K + ɛ, ɛ > 0, then L is Hilbert-Schmidt and L 2 HS Tr(K).

23 Thank you for your attention and, last but not least, HAPPY BIRTHDAY HERBERT!