Mean field Heisenberg models and random permutations Jakob E. Bj - PowerPoint PPT Presentation

Mean field Heisenberg models and random permutations Jakob E. Bj¨ ornberg University of Gothenburg, Sweden

Plan ◮ Heisenberg model ◮ probabilistic description ◮ results for complete graph

Heisenberg model (ferromagnet) Finite graph G = ( V , E ), for example [ − n , n ] d ⊆ Z d or K n

Heisenberg model (ferromagnet) Finite graph G = ( V , E ), for example [ − n , n ] d ⊆ Z d or K n � H = − 2 S x · S y xy ∈ E

Heisenberg model (ferromagnet) Finite graph G = ( V , E ), for example [ − n , n ] d ⊆ Z d or K n � H = − 2 S x · S y xy ∈ E where S x · S y = � 3 j =1 S j x S j y and � 0 � � 0 � � 1 � 1 − i 0 S 1 = 1 S 2 = 1 S 3 = 1 , , 2 2 2 1 0 0 0 − 1 i x = S j ⊗ Id V \{ x } S j

Heisenberg model (ferromagnet) Finite graph G = ( V , E ), for example [ − n , n ] d ⊆ Z d or K n � H = − 2 S x · S y xy ∈ E where S x · S y = � 3 j =1 S j x S j y and � 0 � � 0 � � 1 � 1 − i 0 S 1 = 1 S 2 = 1 S 3 = 1 , , 2 2 2 1 0 0 0 − 1 i x = S j ⊗ Id V \{ x } S j Conjecture. On Z d with d ≥ 3 there is a phase transition.

Heisenberg model (ferromagnet) Finite graph G = ( V , E ), for example [ − n , n ] d ⊆ Z d or K n � H = − 2 S x · S y xy ∈ E where S x · S y = � 3 j =1 S j x S j y and � 0 � � 0 � � 1 � 1 − i 0 S 1 = 1 S 2 = 1 S 3 = 1 , , 2 2 2 1 0 0 0 − 1 i x = S j ⊗ Id V \{ x } S j Conjecture. On Z d with d ≥ 3 there is a phase transition. Results: ◮ No phase-transition if d ≤ 2 (Mermin–Wagner) ◮ Phase-transition if G = K n ← this talk!

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently G = { 1 , 2 , . . . , 10 } ⊆ Z : 2 3 9 10 1 ω = process of ‘crosses’ 1 2 3 4 5 6 7 8 9 10

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently 1 G = { 1 , 2 , . . . , 10 } ⊆ Z : 2 3 9 10 1 ω = process of ‘crosses’ 1 2 3 4 5 6 7 8 9 10

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently 3 1 G = { 1 , 2 , . . . , 10 } ⊆ Z : 2 3 9 10 1 ω = process of ‘crosses’ 1 2 3 4 5 6 7 8 9 10

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently 3 1 2 G = { 1 , 2 , . . . , 10 } ⊆ Z : 2 3 9 10 1 ω = process of ‘crosses’ 1 2 3 4 5 6 7 8 9 10

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently 3 4 1 7 5 2 6 9 10 8 G = { 1 , 2 , . . . , 10 } ⊆ Z : 2 3 9 10 1 ω = process of ‘crosses’ 1 2 3 4 5 6 7 8 9 10

Interchange process (Harris ’72) ◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently 3 4 1 7 5 2 6 9 10 8 G = { 1 , 2 , . . . , 10 } ⊆ Z : 2 3 9 10 1 ω = process of ‘crosses’ σ t ( ω ) = (1 , 3)(2 , 6 , 7 , 4)(5)(8 , 10 , 9) 1 2 3 4 5 6 7 8 9 10

Coloured/weighted interchange process Bicolour particles uniformly at random:

Coloured/weighted interchange process Bicolour particles uniformly at random: 1 2 3 4 5 6 7 8 9 10

Coloured/weighted interchange process Bicolour particles uniformly at random: 1 2 3 4 5 6 7 8 9 10 M = { all cycles monochromatic }

Coloured/weighted interchange process Bicolour particles uniformly at random: 1 2 3 4 5 6 7 8 9 10 M = { all cycles monochromatic } σ t ( ω ) = (1 , 3)(2 , 6 , 7 , 4)(5)(8 , 10 , 9) ⇒ M fails.

Coloured/weighted interchange process Bicolour particles uniformly at random: 1 2 3 4 5 6 7 8 9 10 M = { all cycles monochromatic } σ t ( ω ) = (1 , 3)(2 , 6 , 7 , 4)(5)(8 , 10 , 9) ⇒ M holds.

Coloured/weighted interchange process Condition on ω : � 2 ) | γ | � � = 2 −| V | E [2 ℓ ( ω ) ] 2( 1 P ( M ) = E cycles γ where ℓ ( ω ) = number of disjoint cycles.

Coloured/weighted interchange process Condition on ω : � 2 ) | γ | � � = 2 −| V | E [2 ℓ ( ω ) ] 2( 1 P ( M ) = E cycles γ where ℓ ( ω ) = number of disjoint cycles. Event A depending only on σ : I A 2 ℓ ( ω ) ] P ( A | M ) = E [ 1 =: P 2 ( A ) . E [2 ℓ ( ω ) ]

Interpretation as quantum spin system Colours = + , − x ∈ V C 2 Colouring ↔ basis vector ⊗ x ∈ V | σ x � of �

Interpretation as quantum spin system Colours = + , − x ∈ V C 2 Colouring ↔ basis vector ⊗ x ∈ V | σ x � of � where | + � = ( 1 0 ) and |−� = ( 0 1 )

Interpretation as quantum spin system Colours = + , − x ∈ V C 2 Colouring ↔ basis vector ⊗ x ∈ V | σ x � of � where | + � = ( 1 0 ) and |−� = ( 0 1 ) Can check: T xy := 2( S x · S y ) + 1 2 acts by T xy ⊗ z ∈ V | σ z � = ⊗ z ∈ V | σ τ ( z ) � τ = ( x , y ) transposition .

Interpretation as quantum spin system Colours = + , − x ∈ V C 2 Colouring ↔ basis vector ⊗ x ∈ V | σ x � of � where | + � = ( 1 0 ) and |−� = ( 0 1 ) Can check: T xy := 2( S x · S y ) + 1 2 acts by T xy ⊗ z ∈ V | σ z � = ⊗ z ∈ V | σ τ ( z ) � τ = ( x , y ) transposition . Heisenberg Hamiltonian � H = − ( T xy − 1) xy ∈ E

Matrix exponential Lemma � � � � ∗ � exp β � xy ∈ E ( T xy − 1) = E ( xy , t ) ∈ ω T xy .

Matrix exponential Lemma � � � � ∗ � exp β � xy ∈ E ( T xy − 1) = E ( xy , t ) ∈ ω T xy . Proof. ∞ β k � k � � � � � = e − β | E | � exp β ( T xy − 1) T xy k ! xy ∈ E k =0 xy ∈ E

Matrix exponential Lemma � � � � ∗ � exp β � xy ∈ E ( T xy − 1) = E ( xy , t ) ∈ ω T xy . Proof. ∞ β k � k � � � � � = e − β | E | � exp β ( T xy − 1) T xy k ! xy ∈ E k =0 xy ∈ E e − β β k e �� e k e ! � � � = T b k T b k − 1 · · · T b 1 k e ! k ! b ∈ E N e ∈ E k e = #copies of e in b .

Matrix exponential Lemma � � � � ∗ � exp β � xy ∈ E ( T xy − 1) = E ( xy , t ) ∈ ω T xy . Proof. ∞ β k � k � � � � � = e − β | E | � exp β ( T xy − 1) T xy k ! xy ∈ E k =0 xy ∈ E e − β β k e �� e k e ! � � � = T b k T b k − 1 · · · T b 1 k e ! k ! b ∈ E N e ∈ E k e = #copies of e in b .

Matrix exponential Lemma � � � � ∗ � exp β � xy ∈ E ( T xy − 1) = E ( xy , t ) ∈ ω T xy . Proof. ∞ β k � k � � � � � = e − β | E | � exp β ( T xy − 1) T xy k ! xy ∈ E k =0 xy ∈ E e − β β k e �� e k e ! � � � = T b k T b k − 1 · · · T b 1 k e ! k ! b ∈ E N e ∈ E k e = #copies of e in b . Configuration ω ↔ random product � ∗ ( xy , t ) ∈ ω T xy .

Matrix exponential Lemma � � � � ∗ � exp β � xy ∈ E ( T xy − 1) = E ( xy , t ) ∈ ω T xy . Proof. ∞ β k � k � � � � � = e − β | E | � exp β ( T xy − 1) T xy k ! xy ∈ E k =0 xy ∈ E e − β β k e �� e k e ! � � � = T b k T b k − 1 · · · T b 1 k e ! k ! b ∈ E N e ∈ E k e = #copies of e in b . Configuration ω ↔ random product � ∗ ( xy , t ) ∈ ω T xy . β ↔ time

T´ oth’s representation of Heisenberg ferromagnet ( σ, σ )-diagonal element: � 1 if M happens , �� ∗ T xy � � σ | | σ � = 0 otherwise.

T´ oth’s representation of Heisenberg ferromagnet ( σ, σ )-diagonal element: � 1 if M happens , �� ∗ T xy � � σ | | σ � = 0 otherwise. e β � ( T xy − 1) � = E [2 ℓ ( ω ) ] = 2 | V | P ( M ) . � So tr

T´ oth’s representation of Heisenberg ferromagnet ( σ, σ )-diagonal element: � 1 if M happens , �� ∗ T xy � � σ | | σ � = 0 otherwise. e β � ( T xy − 1) � = E [2 ℓ ( ω ) ] = 2 | V | P ( M ) . � So tr Theorem (T´ oth ’93) xy S x · S y and Z ( β ) = tr ( e − β H ) . Then the Let H = − 2 � correlation function S 3 x S 3 y e − β H � � y � = tr � S 3 x S 3 = 1 4 P 2 ( x ↔ y ) Z ( β )

The Heisenberg ferromagnet γ 0 = cycle containing the origin Conjecture. Let G = [ − n , n ] d ⊆ Z d , d ≥ 3, and m ( β ) = lim k →∞ lim n →∞ P 2 ( | γ 0 | > k )

The Heisenberg ferromagnet γ 0 = cycle containing the origin Conjecture. Let G = [ − n , n ] d ⊆ Z d , d ≥ 3, and m ( β ) = lim k →∞ lim n →∞ P 2 ( | γ 0 | > k ) Then there is β c s.t. � = 0 for β < β c , m ( β ) > 0 for β > β c .

Higher spins Cycle-weighted interchange processes: I A θ ℓ ( ω ) ] P θ ( A ) = E [ 1 E [ θ ℓ ( ω ) ] , θ > 0

Higher spins Cycle-weighted interchange processes: I A θ ℓ ( ω ) ] P θ ( A ) = E [ 1 E [ θ ℓ ( ω ) ] , θ > 0 Integer θ ↔ spin S = ( θ − 1) / 2