Reflection Positivity and Phase Transitions · 2016. 5. 30. · Re ection Positivity and Phase Transitions Daniel Ueltschi Department of Mathematics, University of Warwick \Positive

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Reflection Positivity and Phase Transitions

Daniel Ueltschi

Department of Mathematics, University of Warwick

“Positive Reflections”, in memory of Robert SchraderETH Zurich, 23 May 2016

D. Ueltschi (Univ. Warwick) RP and Phase Transitions Zurich, 23.5.16 1 / 16

Lattice spin systems

Condensed matter: Atoms form a regular latticeand electrons move in this periodic potential

Each electron carries a “spin”, i.e. an intrinsicmagnetic moment

Crystalline structure ofsodium chloride (table salt)

Source: Wikipedia

Heisenberg model: There is exactly one localised electron on eachatom, that is characterised by its spin

Interactions are nearest-neighbour only, and they are invariant underspin rotations

Important model for magnetism, and more generally for condensedmatter physics and quantum information

Further simplification: classical spin models


Classical spin models

Lattice Λ = {−L2 + 1, . . . , L2 }

d with periodic boundary conditions

State space: (SN−1)Λ

Hamiltonian: HΛ(σ) =∑

xy∈Λ(σx − σy)2

Special cases: N = 0: self-avoiding random walk; N = 1: Ising;N = 2: XY model; N = 3: classical Heisenberg

Ising model: Phase transitions in dimension 2 and higher; exactly twoextremal Gibbs states at low temperatures

For N > 2: model has continuous symmetry, that prevents ordering in2D (Mermin-Wagner); spontaneous magnetisation is expected in d > 3

N = 2: [Frohlich, Pfister ’83] prove that, whenever the free energy isdifferentiable in β, there is either a unique extremal Gibbs state, or allextremal states are labelled by the elements of the symmetry groupSO(2)


Results from RP & IRB

Spectacular results come from the method of reflection positivityand infrared bounds: the model has spontaneous magnetisation atlow enough temperatures!

Finite volume Gibbs state: 〈·〉 = 1Z(Λ)

∫(SN−1)Λ · e−βHΛ(σ) dσ

Theorem [Frohlich, Simon, Spencer ’76]

1

|Λ|2∑x,y∈Λ

〈σxσy〉 > 1− 3

β|Λ|∑

k∈Λ∗\{0}

1

ε(k)

Here, Λ∗ = 2πL {−

L2 + 1, . . . , L2 }

d and ε(k) = 2∑d

i=1(1− cos ki)

As a consequence,

lim infL→∞

⟨( 1

|Λ|∑x∈Λ

σx

)2⟩> 1− 3

(2π)dβ

∫[−π,π]d

dk

ε(k)

Hence there cannot be a single extremal state! (d > 3, β large)D. Ueltschi (Univ. Warwick) RP and Phase Transitions Zurich, 23.5.16 4 / 16

Overview of the method of proof

Let κ(x) = 〈σ30σ

3x〉 the two-point fct and κ(k) its Fourier transform:

κ(k) =∑x∈Λ

e−ikx κ(x), k ∈ Λ∗

Notice that

κ(x) =1

|Λ|∑k∈Λ∗

eikx κ(k)

The goal is to show that1

|Λ|κ(0) > c > 0

For v ∈ RΛ, let

Z(v) =

∫(SN−1)Λ

exp{−β

∑xy∈Λ

(σx + vx − σy − vy)2}

dσ

Notice that Z(v = 0) = Z(Λ)D. Ueltschi (Univ. Warwick) RP and Phase Transitions Zurich, 23.5.16 5 / 16

Overview of the method of proof

One proves a series of 4 lemmas:

Lemma 1: Reflection positivity

Z(v1, v2)2 6 Z(v1, Rv1)Z(Rv2, v2)

Pv1 v2

L2: RP ⇒ gaussian domination

Z(v) 6 Z(0)

L3: GD ⇒ infrared bound

κ(k) 61

βε(k)

Lemma 4: IRB ⇒ theorem1

|Λ|κ(0) >

1

3− 1

β|Λ|∑

k∈Λ∗\{0}

1

ε(k)


Reflection positivity

Lemma 1: Reflection positivity

Z(v1, v2)2 6 Z(v1, Rv1)Z(Rv2, v2)

Pv1 v2

Z(v1,v2) =

∫Λ1

dσ1

∫Λ2

dσ2 e−F (σ1,v1)−F (σ2,v2)

exp{−β

∑xy∈P

d∑i=1

(σi1x + vi1x − σi2y − vi2y)2}

=( ∏xy∈P

d∏i=1

∫ ∞−∞

dξixy√2π

e−12 (ξixy)2

)

·∫

Λ1

dσ1 e−F (σ1,v1) exp{

i√

2β∑xy∈P

d∑i=1

ξixy(σi1x − vi1x)

}∫Λ2

dσ2 . . .

Lemma follows from Cauchy-SchwarzD. Ueltschi (Univ. Warwick) RP and Phase Transitions Zurich, 23.5.16 7 / 16

From reflection positivity to gaussian domination


Z(v) 6 Z(0)

Suppose (v1, v2) is maximiser. Then (v1, Rv1) is also maximiser by RP

Pv1 v2

(Periodic boundary conditions are not shown here)D. Ueltschi (Univ. Warwick) RP and Phase Transitions Zurich, 23.5.16 8 / 16



Z(v) 6 Z(0)


Pv1 v2




Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)





Z(v) 6 Z(0)


There is a space-invariant maximiser!


From gaussian domination to the infrared bound

L3: GD ⇒ infrared bound

κ(k) 61

βε(k)

Z(v) =

∫(SN−1)Λ

dσ exp{−β∑

(σx−σy)2−2β(σ,∆v)−β∑

(vx−vy)2}

Z(v) 6 Z(0) ⇔ 〈 e2β(σ,∆v) 〉 6 e−β(v,∆v)

Take vx = cos kx, for fixed k ∈ Λ∗. Then −∆v = ε(k)vFor small field ηv, the inequality becomes

1 + η2β2ε(k)2κ(k)‖v‖2 +O(η4) 6 1 + η2βε(k)‖v‖2 +O(η4)

This implies the IRBD. Ueltschi (Univ. Warwick) RP and Phase Transitions Zurich, 23.5.16 9 / 16

From the infrared bound to spontaneous magnetisation

Lemma 4: IRB ⇒ theorem1

|Λ|κ(0) >

1

3− 1

β|Λ|∑

k∈Λ∗\{0}

1

ε(k)

We have 〈(σ30)2〉 = 1

3 ; by the inverse Fourier transform,

1

3= κ(0) =

1

|Λ|κ(0) +

1

|Λ|∑

k∈Λ∗\{0}

κ(k)

The lemma follows immediately


Quantum Heisenberg models

Lattice Λ = {−L2 + 1, . . . , L2 }

d with periodic b.c. as beforeHilbert space HΛ =

⊗x∈Λ C2S+1, S ∈ 1

2N

Spin operators S1, S2, S3 on C2 such that [S1, S2] = iS3, etc...Six = Si ⊗ IdΛ\{x}Hamiltonian:

H(u)Λ = −

∑{x,y}∈E

(S1xS

1y + uS2

xS2y + S3

xS3y

)u = 1: Heisenberg ferromagnet

u = −1: unitarily equivalent to Heisenberg antiferromagnet

u = 0: quantum XY model

Gibbs state 〈a〉 = Tr a e−βH(u)Λ

/Tr e−βH

(u)Λ


Difficulties in the quantum case

Let κ(x) = 〈S30S

3x〉. The infrared bound cannot hold!

Indeed, in the case S = 12 ,

〈~S0 · ~S0︸︷︷︸=3/4

〉 − 〈~S0 · ~Sx︸︷︷︸6 1/4

〉 > 12

However, assuming the infrared bound to hold,∣∣〈~S0 · ~S0〉 − 〈~S0 · ~Sx〉∣∣ =

1

|Λ|

∣∣∣∑k∈Λ∗

(1− eikx )κ(k)∣∣∣

61

|Λ|const

β

∑k∈Λ∗\{0}

|1− eikx |ε(k)

which goes to 0 as β →∞, contradiction


Extension to quantum systems

Theorem [Dyson, Lieb, Simon ’78] Assume that u ∈ [−1, 0]. Then

1

|Λ|2∑x,y∈Λ

〈S3xS

3y〉 > 1

3S(S + 1)− 1√2|Λ|

∑k∈Λ∗\{0}

√εu(k)

ε(k)

− 1

2β|Λ|∑

k∈Λ∗\{0}

1

ε(k)

where εu(k) =∑d

i=1

((1− u cos ki)〈S1

0S1ei〉+ (u− cos ki)〈S2

0S2ei〉)

Since εu(k) 6 S(S + 1), the lower bound is positive for S large enough

A better lower bound was proposed in [Kennedy, Lieb, Shastry ’88]


Changes in the quantum case

The proof follows [Frohlich, Simon, Spencer ’76] but with some changes

First, the infrared bound holds for the Duhamel two-point function:

(S30 , S

3x) =

1

Z(Λ)

∫ β

0TrS3

0 e−sH(u)Λ S3

x e−(β−s)H(u)Λ

Unlike 〈S30S

30〉, we do not have (S3

0 , S30) = const

Solution: Falk-Bruch inequality [Falk, Bruch ’69]. WithΦ(s) =

√s coth 1√

s,

2〈A∗A+AA∗〉〈[A∗, [H,A]]〉

6 Φ( 4(A,A)

〈[A∗, [H,A]]〉

)This allows to transfer the infrared bound for the Duhamel function, toa lower bound for the usual two-point function


Reflection positivity for quantum systems

LemmaHilbert space H = H1 ⊗H2, finite-dimensional, with H1 ' H2.Matrices A,B,Ci, Di in small space H1. Then∣∣∣TrH eA⊗1l+1l⊗B−

∑k(Ci⊗1l−1l⊗Di)2∣∣∣2 6 TrH eA⊗1l+1l⊗A−

∑k(Ci⊗1l−1l⊗Ci)2

· TrH eB⊗1l+1l⊗B−∑k(Di⊗1l−1l⊗Di)

2

Here, A is the complex conjugate of A

This gives the restriction u ∈ [−1, 0], which excludes the quantumHeisenberg ferromagnet!

More general approach in [Frohlich, Israel, Lieb, Simon ’78–80]


Conclusion

Reflection positivity allows to get infrared bound on spincorrelations, which allows to prove the occurrence of phasetransitions in models with continuous symmetries

Applies to classical and quantum models

Does not apply to irregular lattices, nor to quantum Heisenbergferromagnet

Gives interesting results for quantum systems in 2D [Neves, Perez

’86]

Further developments in statistical physics: chessboard estimates;matrix models (spin nematics); random loop models

Our knowledge of statistical physics would be much poorerwithout the methods of reflection positivity!

THANK YOU!


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Reflection Positivity and Phase Transitions · 2016. 5. 30. · Re ection Positivity and Phase Transitions Daniel Ueltschi Department of Mathematics, University of Warwick \Positive