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SRQ seminar – September 14th, 2018

Notes from a talk in the SRQ series.

INI Seminar 20180914. Brydges

1Chessboard estimates

Let \(\mu\) be a OS positive measure and let \(A, B\) be two observables which are measurable w.r.t. fields supported in two adjacent regions which are mapped on each other via \(\theta\). Then we write graphically

\(\displaystyle \begin{array}{|l|l|} \hline A & B\\ \hline \end{array} := \int \overline{A (\phi)} B (\phi) \mu (\mathrm{d} \phi) = \langle \theta A, B \rangle_{\mathcal{H}}\)

The drawing on the left emphasises the geometric relative location of the supports of the various functions: \(A\) is measurable with respect to fields \(\phi (f)\) with \(f\) supported in the left hand box and \(B\) is measurable with respect to fields \(\phi (f)\) with \(f\) supported in the right hand box. Recall that \(\langle F, G \rangle_{\mathcal{H}} = \langle \theta A, B \rangle_{L^2 (\mu)}\). By the Cauchy–Schwartz inequality we have

\(\displaystyle \left| \begin{array}{|l|l|} \hline A & B\\ \hline \end{array} \right| = | \langle \theta A, B \rangle_{\mathcal{H}} | \leqslant \langle \theta A, \theta A \rangle_{\mathcal{H}}^{1 / 2} \langle B, B \rangle_{\mathcal{H}}^{1 / 2} = \left( \hspace{0.17em} \begin{array}{|l|l|} \hline A & \theta A\\ \hline \end{array} \hspace{0.17em} \right)^{1 / 2} \left( \hspace{0.17em} \begin{array}{|l|l|} \hline \theta B & B\\ \hline \end{array} \hspace{0.17em} \right)^{1 / 2} .\)

since \(\theta A\) is supported in the right square. With two reflections \(\theta_1, \theta_2\) about orthogonal hyperplanes one can use Cauchy–Schwartz twice to obtain

\(\displaystyle \begin{array}{rl} & \left| \begin{array}{|l|l|} \hline A & B\\ \hline D & C\\ \hline \end{array} \right| \leqslant\\ & \left( \hspace{0.27em} \begin{array}{|l|l|} \hline A & \theta_1 A\\ \hline \theta_2 A & \theta_1 \theta_2 A\\ \hline \end{array} \hspace{0.27em} \right)^{1 / 4} \left( \hspace{0.27em} \hspace{0.17em} \begin{array}{|l|l|} \hline \theta_1 B & B\\ \hline \theta_2 \theta_1 B & \theta_2 B\\ \hline \end{array} \hspace{0.27em} \right)^{1 / 4} \left( \hspace{0.27em} \begin{array}{|l|l|} \hline \theta_1 D & \theta_1 \theta_2 D\\ \hline D & \theta_1 D\\ \hline \end{array} \hspace{0.27em} \right)^{1 / 4} \left( \hspace{0.27em} \begin{array}{|l|l|} \hline \theta_1 \theta_2 C & \theta_2 C\\ \hline \theta_1 C & C\\ \hline \end{array} \hspace{0.27em} \right)^{1 / 4} . \hspace{3cm} \text{(1)} \end{array}\)

In order to realise the above example consider the \(\mathbb{Z}_4\) invariant graph with four vertices \(\{a, b, c, d\}\) and edges \(\{ab, bc, cd, da\}\). This is invariant under \(\theta_1\) that maps \(a\) to \(b\) and \(d\) to \(c\) and \(\theta_2\) that maps \(a\) and \(b\) to \(d\) and \(c\). The \(\mathbb{Z}_4\) invariant measure

\(\displaystyle \mu (d \phi) = \exp \left[ - \frac{1}{2} \sum_{xy} (\phi_x - \phi_y)^2 \right] \prod_x \rho (d \phi_x) .\)

has the required reflection positivity and we can apply the above estimate with \(A\) be a function of \(\phi_a\), \(B\) a function of \(\phi_b\) etc. We have the required O.S. positive reflections because \(\exp [- \frac{1}{2} \phi^2]\) is a positive definite function (Exercise). This is an example of reflection positivity through hyperplanes between lattice sites.

Example. Let \(\Lambda\) be a lattice torus \((\mathbb{Z}/\mathbb{Z}_N)^d\) with \(N\) even. By the chessboard estimate

\(\displaystyle \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi, \nabla \phi)} \prod_{x \in \Lambda} A_x (\phi_x) \mathrm{d} \phi_x \leqslant \prod_{y \in \Lambda} \left( \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi, \nabla \phi)} \prod_{x \in \Lambda} A_y (\phi_x) \mathrm{d} \phi_x \right)^{1 / | \Lambda |} \)

(2)

Problem 1. (1) Verify this. (2) Does (2) hold if \(N\) is odd and all \(A_x\) are positive?

The estimate of this example was first proved in [FSS76]. As (1) suggests, it can be generalised to allow functions that depend on many fields or continuum fields in place of the single field functions \(A_x\) . Such generalisations were developed in detail in [FILS78].

2Infrared bounds

In this section we define the \(O (n)\) sigma model and prove that it has long range order. Our argument is taken from [FSS76], which was the first proof of existence of continuous symmetry breaking in these models. To this day it is the only simple proof for \(n > 2\). For \(n = 2\) there are other easy proofs. See for example [KK86]. Since the proof below relies on O.S. positivity it is a very delicate proof: if for example we make the left hand side of (2) more ferromagnetic by adding ferromagnetic next-to-nearest neighbour interaction then the above proof no longer applies. Yet one would expect that making the model more ferromagnetic would increase only its tendency to correlate the direction of all its spins. In a sequence of papers that starts with [Bał95] Balaban has developed a low temperature renormalisation group expansion that is not delicate and gives detailed information.

Theorem 1. (IR bounds) Take \(A \in C_0^{\infty} (\mathbb{R})\). Define

\(\displaystyle \langle F \rangle_A := Z_A^{- 1} \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi, \nabla \phi)} F (\phi) \prod_{x \in \Lambda} A (\phi_x) \mathrm{d} \phi_x\)

then for \(g : \Lambda \rightarrow \mathbb{R}\)

\(\displaystyle \langle e^{- (\nabla \phi, \nabla g)} \rangle_A \leqslant e^{\frac{1}{2} (\nabla g, \nabla g)} . \) (3)

(a Gaussian upper bound).

Proof. Complete the square in \(\langle e^{- (\nabla \phi, \nabla g)} \rangle_A\):

\(\displaystyle \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi, \nabla \phi) - (\nabla \phi, \nabla g)} \prod_{x \in \Lambda} A (\phi_x) \mathrm{d} \phi_x = e^{\frac{1}{2} (\nabla g, \nabla g)} \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi + \nabla g, \nabla \phi + \nabla g)} \prod_{x \in \Lambda} A (\phi_x) \mathrm{d} \phi_x\)

now translate \(\phi_x \rightarrow \phi_x - g_x\) to get

\(\displaystyle = e^{\frac{1}{2} (\nabla g, \nabla g)} \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi, \nabla \phi)} \prod_{x \in \Lambda} A (\phi_x - g_x) \mathrm{d} \phi_x\)

and use chessboard estimate (2) to bound the integral as

\(\displaystyle \leqslant e^{\frac{1}{2} (\nabla g, \nabla g)} \prod_{y \in \Lambda} \left( \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi, \nabla \phi)} \prod_{x \in \Lambda} A (\phi_x - g_y) \mathrm{d} \phi_x \right)^{1 / | \Lambda |} .\)

Translate back in the integral \(\phi_x \rightarrow \phi_x + g_y\) which does not changes the energy of the configuration because \(g_y\) is constant in \(x\):

\(\displaystyle \leqslant e^{\frac{1}{2} (\nabla g, \nabla g)} \prod_{y \in \Lambda} \left( \int_{\mathbb{R}^{\Lambda}} e^{- \frac{1}{2} (\nabla \phi, \nabla \phi)} \prod_{x \in \Lambda} A (\phi_x) \mathrm{d} \phi_x \right)^{1 / | \Lambda |} = e^{\frac{1}{2} (\nabla g, \nabla g)} Z_A .\)

Divide both sides by \(Z\) and this finishes the proof.\(\Box\)

Problem 2.

Generalise the IR bound to allow \(\phi : \Lambda \rightarrow \mathbb{R}^n\) where \((\nabla \phi, \nabla \phi) = \sum_{x \sim y} | \phi_x - \phi_y |^2\).
The group \(O (n)\) is by definition the set of all orthogonal linear transformations of \(\mathbb{R}^n\) to itself. Invariant models are obtained when \(\phi : \Lambda \rightarrow \mathbb{R}^n\) and \(A (\phi)\) is \(O (n)\) invariant, namely \(A (\phi) = a (| \phi |)\). The \(O (n)\) sigma model is the special choice
\(\displaystyle A (\phi_x) \mathrm{d} \phi_x \rightarrow \sigma_{S_R^{n - 1}} (\mathrm{d} \phi_x)\)
where \(\sigma_{S^{n - 1}}\) is the surface measure on the sphere \(S_R^{n - 1} = \{|x| = R\} \subseteq \mathbb{R}^n\). For example, when \(n = 1\) we have the Ising model.

The IR bound implies:

Corollary 2. For \(g\) orthonal to constants we have

\(\displaystyle \langle \phi (g)^2 \rangle_A \leqslant (g, (- \Delta)^{- 1} g) .\)

Proof. Replace \(g\) with \(tg\) in (3) and take \(t \rightarrow 0\) to get

\(\displaystyle 1 + t \langle (\nabla \phi, \nabla g) \rangle_A + \frac{1}{2} t^2 \langle (\nabla \phi, \nabla g)^2 \rangle_A + \cdots \leqslant 1 + \frac{t^2}{2} (\nabla g, \nabla g) + \cdots\)

which implies \(\langle (\nabla \phi, \nabla g) \rangle_A = 0\) and

\(\displaystyle \langle (\phi, \Delta g)^2 \rangle_A \leqslant (\nabla g, \nabla g)\)

so replacing \(g \rightarrow (- \Delta)^{- 1} g\) we have

\(\displaystyle \langle (\phi, g)^2 \rangle_A \leqslant (g, (- \Delta)^{- 1} g) .\) \(\Box\)

Theorem 3. Let \(d > 2\). The \(\infty\) volume limit of the \(O (n)\) sigma model has long range order (LRO) in the sense that for \(R \gg 1\) (radius) we have

\(\displaystyle \liminf_{| \Lambda | \rightarrow \infty} \left\langle \left| \frac{1}{| \Lambda |} \sum_{x \in \Lambda} \phi_x \right|^2 \right\rangle_{\Lambda} > 0.\)

This bound implies that for \(R\) large no infinite volume limit measure can be ergodic under translations because if it were then the spatial average \(\frac{1}{| \Lambda |} \sum_{x \in \Lambda} \phi_x\) would converge pointwise as \(\Lambda\) increases to the expectation of \(\phi_0\) which is zero because this expectation is the limit of the finite volume expectations which are zero since the finite volume models are \(O (n)\) symmetric. Yet Theorem 3 says that the variance of the limit of \(\frac{1}{| \Lambda |} \sum_{x \in \Lambda} \phi_x\) is not zero. The physical interpretation of Theorem 3 is that the spins have an average direction and the measure has split into ergodic components labeled by the average direction, but I do not think this has been proved.

Proof. The infinite volume limit through torii of size \(N\) with \(N \rightarrow \infty\). Consider

\(\displaystyle e_k (x) := \frac{1}{\sqrt{| \Lambda |}} e^{ik \cdot x}, \qquad k \in \Lambda^{\ast} = \{ k : e^{ik \cdot x} \text{ is periodic of period } N \}\)

which is an orthonormal basis of \(\ell^2 (\Lambda)\). Write

\(\displaystyle R^2 = \frac{1}{| \Lambda |} \sum_{x, y \in \Lambda} \langle \phi_x \cdot \phi_y \rangle_{\Lambda} \delta_{x, y} = \frac{1}{| \Lambda |} \sum_{k \in \Lambda^{\ast}} \langle \phi (e_{- k}) \cdot \phi (e_k) \rangle_{\Lambda}\)

where \(\phi (e_k) = \sum_x \phi_x e_k (x) = | \Lambda |^{- 1 / 2} \sum_x \phi_x \cos (k \cdot x) + i | \Lambda |^{- 1 / 2} \sum_x \phi_x \sin (k \cdot x)\) so

\(\displaystyle R^2 = \frac{1}{| \Lambda |} \sum_{k \in \Lambda^{\ast}} (\langle \phi (\operatorname{Re}(e_k)) \cdot \phi (\operatorname{Re}(e_k)) \rangle_{\Lambda} + \langle \phi (\operatorname{Im}(e_k)) \cdot \phi (\operatorname{Im}(e_k)) \rangle_{\Lambda})\) \(\displaystyle \leqslant \underbrace{\frac{1}{| \Lambda |} \sum_{k \in \Lambda^{\ast}, k \neq 0} (e_k, (- \Delta)^{- 1} e_k)}_{\simeq (2 \pi)^{- d} \int_{[- \pi, \pi]^d} \frac{\mathrm{d} k}{|k|^2} \text{(prove it, as an upper bound)} <_{d > 2} \infty .} + \underbrace{\frac{1}{| \Lambda |} \langle | \phi (e_0) |^2 \rangle_{\Lambda}}_{= \left\langle \left| \frac{1}{| \Lambda |} \sum_{x \in \Lambda} \phi_x \right|^2 \right\rangle_{\Lambda}}\)

Therefore for all \(C > 0\) there exists \(R\) such that

\(\displaystyle \left\langle \left| \frac{1}{| \Lambda |} \sum_{x \in \Lambda} \phi_x \right|^2 \right\rangle_{\Lambda} \geqslant C\)

uniformly in \(\Lambda\).\(\Box\)

In particular the limiting measure should not be unique (and should decompose into ergodic components parametrized by the overall direction of the spin field).

\(\displaystyle \)

Bibliography

[KK86]: Tom Kennedy and Chris King. Spontaneous symmetry breakdown in the abelian Higgs model. Comm. Math. Phys., 104(2):327–347, 1986.
[FSS76]: J. Fröhlich, B. Simon, and T. Spencer. Infrared bounds, phase transitions, and continuous symmetry breaking. Commun. Math. Phys., 50:79–95, 1976.
[FILS78]: J. Fröhlich, R. Israel, E.H. Lieb, and B. Simon. Phase transitions and reflection positivity. I. General theory and long range lattice models. Commun. Math. Phys., 62:1–34, 1978.
[Bał95]: Tadeusz Bałaban. A low temperature expansion for classical \(n\)-vector models. I. A renormalization group flow. Commun. Math. Phys., 167(1):103–154, 1995.