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SRQ seminar – November 9th, 2018

Notes from a talk in the SRQ series.

SRQ Seminar 20181109 Brydges

The Higgs mechanism

References

Setup

Box \(\Lambda \subset \mathbb{Z}^d\). Higgs field \(\phi : \Lambda \rightarrow \mathbb{C}\), \(\phi \in \mathbb{C}^{\Lambda}\). Yang–Mills field \(A : E \rightarrow \mathbb{R}\) where \(E\) is the edge set of \(\mathbb{Z}^d\) and \(A_{x y} = - A_{y x}\) where we choose a standard orientation for each edge.

We want to compactify the plane by adding a point at \(\infty\) calling it \(\ast\) and joining it with an edge to every bounday point of \(\Lambda\). We give \(\phi_{\ast}\) a fixed value chosen later. We consider \(E\) including these additional edges.

Higgs action

\(\displaystyle S_{\operatorname{Higgs}} (A, \phi) := \frac{1}{2} \sum_{e \in E} | \phi_x - e^{i A_{x y}} \phi_y |^2 + \sum_{x \in \Lambda} V (\phi_x) .\)

The potential \(V (\phi_x) = v (| \phi_x |)\) looks like

\(\displaystyle V (\phi_x) := \lambda | \phi_x |^4 - \frac{1}{4} \mu^2 | \phi_x |^2 - E\)

YM action

For \(p \in \{ \text{faces} \} = : F\) (plaquettes). They come with two orientations. We pick a standard orientation for each plaquette. We denote by \(- p\) the plaquette \(p\) taken with the opposite orientation.

Given \(A : E \rightarrow \mathbb{R}\) (a \(1\)–form) we let

\(\displaystyle (\partial A)_p := \sum_{x y \in \partial p} A_{x y}, \qquad (\partial A)_{- p} = - (\partial A)_p\)

This is a 2–form. The YM action is given by

\(\displaystyle S_{Y M} (A) := \frac{1}{2 e^2} \sum_{p \in F} (\partial A_p)^2,\)

where \(e\) is the electric charge.

Gauge invariance

The measure

\(\displaystyle e^{- S_{\text{YM}} (A) - S_{\text{Higgs}} (A, \phi)} \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)

is not normalizable, because there is a non-compact group of symmetries for it. Gauge transformations:

\(\displaystyle \phi_x \rightarrow e^{i \theta_x} \phi_x, \qquad A_{x y} \rightarrow A_{x y} + (\partial \theta)_{x y}\)

with \((\partial \theta)_{x y} := \theta_x - \theta_y\), \(\theta_x \in \mathbb{R}\) for all \(x \in \Lambda\) and \(\theta_{\ast} = 0\). This transformation leaves the measure invariant.

Gauge fixing

We need a gauge fix to remove this symmetry. We fix the total action to be:

\(\displaystyle S = S_{\text{YM}} (A) + S_{\text{Higgs}} (A, \phi) + G\)

where \(G = G (A)\) is such that

\(\displaystyle \int_{\mathbb{R}^{\Lambda}} e^{- G (A + \partial \theta)} \mathrm{d}^{\Lambda} \theta = 1. \) (1)

Example: Landau gauge fixing for \(\alpha \rightarrow 0\) and \(\alpha = 1\) is Feynman.

\(\displaystyle G (A) = \frac{1}{2 \alpha^2} \underbrace{(\partial^{\ast} A, \partial^{\ast} A)}_{:= \sum_{x \in \Lambda} (\partial^{\ast} A)_x^2} + \text{const}, \qquad \quad \raisebox{-0.5\height}{\includegraphics[width=14.8909550045914cm,height=8.92450478814115cm]{image-1.pdf}}\)

where \((\partial^{\ast} A)_x = \sum_{y : y \sim x} A_{x y}\) is the discrete analog of the divergence and it is the adjoint of \(\partial\) for the inner–product \(({,})\). The base point \(\ast\) is introduced to deal with the invertibility of \(\partial^{\ast} \partial\).

Exercise: check (1).

Observables

Gauge invariant, periodic in \(A\), finite \(\#\) of fields. Example: \(\bar{\phi}_x e^{i A_{x y}} \phi_y\), \(x y \in E\). Let

\(\displaystyle \langle F \rangle_{\Lambda} := Z_{\Lambda}^{- 1} \int_{\mathbb{R}^E \times \mathbb{C}^{\Lambda}} F (A, \phi) e^{- S_{\text{YM}} (A) - S_{\text{Higgs}} (A, \phi) - G (A)} \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi .\)

Theorem 1. (\(d \geqslant 2\)) Fix \(\mu\) and \(\mu e / \sqrt{8 \lambda}\). For \(\lambda\) sufficiently small, for observables \(F, G\):

  1. \(\langle F \rangle = \lim_{\Lambda \nearrow \mathbb{Z}^d} \langle F \rangle_{\Lambda}\).

  2. There exists an \(m > 0\) such that if we denote by \(G_x\) the translation by \(x \in \mathbb{Z}^d\) of \(G\) we have

    \(\displaystyle | \langle F G_x \rangle - \langle F \rangle \langle G_x \rangle | \leqslant C e^{- m | x |}, \qquad x \in \mathbb{Z}^d .\)

(and the Kennedy–King paper discusses also the massless phase of this model)

This theorem says tha the gauge particle (whose field is \(A\)) is a massive particle even if it is related to a gauge symmetry. So gauge theories are not incompatible with massive gauge particles.

Equivalence to compact \(U (1)\) theory.

Set \(e = 1\).

Theorem 2.

\(\displaystyle \int_{\mathbb{R}^E \times \mathbb{C}^{\Lambda}} e^{- S (A, \phi)} F (A, \phi) \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)

\(\displaystyle = \int_{[- \pi, \pi]^E \times \mathbb{C}^{\Lambda}} \left[ \sum_{v : \partial v = 0} e^{- \frac{1}{2} \sum_{p \in F} (\partial A_p + v_p)^2} \right] e^{- S_{\text{Higgs}} (A, \phi)} F (A, \phi) \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)

where \(v \in (2 \pi \mathbb{Z})^F\). (\(v\) is a vortex variable)

In \(d = 3\): Given \(v \in (2 \pi \mathbb{Z})^F\) and a \(3\)–cell \(c\), \((\partial v)_c = \sum_{p \in \partial c} v_p\).

\(\displaystyle \raisebox{-0.5\height}{\includegraphics[width=14.8909550045914cm,height=8.92450478814115cm]{image-1.pdf}}\)

Therefore \(\partial v = 0\) requires to identify closed loops which carries currents around without letting it splill.

In \(d = 2\) we have \(\sum_p v_p = 0\). (one has to consider the point \(\ast\) to show this).

The expression

\(\displaystyle \left[ \sum_{v : \partial v = 0} e^{- \frac{1}{2} \sum_{p \in F} (\partial A_p + v_p)^2} \right]\)

is \(2 \pi\) periodic in each \(A_{x y}\). (This will allow us to introduce \(U (1)\) invariance).

\(\displaystyle A_{x y} \rightarrow A_{x y} + \partial \theta_{x y} \quad \text{(mod $2 \pi$)} .\)

Why is periodic. A change \(A_{x y} \rightarrow A_{x y} + 2 \pi\) can be absorbed by changing \(v\).

Higgs mechanism

Make a change of variables

\(\displaystyle \phi_x = \rho_x e^{i \theta_x}, \qquad A_{x y} \rightarrow A_{x y} + \partial \theta_{x y}\)

with \(\rho_x \geqslant 0\). Now the rhs of Theorem 2 (putting back the \(e\)) has the form

\(\displaystyle = \int_{[- \pi / e, \pi / e]^E \times \mathbb{C}^{\Lambda}} \left[ \sum_{v : \partial v = 0} e^{- \frac{1}{2} \sum_{p \in F} (\partial A_p + v_p)^2} \right] e^{- \frac{1}{2} \sum_{e \in E} | \rho_x - e^{i e A_{x y}} \rho_y |^2} F (A, \rho e^{i \theta}) \mathrm{d}^E A \prod_{x \in \Lambda} e^{- V (\rho_x)} \rho_x \mathrm{d} \rho_x\)

and approximating \(\rho_x\) with \(\mu / (8 \lambda)^{1 / 2}\) (due to the potential term) we have

\(\displaystyle - \frac{1}{2} \sum_{e \in E} | \rho_x - e^{i A_{x y}} \rho_y |^2 \approx - \frac{1}{2} \sum_{e \in E} \frac{\mu^2}{(8 \lambda)} | 1 - e^{i e A_{x y}} |^2 \approx - \frac{1}{2} \sum_{e \in E} \underbrace{\frac{\mu^2 e^2}{(8 \lambda)}}_{m_A} | A_{x y} |^2\)

where \(m_A\) is called classical mass. Taking into account flucutations one can prove that \(\rho_x\) behaves like a Gaussian field fluctuating around \(\mu / (8 \lambda)^{1 / 2}\) and \(A\) is a massive gaussian. And the vortices provide only rare fluctuations which can be rigorously estimated and do not change this leading order behaviour. The rigorous proof goes via cluster expansion.

Proof of equivalence in Theorem 2.

\(\displaystyle \int_{\mathbb{R}^E \times \mathbb{C}^{\Lambda}} e^{- S_{\text{YM}} (A) - S_{\text{Higgs}} (A, \phi)} F (A, \phi) \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)
\(\displaystyle = \int_{\mathbb{R}^E \times \mathbb{C}^{\Lambda}} e^{- S_{\text{YM}} (A) - S_{\text{Higgs}} (A, \phi)} F (A, \phi) \left[ (2 \pi^{- \Lambda}) \int_{[- \pi, \pi]^{\Lambda}} e^{- G (A + \partial \theta)} \mathrm{d}^{\Lambda} \theta \right] \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)
\(\displaystyle = \sum_{a \in (2 \pi \mathbb{Z})^E} \int_{[- \pi, \pi]^E} e^{- S_{\text{YM}} (A + a) - S_{\text{Higgs}} \left( A + \not{a}, \phi \right) } F \left( A + \not{a}, \phi \right) \left[ (2 \pi^{- \Lambda}) \int_{[- \pi, \pi]^{\Lambda}} e^{- G (A + a + \partial \theta)} \mathrm{d}^{\Lambda} \theta \right] \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)
\(\displaystyle = \sum_{v : \partial v = 0} \sum_{a : \partial a = v} \int_{[- \pi, \pi]^E} e^{- S_{\text{YM}} (A + a) - S_{\text{Higgs}} \left( A + \not{a}, \phi \right) } F \left( A + \not{a}, \phi \right) \times\)
\(\displaystyle \times \left[ (2 \pi^{- \Lambda}) \int_{[- \pi, \pi]^{\Lambda}} e^{- G (A + a + \partial \theta)} \mathrm{d}^{\Lambda} \theta \right] \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)

(here we use the fact that we have an exact sequence). Moreover again by exactness we have that \(\partial a = v\) iff \(a = a_v + \partial b\). Therefore the sum over \(a\) is equivalent to the sum over \(b\). So

\(\displaystyle = \sum_{v : \partial v = 0} \sum_b \int_{[- \pi, \pi]^E} e^{- \frac{1}{2} \sum_{p \in F} (\partial A_p + v_p)^2 - S_{\text{Higgs}} (A, \phi) } F (A, \phi) \times\)
\(\displaystyle \times \left[ (2 \pi^{- \Lambda}) \int_{[- \pi, \pi]^{\Lambda}} e^{- G (A + a_v + \partial (\theta + b))} \mathrm{d}^{\Lambda} \theta \right] \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)
\(\displaystyle = \sum_{v : \partial v = 0} \int_{[- \pi, \pi]^E} e^{- \frac{1}{2} \sum_{p \in F} (\partial A_p + v_p)^2 - S_{\text{Higgs}} (A, \phi) } F (A, \phi) \times\)
\(\displaystyle \times \sum_b \left[ (2 \pi^{- \Lambda}) \int_{[- \pi, \pi]^{\Lambda}} e^{- G (A + a_v + \partial (\theta + b))} \mathrm{d}^{\Lambda} \theta \right] \mathrm{d}^E A \mathrm{d}^{\Lambda} \phi\)

and now

\(\displaystyle \sum_b (2 \pi^{- \Lambda}) \int_{[- \pi, \pi]^{\Lambda}} e^{- G (A + a_v + \partial (\theta + b))} \mathrm{d}^{\Lambda} \theta = (2 \pi^{- \Lambda}) \int_{\mathbb{R}^{\Lambda}} e^{- G (A + a_v + \partial \theta)} \mathrm{d}^{\Lambda} \theta = 1!!\)

so we are done.

The exact sequence we used is (for \(d = 3\))

\(\displaystyle 0 \longrightarrow (2 \pi \mathbb{Z})^{\Lambda \cup \{ \ast \}} / (2 \pi \mathbb{Z}) \overset{\partial}{\longrightarrow} (2 \pi \mathbb{Z})^E \overset{\partial}{\longrightarrow} (2 \pi \mathbb{Z})^F \overset{\partial}{\longrightarrow} (2 \pi \mathbb{Z})^{\{ \text{$3$} -\operatorname{cells} \}} \overset{\partial}{\longrightarrow} (2 \pi \mathbb{Z})^{\{ \text{$4$} -\operatorname{cells} \}} \longrightarrow 0.\)