Deriving the Yang-Mills-Higgs Equations of Motion: A Step-by-Step Guide

[Charles_Henry]

Homework Statement

1. Derive the Yang-Mills-Higgs equations of motion

$(D_{v}F^{\mu, v})^{a} = - \epsilon^{abc}\phi^{b}(D^{\mu}\phi)^{c}$ and $(D_{\mu}D^{\mu}\phi)^{a} = - c(|\phi|^{2} - v^{2}) \phi^{a}$

from $\mathcal{L} = - \frac{1}{4} F^{a}_{\mu v}F^{\mu v a} + \frac{1}{2}D_{\mu}\phi^{a}D^{\mu}\phi^{a} - U(\phi)$

where is c is a constant.

2. Show that in SU(2) Yang-Mills-Higgs theory the general solution to the equation $D_{i}\hat{\phi}=0$ with $|\hat{\phi}| = 1$ is

$A^{a}_{i} = -\epsilon^{abc}\partial_{i}\hat{\phi}^{b}\hat{\phi}^{c} + k_{i}\hat{\phi}^{a}$


The attempt at a solution

1st question:

Assume finiteness of the energy by

$\phi \rightarrow v$ $D_{mu} \phi \rightarrow 0$ $F_{{\mu}v} \rightarrow 0$ as $r \rightarrow \infty$

Let

$F = \frac{1}{2} F_{{\mu},v}dx^{\mu} \wedge dx{v} = dA + A \wedge A$ be

the gauge field of A.

with components

$F_{{\mu}v} = \partial_{\mu}A_{v} - \partial_{v}A_{\mu} + [ A_{\mu}, A_{v}] = [D_{\mu}, D_{v}]$

We define gauge identities $(A, A')$ and $(F, F')$ where

$A' = gAg^{-1} - dgg^{-1}$ $F' = gFg^{-1}$ and $g = g(x^{\mu}) \in G$

which satisfies the bianchi identity

$DF = dF + [A,F] = d^2 A + dA \wedge A - A \wedge dA + A \wedge dA + A^3 - dA \wedge A - A^3 = 0$

The field

$\phi: \mathbb{R}^{D+1} \rightarrow \mathfrak{g}$

$D\phi = d\phi + [A, \phi]$ with gauge transformation,

$\phi' = g \phi g^{-1}$

if $G = SU(2)$ choose a basis $T_{a} , a =1,2,3$

for the algebra of anti hermitian (2x2 matrices), such that

$[T_{a},T_{b}] = - \epsilon_{abc}T_{c} T_{a} = \frac{1}{2} i\sigma_{a}$ where $\sigma_{a}$ are pauli matrices.

$Tr (T_{a}T_{b}) = - \frac{1}{2} \delta_{ab}$

a general group element is $g = exp (\alpha^{a}T_{a})$ with $\alpha^{a}$ being real. The components of $D/phi$ and $F$ with respect to the basis are given by

$(D_{\mu} \phi)^{a} = \partial_{\mu} \phi^{a} - \epsilon^{abc} A^{b}_{\mu} \phi^{c}$ and $F^{a}_{{\mu}v} = \partial_{\mu}A^{a}_{v} - \partial_{v}A^{a}_{\mu} - \epsilon^{abc}A^{b}_{\mu}A^{c}_{v}$

when $D+1 = 4$ the dual of the field tensor is

$(\ast F)_{{\mu}v} = \frac{1}{2} \epsilon_{{\mu}v{\alpha}{\beta}}F^{{\alpha}{\beta}}$

we introduce a two form $F = (\frac{1}{2}) F_{{\mu}v}dx^{\mu} \wedge dx^{v}$ (Self-Dual)

or $\ast F = F$ $\ast F = - F$ (Anti-Self-Dual)

$- Tr (F \wedge \ast F) = - \frac{1}{2} Tr (F_{{\mu}v} F^{{\mu}v})d^{4}x \frac{1}{4} F^{a}_{{\mu}v}F^{{\mu}va}d^{4}x$

where $d^{4}x = \frac{1}{24}\epsilon_{{\mu} v {\alpha} {\beta} } dx^{\mu} \wedge dx^{v} \wedge dx^{\alpha} \wedge dx^{\beta}$

to be continued...

 

[fzero]

For part 1, you will want to use the Euler-Lagrange equations. Gauge transformations and differential forms don't directly figure into deriving equations of motion from a Lagrangian.

 

[Charles_Henry]

thank you :)

Ok, so a YM over 'space-time' $\mathbb{R}^{3}$ x $\mathbb{R}$ with a euclidean metric,

$Y = \mathbb{R}^{3}$ with t a local co-ordinate on $\mathbb{R}$

$A_{y} dt + A$ when $A_{y} = 0$

therefore,

$F_{ni} = \frac{{\partial} A_{i}}{{\partial}_{t}}$

such that

$\frac{{\partial}A_{i}}{dt} = \frac{{\delta}W[A]}{{\delta}A_{i}}$

where $W[A] = \int_{y} Tr (A \wedge dA + \frac{2}{3} A \wedge A \wedge A)$ is

$\int_{\mathbb{R}} (\frac{1}{2} ||\hat{A}||^{2} - V [A]) dt$

where $v[A]$ is the magnetic part of the YM curvature, that is

$||A||^2 = \int_{\mathbb{R}} Tr (\frac{dA_{i}}{dt} \frac{dA_{i}}{dt}) d^{n}x$

$v[A] = \frac{1}{4} \int_{\mathbb{R}^{n}} Tr (F_{ij}F_{ij})d^nx$

$\frac{1}{2} \int_{\mathbb{R}^{n}} Tr \frac{{\delta}W}{{\delta}A_{i}} \frac{{\delta}W}{{\delta}A_{i}})d^{n}x$

$= \frac{1}{2} |\frac{{\delta}W}{{\delta}A}|^{n}$

if n above is 2

$\frac{1}{2} |\Delta W |^{2} + V(q) = 0$

adding a constant E.

$\frac{1}{2} |\Delta W|^{2} + V(q) = E$

we can approximate the system by classical motion along gradient lines with reversed potential.

$\frac{1}{2} |\Delta W|^{2} = V(q) - E$

Moreover, when $\mathbb{R}^{3}$

$\int_{\mathbb{R}^{3}} \frac{1}{4} F^{a}_{ij} F^{a}_{ij} + \frac{1}{2} (D_{k} \phi)^{a} (D_{k} \phi)^{a} ] d^{3}x$

$\frac{1}{2} \int_{\mathbb{R}^3} (B_{k} - D_{k} \phi)^{a}$

$(B_{k}-D_{k}\phi)^{a} d^{3}x$ + $\int_{\mathbb{R}^{3}} {B^{a}_{k}} (D_{k} \phi)^{a} d^{3}x$

$= E_{1} + E_{2}$

is this the right path?

 

[fzero]

Not really. You gave the Lagrangian $\mathcal{L}$ in the statement of part 1, so you just have to use the Euler-Lagrange equations. Schematically, these are

$$\left(\partial_\mu \frac{\delta }{\delta (\partial_\mu \psi)} - \frac{\delta}{\delta \psi}\right) \mathcal{L}(\psi,\partial_\mu\psi) = 0 .$$

You just have to adapt this to the case where the fields are $\psi = ( A^a_\mu , \phi^a)$. If you are having trouble, make sure that you know how this works for the simpler cases of the free scalar field and U(1) gauge field without matter.

 

[Charles_Henry]

Ok,

A lagrangian density $\mathcal{L} = \mathcal{L}(\phi^{a}, \partial_{\mu}\phi^{a})$ where $\partial_{\mu} = \frac{\partial}{\partial_{\mu}}$ depending on the fields and first derivatives?[itex][/itex]

$S = \int_{\mathbb{R}^{D}} \mathcal{L} d^{D} xdt$

and from the least action principle

$\frac{\partial \mathcal{L}}{\partial \phi^{a}} - \frac{\partial}{\partial x^{\mu}} \frac{\partial \mathcal{L}}{(\partial (\partial_{\mu}\phi^{a}}) = 0$

$\partial^{\mu}\partial_{\mu} \phi^{A} = - \frac{\partial U}{\partial \phi^{a}}$

find $\phi_{s} = \phi_{s}(x,y)$ ?

from hereon, the literature derives the lorentz-invariance, and then proceeds to calculate the corresponding infinitesimal change in lagrangian density. Euler-Lagrange equations are then used to cancel out remaining terms, and the equation now fits the conserved current in accordance to Noether (Q).

 

[Charles_Henry]

trying different things

$U(\phi)$ is gauge invariant

$\partial^{\mu} F^{a}_{{\mu} v} + g \epsilon^{abc}D^{{\mu} b} F^{c}_{{\mu} v} = 0$

$F_{\mu v} = T^{a} F^{a}_{\mu v}$

$(D^{\mu} F_{v \mu})^{a} + (D_{k} F_{\mu v})^{a} + (D_{v} F_{k \mu})^{a} = 0$

$\partial^{\mu} F^{a}_{\mu v} + g \epsilon^{abc} A^{\mu b} F^{c}_{\mu v} = -J^{a}_{v}$

$c = (A, \phi)$ $D_{A} \ast F_{A} + [ \phi, D_{A} \phi ] = 0$

$D_{A} \ast D_{A} \phi = 0$

With boundary conditions we have

$lim _{|x| \rightarrow \infty} | \phi| = 1$

$A( A, \phi) = (\frac{1}{2} \int_{\mathbb{R}^3} |F_{A}|^2 (x) + |D_{A} \phi|^2 (x) ) d^{3} x$

Alternatively,

the potential $A_{\mu} = \frac{1}{2} \tau_a A^{a}_{\mu}$

$F_{{\mu} v} = \partial_{\mu} A_{v} - \partial_{v} A_{\mu} + i [A_{\mu}, A_{v}]$

the covariant higgs field $\phi = \tau_{a} \phi^{a}$

$D_{\mu} \phi = \partial_{\mu} \phi + i [A_{\mu}, \phi]$

$A_{\mu} \rightarrow gA_{\mu}g^{-1} + i \partial_{\mu}gg^{-1}$

$\phi \rightarrow g \phi g^-{1}$

 

[fzero]

Ok,

A lagrangian density $\mathcal{L} = \mathcal{L}(\phi^{a}, \partial_{\mu}\phi^{a})$ where $\partial_{\mu} = \frac{\partial}{\partial_{\mu}}$ depending on the fields and first derivatives?[itex][/itex]

$S = \int_{\mathbb{R}^{D}} \mathcal{L} d^{D} xdt$

and from the least action principle

$\frac{\partial \mathcal{L}}{\partial \phi^{a}} - \frac{\partial}{\partial x^{\mu}} \frac{\partial \mathcal{L}}{(\partial (\partial_{\mu}\phi^{a}}) = 0$

$\partial^{\mu}\partial_{\mu} \phi^{A} = - \frac{\partial U}{\partial \phi^{a}}$

Yes, that's how it works for the scalar without the gauge field. The least action principle leads to the Euler-Lagrange equations which are the equations of motion you're looking for. Just go back and apply them to the Lagrangian with the gauge field. The rest of the equations you keep quoting are irrelevant for the problem.

 

[Charles_Henry]

oh, I see...

The least-action principle does not hold on this occasion, because we're gauging (localizing).

$\psi \rightarrow e^{i e \theta \psi}$

and

$\theta \neq$ constant, therefore $\theta = \theta (x^{\mu})$

$\frac{\partial}{\partial x^{\mu}}$ changes to $D_{\mu} = \partial_{\mu} - ieA_{\mu}$ $A_{\mu}$ transitions to $A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \theta$

so $\mathcal{L} = \int_{\mathbb{R}} \frac{1}{2} D_{\mu}D^{\mu} \phi^{2}$

$\mathcal{L} = (\phi^{a}, D_{\mu} \phi^{a})$

$D_{\mu} = \partial_{\mu} - ieA_{\mu}$

$\mathcal{L} = \frac{1}{2} D_{\mu} \phi^{a} D^{\mu} \phi^{a} - \frac{1}{4}c(|\phi|^2 - v^2)^{2}$

$D^{\mu} D_{\mu} \phi^{a} = -\frac{\partial}{\partial \phi ^{a}}$

$\mathcal{L} \rightarrow \mathcal{L} + \epsilon D_{\mu}B^{\mu}$

$\mathcal{L} \rightarrow \mathcal{L} + \epsilon \partial_{\mu} - ie A_{\mu} B^{\mu}$

$\phi^{a} (x) \rightarrow \phi^{a} (x) + \epsilon W^{a} (x)$

closer?

 

[Charles_Henry]

ok in full force now,

the matter lagrangian is complemented by adding a gauge term $F_{\mu v} F^{\mu v}$

$F_{\mu v} = \partial_{\mu}A_{v} - \partial_{v}A_{\mu}$ (gauge field)

now,

$\frac{\partial}{\partial x^{\mu}}$ transforms to $D_{\mu} = \partial_{\mu} - ie A_{\mu}$

$A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \theta$

$(D_{\mu} \phi)^a = \partial_{\mu} \phi^a - \epsilon^{abc} A_{\mu}^{b} A_{v}^{c}$

and finally,

$\mathcal{L} = \mathcal{L} = - \frac{1}{4}(\partial_{\mu} A^{a}_{v} - \partial_{v}A^{a}_{\mu} - \epsilon^{abc} A^{b}_{\mu} A^{c}_{v}) F^{\mu v a } + \frac{1}{2}(\partial_{\mu} - ieA_{\mu}) \phi^a D^{\mu} \phi^{a} - U({\phi}) = 0$

then simplify from hereon and solve...