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Yang-Mills Instantons (I)

By M. E. Irizarry-Gelpí

In this series of post I would like to scratch the surface of an enormous iceberg called “instanton physics”. First I would like to mention some references:

  1. arXiv:0802.1862 – Lectures on instantons by Vandoren and van Nieuwenhuizen,
  2. arXiv:hep-th/0206063 – The calculus of many instantons by Dorey, Hollowood, Khoze and Mattis,
  3. arXiv:hep-th/0004186 – Yang-Mills- and D-instantons by Belitsky, Vandoren and van Nieuwenhuizen.

Most of these posts are going to be very loyal to the first item, meaning that I will only discuss one instanton cases. As the title suggests, the second item deals with the case of many instantons.

So what is an instanton? A Yang-Mills instanton is a solution to the classical field equations in Euclidean space that give a finite action. Next you might ask, why finite action? Recall that for (classical) Yang-Mills the path integral has the form

Z_{E}\left(A\right) = \displaystyle\int DA \exp{\left(-\frac{S_{cl}}{\hbar}\right)}.

A finite action then gives the leading contribution to Z_{E}\left(A\right). We can distinguish between regular instantons, ones that have a singularity at Euclidean infinity, and singular instanton, which don’t have singularity at infinity but at some point in space x_{0}^{m}. It turns out that we can map singular to regular solutions by a singular gauge transformation.

Later we will consider systems in the background of an instanton. We can achive this by the usual ways of minimal coupling,

\partial \rightarrow \partial + A .

When we have such a background one has to be careful with zero modes. These are solutions of the linearized field equations that are normalizable. Alternatively, zero modes are eigenfunctions of the quantum operator with zero eigenvalue. The quantum operator is (I think) the operator that appears in the action when one integrates by part the Lagrangian. For example,

S = \int d^4 x \partial X \partial X \rightarrow -\int d^4 x X \partial^{2} X.

In this case the quantum operator corresponds to \partial^2. Zero modes have their own measure in the path integral and sometimes they are the only contribution (e.g. in supersymmetric theories the non-zero modes cancel).

Let us be a bit precise. Let us consider Yang-Mills gauge theory in 4 Euclidean space dimensions with gauge group SU(N). The Lie algebra has generators T_{a} that are traceless, anti-hermitian N \times N matrices with the normalization

tr\left(T_{a}T_{b}\right) = -\displaystyle\frac{1}{2}\delta_{ab}.

The action for Yang-Mills theory is

S = \displaystyle\frac{-1}{2 g^{2}} \int_{\mathbb{R}^{4}} d^{4}x \left[ tr\left(F^{mn}F_{mn}\right) \right],

with the field strength given by

F_{mn} = \partial_{m} A_{n} - \partial_{n}A_{m} + \left[A_{m} , A_{n}\right].

The classical field equations for A_{m} are found from the Euler-Lagrange equations:

D_{m} F^{mn} = 0 \qquad D_{m} \cdot = \partial_{m} \cdot + \left[A_{m}, \cdot \right] .

Since instantons are solutions to this equation but have finite action, we expect the field strength to vanish very far away from the origin. Since F appears quadratic in the action, it should vanish faster than r^{2}. The statement that the field strength vanishes leads us to looking for gauge potentials that are pure gauge, that is, they have the form

A_{m} = U^{-1} \partial_{m} U \qquad U \in SU(N).

[To be continued...]

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This entry was posted on October 15, 2008 at 2:15 PM and is filed under Gauge theory. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Yang-Mills Instantons (I)”

  1. Eitan Says:
    October 16, 2008 at 11:10 AM | Reply

    Hey cool. Also, check out this note by S.S. Chern http://psroc.phys.ntu.edu.tw/cjp/v30/949.pdf

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