research32 min studyLesson 702 in Physics

Gauge transformations and parallel transport

Track endpoint covariance and path ordering for finite matrix-valued transport in a fixed convention.

Before you begin

On this page

The question for today

What you’ll understand

Derive the connection transformation from covariant matter evolution.
Multiply segment transporters in the correct path order.
Distinguish an open transporter from a traced closed-loop observable.

A gauge transformation changes the local internal basis used to describe matter and the connection. Parallel transport compares those bases along a specified oriented path. In a non-Abelian theory the sequence of matrix multiplications matters, so an integral of the connection cannot generally be exponentiated as one commuting number. Endpoint transformations and path orientation provide useful checks: they reveal whether a proposed expression is a physical invariant, an appropriately covariant object, or simply an inconsistent multiplication order.

Transform the derivative as one object

Let a matter multiplet transform as $\psi'=U\psi$ , with unitary U depending on spacetime. Require its covariant derivative to transform the same way. Moving the derivative through U gives an inhomogeneous term in the transformed connection. This term compensates the change of local basis; it is why the connection itself does not transform like curvature. In the Abelian subgroup the formula reduces to an ordinary gradient shift.

For a small transformation written $U=e^{ig\alpha}$ , the connection change is the adjoint covariant derivative of alpha. Here alpha is a Hermitian Lie-algebra-valued parameter with dimension zero. One may instead put no g in the exponent and rescale the parameter, but then the infinitesimal formula changes. A pure-gauge connection obtained by transforming zero has zero curvature locally even when its matrix components are nonzero.

D_\mu'=UD_\mu U^{-1},\qquad A_\mu'=UA_\mu U^{-1}+\frac{i}{g}U\partial_\mu U^{-1},\qquad \delta A_\mu=D_\mu\alpha

The finite formula uses the same D=partial-minus-i-g-A convention as the previous lesson. The infinitesimal formula assumes U=exp(i g alpha), and U inverse equals its Hermitian adjoint.↔ Scroll wide equations horizontally.

Later path segments multiply on the left

Parameterize an oriented path from an initial point x to a final point y. Covariantly constant transport obeys an ordinary matrix differential equation with initial transporter identity. Its iterative solution is a path-ordered exponential, with matrices encountered later on the path placed on the left. This convention makes the full transporter act on the starting matter column to produce the final column. Reversing the path gives the inverse transporter.

For a piecewise constant connection along two consecutive segments, first compute each segment exponential. The total is the second segment matrix times the first. Commuting segment matrices can be combined into a simple exponential of their sum; noncommuting ones cannot without extra commutator corrections. The difference is a finite version of the local curvature mismatch. Relative orientation and matrix order are part of the definition, not decorative notation that can be rearranged cyclically before taking a trace.

\frac{dW(s)}{ds}=ig\dot x^\mu A_\mu(x(s))W(s),\quad W(0)=I,\qquad W(y,x)=\mathcal P\exp\!\left(ig\int_x^y A_\mu\,dx^\mu\right)

The exponent is dimensionless. W is in the chosen unitary representation, and the definition fixes later matrices on the left; it is not ordinary scalar integration.↔ Scroll wide equations horizontally.

Keep endpoint covariance separate from invariance

An open transporter changes by U at the final endpoint and inverse U at the initial endpoint. It is therefore covariant, not a standalone gauge-invariant number. It can connect matter operators whose endpoint transformations cancel those factors. A closed transporter based at one point transforms by conjugation, so its trace in a specified representation is invariant. Moving the starting point around a smooth closed path also conjugates its holonomy.

Curvature zero in a local patch does not settle all global holonomy questions. Noncontractible paths and allowed boundary transformations can retain information beyond a local derivative test. Conversely, changing only endpoint bases does not create physical curvature. These distinctions are essential when interpreting Wilson loops, boundary charges, and comparisons of gauges. A pure two-level state lab can display the action of a particular SU(2) matrix, but cannot establish the connection or gauge orbit that produced it.

W(y,x)'=U(y)W(y,x)U^{-1}(x),\qquad W(x,y)=W(y,x)^{-1}

The inverse relation refers to the same geometric path with opposite orientation. Gauge-invariant closed-loop observables additionally require a trace and a declared representation.↔ Scroll wide equations horizontally.

Work it through

Two ordered SU(2) segments

In the fundamental representation let the first segment give $W_1=(I+i\sigma_1)/\sqrt2$ and the second $W_2=(I+i\sigma_2)/\sqrt2$ . Multiply them in path order and compare with reversing the segment order.

With the declared column-vector convention the original path gives $W_2W_1$ . Expanding yields $(I+i\sigma_1+i\sigma_2-\sigma_2\sigma_1)/2$ .
Pauli multiplication gives $\sigma_2\sigma_1=-i\sigma_3$ , so the transporter is $(I+i\sigma_1+i\sigma_2+i\sigma_3)/2$ . The two segment matrices are unitary rotations with integrated generator angle pi over two.
Reversing the segment order gives $(I+i\sigma_1+i\sigma_2-i\sigma_3)/2$ . Their difference is $i\sigma_3$ , explicitly nonzero. This comparison changes the ordered path, rather than merely rewriting the same transporter.

The result

The two orders differ in the sign of the third-generator term; finite non-Abelian parallel transport cannot discard path ordering.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

A transport-sign check on one state

For a commuting T-three segment with angle sixty degrees, transport exp(plus i angle T-three) adds a minus-sixty-degree relative phase to a plus state. Reset and set mixing angle ninety degrees and phase minus sixty degrees; predict Bloch coordinates $(1/2,-\sqrt3/2,0)$ . Compare a fresh Reset run at ninety degrees and phase zero. This prepares pure-state geometry, not a path, connection, or gauge orbit.

Bloch vector / P(0)P(1)Phase latitude

Probability |0⟩: 71.1%
Probability |1⟩: 28.9%
Bloch (x, y, z): 0.78, 0.45, 0.42

Mixing angle θ65°Relative phase φ30°

t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTransport signs follow the chosen derivative convention; the lab does not perform non-Abelian path ordering.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The first segment acts first on the column vector, and the second then acts on that result. This gives W-two W-one, with path order fixed before any trace is taken.

Your turn

Try it yourself

For one commuting segment take $W=e^{i(\pi/3)T^3}$ with $T^3=\sigma_3/2$ . Apply it to plus, compute the relative phase and Bloch coordinates, then find the reverse-path action.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The two diagonal phases are opposite half-angles. Remove only a common global phase, not their relative phase.

Compare with the solution

The diagonal entries are $e^{i\pi/6}$ and $e^{-i\pi/6}$ . Acting on plus gives a relative one-to-zero phase $e^{-i\pi/3}$ after removing the common first phase. Thus the Bloch vector is $(1/2,-\sqrt3/2,0)$ and both basis probabilities remain one half. The reversed path uses $W^{-1}$ and adds relative phase plus pi over three, giving $(1/2,\sqrt3/2,0)$ . These are transport matrices in the declared convention, not changes of a measured gauge-invariant color vector.

Keep the idea

Take it with you

Connections transform inhomogeneously while curvature transforms by conjugation.
Path orientation and segment order fix finite matrix transport.
An open transporter is covariant; a closed traced holonomy can be invariant.

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