Gauge? Who is he?

No, gauge does not indicate the name of a famous theorist. The word is to be considered a historical misnomer for phase.

Gauge theories are a special class of relativistic quantum field theories based on the idea that some transformations (e.g. phase transformations) leave unchanged the Lagrangian of the system (symmetries).

Gauge invariance (or gauge symmetry) is the property of a field theory in which different configurations of the underlying fields result in identical observable quantities.

A transformation from one such field configuration to another is called gauge transformation. Gauge transformations can be either global and local.

Let’s see some well-known cases.

Classical electrodynamics

In classical electromagnetism it is convenient to introduce the vector potential $A_\mu (x)$ in place of the fields $\vec{E}$ and  $\vec{B}$:

$\vec{B} =\vec{\triangledown}\times \vec{A}$
$\vec{E} =-\vec{\triangledown}V-\frac{\partial\vec{A}}{\partial t}$.

These equations define the three dimentions vector potential $\vec{A}$ and the scalar potential V.

If we now consider a transformation like

$\vec{A}\rightarrow\vec{A}' = A + \triangledown\chi$,
$V\rightarrow{V}' =V-\frac{\partial \chi}{\partial t}$,

where $\chi$ is an arbitrary function, both$\vec{E}$ and  $\vec{B}$ stay unchanged.

It is now possibile to introduce the four-vector potential $A^{\mu}=(V,\vec{A})$ which allows to combine the previous equations into a single tensor equation

$A^{\mu}\rightarrow A'^{\mu}=A^{\mu}-\partial^{\mu} \chi$,

and to rewrite the Maxwell’s equations in a manifestly covariant form

$\partial_\mu F^{\mu\nu}=j^{\nu}_{em}$

by introducing the four-current $j^{\mu}_{em}=(\rho_{\mu},\vec{j}_{em})$ and the field strength tensor $F^{\mu\nu}\equiv\partial^\mu A^\nu - \partial ^\nu A^\mu$. It is worthy to note that $F$ remains unchanged under the gauge transformation stated above ($A^{\mu}\rightarrow A'^{\mu}=A^{\mu}-\partial^{\mu} \chi$) and thus it is said gauge invariant, like the Maxwell’s equations in the form $\partial_\mu F^{\mu\nu}=j^{\nu}_{em}$.  At this point it is possibile to write the Lorentz-covariant and gauge-invariant field equations as

$\Box A^\nu-\partial\nu(\partial_\nu A^\mu)=j^\nu_{em}$,

but this is another story. What matters here is that $\vec{B}$ and $\vec{E}$ are gauge-invariant and classical electrodynamics is a gauge theory.

Quantum mechanics

A gauge transform in quantum mechanics assumes a different form.

Let’s consider a wave function $\psi$. Observables corresponds to $| \psi |^2$. Thus, the theory stays invariant under transformations like

$\psi\rightarrow \psi ' =e^{-i\alpha}\psi$

where $\alpha$ is constant and indicates a phase. The transformation above is a global gauge transformation since $\psi (\vec{r},t)$ is everywhere invariant i.e. $\alpha$ do not depend by $x^\mu$.

Invariance under a constant change in phase is an example of a global invariance.

To obtain a theory which is dependent on $x^\mu$ we need a constant depending on the point. The transformation will be like the following

$\psi(\vec{r},t)\rightarrow\psi '(\vec{r},t)=e^{-i\chi(\vec{r},t)} \psi(\vec{r},t)$

which expresses a local gauge transformation.

If the original wavefunction $\psi(\vec{r},t)$ satisfied the free-particle Schrödinger equation,

$\frac{-i}{2m}\triangledown^2\psi(\vec{x},t)=i\frac{\partial}{\partial t}\psi(\vec{x},t)$

then the transformed wavefunction $\psi '$ given by the local gauge transformation will not, since both $\triangledown$ and $\partial/\partial t$ now act on $\chi$ in the phase factor.

To satisfy the local phase invariance one should change the Schrödinger equation but a such modified equation will no longer describe a free particle. The covariance of the new equation will be showed in the inability to distinguish between a local change in phase convention and the effect of some new field in which the particle moves: the freedom to alter the phase of a charged particle’s wavefunction locally is only possible if some kind of force field is introduced in which the particle moves. In fact, if we modify the equation to

$\frac{1}{2m}(-i\triangledown-qA)^2\psi(\vec{x},t)=\left(i\frac{\partial}{\partial t}-qV\right)\psi(\vec{x},t)$

the local phase invariance $\psi\rightarrow\psi '=e^{i\chi(\vec{x},t)}\psi$ is satisfied if we require that while $\psi$ transforms to $\psi '$, $A$ and $V$ should transform as

$A\rightarrow A' = A+\frac{1}{q}\triangledown \chi$
$V\rightarrow V' = V-\frac{1}{q}\frac{\partial \chi}{\partial t}$.

The two quantities $A$ and $V$ can be considered as time and spatial components of a $A^{\mu}$ four-vector, while $-\triangledown$ and $\frac{\partial}{\partial t}$ are part of the $\partial^\mu$ operator.

The vector field $A^\mu$ introduced to guarantee the local phase invariance and interacting with any particle of charge $q$ is called gauge field.

To be invariant under local gauge transformations, the Schrödinger equation needs a field $A^\mu$. Such field can be expressed in term of creation and annihilation operators. A particle, the photon, is associate to the field due to the quantization. The invariance of the Schrödinger equation phase therefore requires the existence of a photon and the existence of an electromagnetic interaction but, again, this is another story.