In physics a
Gauge theory is a type of field theory in which the Lagrangian is invariant under certain 'Lie' group of local Transformation.Gauge Transformationis the english translation of German word 'Eichinvarianz' ,Where Eich means 'Gauge'.
Classical Gauge Transformation in Electrodynamics:
The Maxwell's equations in free space :
$$ (1) \nabla.\overrightarrow{E} = \frac{\rho}{\epsilon_0} \text{ (Gauess's Law)}$$
$$(2) \nabla\text{.}\overrightarrow{B} = 0 \text{ (No Name i.e no existance of magnetic monopole)} $$
$$(3) \nabla\text{x}\overrightarrow{E}=\frac{\partial\overrightarrow{B}}{\partial{t}} \text{ (Farady's Law)} $$
$$ (4) \nabla\text{x}\overrightarrow{B} = \mu_0\overrightarrow{J}+\mu_0\epsilon_0\frac{\partial{\overrightarrow{E}}}{\partial {t}} \text{(Ampears Law with Maxwell's corrections)}$$
Where $$\mu_0$$ and $$\epsilon_0$$ are free space permitivity and free space permeability respectively.
Suppose we have two sets of potentials $$(V,\overrightarrow{A})$$ and $$\{{\widetilde{V},\widetilde{\overrightarrow{A}}}\}.$$
They are related in such way that electric and magnetic fields are remain same.
Where we introduce two new constants $${\overrightarrow{\alpha} and \beta}$$.
Therefore
$$\overrightarrow{B}=\nabla\text{x}\overrightarrow{A}=\nabla\text{X}\widetilde{\overrightarrow{A}}$$
$$or \nabla \text{x} \overrightarrow{A}=\nabla X (\overrightarrow{A}+\overrightarrow{\alpha)}$$
$$or\> \nabla \text{x}\overrightarrow{\alpha}=0 \text{ --------------(1)}$$
i.e we can say
$$\overrightarrow{\alpha}=\nabla\phi$$
where $\phi $ is a new kind constant .
From third equation of Maxwell's equation we can write
$$\nabla \text{x} \overrightarrow{E}= -\frac{\partial(\nabla \text{X} \overrightarrow{A})}{\partial t}$$
$$\>\nabla \text{X}(\overrightarrow{E}+\frac{\partial\overrightarrow{A}}{\partial t})=0$$
$$\overrightarrow{E}=-\nabla V-\frac{\partial\overrightarrow{A}}{\partial t}$$
Now,as the two sets potential are giving same electric field then we can write
$$-\nabla V - \frac{\partial\overrightarrow{A}}{\partial t}=-\nabla\widetilde{V}-\frac{\widetilde{\overrightarrow{A}}}{\partial t}$$
$$-\nabla V -\frac{\overrightarrow{A}}{\partial t}=-\nabla V -\nabla \beta -\frac{\partial\overrightarrow{A}}{\partial t}-\frac{\overrightarrow{\alpha}}{\partial t}$$
$$-\nabla\beta-\frac{\partial\overrightarrow{\alpha}}{\partial t}=0$$
If we put $\overrightarrow{\alpha}=\nabla\phi$ in the above equation we get
$$\beta+\frac{\partial\phi}{\partial t}=\text{Constant(t)}$$
Therefore we can say
$$\widetilde{\overrightarrow{A}}=\overrightarrow{A}+\nabla\phi \text{;}$$
$$\widetilde{V}=V+\beta=V-\frac{\partial\phi}{\partial t}+\text{Constant(t)}$$
Now in the above second equation,we can ignor constant term as it just added to V nothing can contribute in gradient so we can write again
$$\widetilde{\overrightarrow{A}}=\overrightarrow{A}+\nabla\phi \text{;}$$
$$\widetilde{V}=V-\frac{\partial\phi}{\partial t}$$
In this transformation Electric and Magnetic fields are remain same which is known as "Gauge Transformation"
N.B-We will get same expression if we used Maxwell's equations in a matter.\\ \\
\textbf{Change of Dynamical momentum of a charge in presence of Magnetic field:}
The charge 'e' experience total force
$$\overrightarrow{F}=e(\overrightarrow{E}+\overrightarrow{u}X\overrightarrow{B})$$
$$\frac{d\overrightarrow{p}}{dt}=e\overrightarrow{E}+e\overrightarrow{u}X\overrightarrow{B}$$
$$\frac{d\overrightarrow{p}}{dt}=e(-\nabla V-\frac{\partial\overrightarrow{A}}{\partial t})+e\overrightarrow{u}X(\nabla X \overrightarrow{A})$$
$$\frac{d\overrightarrow{p}}{dt}=-e\nabla V-e\frac{\partial\overrightarrow{A}}{\partial t}+e[\nabla(\overrightarrow{u} . \overrightarrow{A})-(\overrightarrow{u}.\nabla)\overrightarrow{A}]$$
$$\frac{d\overrightarrow{p}}{dt}=-e\nabla V-e\frac{\partial \overrightarrow{A}}{\partial t}+e\nabla(\overrightarrow{u}.\overrightarrow{A})-e(\overrightarrow{u}.\nabla)\overrightarrow{A}$$
$$\frac{d\overrightarrow{p}}{dt}=-e\nabla(V+\overrightarrow{u}.\overrightarrow{A})-\frac{\partial(e\overrightarrow{A})}{\partial t} -e(\overrightarrow{u}.\nabla)\overrightarrow{A}$$
[Here we know
$$ \frac{d\overrightarrow{A}}{dt} = \frac{\partial \overrightarrow{A}}{\partial t}+\frac{\partial \overrightarrow{A}}{\partial x}.\frac{\partial x}{\partial t}+\frac{\partial \overrightarrow{A}}{\partial y}.\frac{\partial y}{\partial t}+\frac{\partial \overrightarrow{A}}{\partial z}.\frac{\partial z}{\partial t}=\frac{\partial \overrightarrow{A}}{\partial t}+(\overrightarrow{u}.\nabla)\overrightarrow{A}$$]
i.e $$\frac{d(\overrightarrow{p}+e\overrightarrow{A})}{dt}=-\nabla(eV-e\overrightarrow{u}.\overrightarrow{A})$$
Now compairing above equation by standard mechanics equation
$ \frac{d\overrightarrow{p_{canonical}}}{dt}=-\nabla U$ we get
$$\overrightarrow{p_{canonical}}=\overrightarrow{p}+e\overrightarrow{A}\Rightarrow \overrightarrow{p}=\overrightarrow{p_{canonical}}-e\overrightarrow{A}$$
where $\overrightarrow{p}$ is dynamical momentum
Corresponding velocity dependent potential $$U=eV-e\overrightarrow{u}.\overrightarrow{A}$$
\section{Gauge transformation in Quantum Mechanics:}
The presence of magnetic field the Hamiltoniun becomes of an Electron
$$H=\frac{(\overrightarrow{p}-e|overrightarrow{A})^2}{2m}+eV$$
Here
$$(\overrightarrow{p}-e\overrightarrow{A})^2=(\overrightarrow{p}-e\overrightarrow{A})(\overrightarrow{p}-e\overrightarrow{A})=\overrightarrow{p}^2-e(\overrightarrow{A}.\overrightarrow{p}+\overrightarrow{p}.\overrightarrow{A})+e^2\overrightarrow{A}^2$$
This must be Hermitian ,so Hamiltoniun is also Hermitian.
If $\phi$ depend only on 't' then
Gauge transform will be
$$\overrightarrow{A}\rightarrow \overrightarrow{A}+\nabla\phi ;; V \rightarrow V-\frac{\partial \phi}{\partial t} ;
|\alpha,t_0;t_0> \rightarrow \widetilde|\alpha,t_0;t_0>.$$
We beleve that it is reasonable to demand that the expectation values in quantum mechanics behave in a manner similar to the corresponding classical quantities under gauge transformation so $$<\overrightarrow{x}> and <\overrightarrow{p}>$$
is expected are not change under gauge transformation, where as $$<\overrightarrow{p}_\text{canonical}$ isexpected to into change.
<\overrightarrow{x}>=<\alpha|\overrightarrow{x}|\alpha>=\widetilde{<\alpha}|\overrightarrow{x}|\widetilde{\alpha}>$$ and $$ <\overrightarrow{x}>=<\alpha|\overrightarrow{p}|\alpha>=\widetilde{<\alpha}|\overrightarrow{p}|\widetilde{\alpha}>$$
Let introducing a operator $$\widetilde{|\alpha>}=\mathscr{G}|\alpha>$$
So then
$$<\overrightarrow{x}>=<\alpha|\overrightarrow{x}|\alpha>=<\widetilde{\alpha}|\overrightarrow{x}|\widetilde{\alpha}>=<\alpha| \mathscr{G^\dagger}\overrightarrow{x}\mathscr{G}|\alpha>$$
i.e $$\overrightarrow{x}=\mathscr{G^\dagger}\overrightarrow{x}\mathscr{G}$$
and similarly $$\overrightarrow{p}=\mathscr{G^\dagger}\overrightarrow{p}\mathscr{G}$$
The mechanical or dynamical momentum ($$\overrightarrow{p}=\overrightarrow{p_\text{canonical}}-e\overrightarrow{A}$) also transform into $\widetilde{\overrightarrow{p}}$$
i.e $$\overrightarrow{p} \rightarrow \widetilde{\overrightarrow{p}}$$
$$\Rightarrow \overrightarrow{p_{canonical}}-e\overrightarrow{A}\rightarrow\overrightarrow{p_{canonical}}-e\widetilde{\overrightarrow{A}}=\overrightarrow{p_{canonical}}-e\overrightarrow{A}-e\nabla\phi$$
If the expectation values are not changing in this two formulation then we can write
$$<\alpha|\overrightarrow{p}-e\overrightarrow{A}|\alpha>=<\alpha|\mathscr{g}(\overrightarrow{p_{canonical}}-e\overrightarrow{A}-e\nabla\phi)\mathscr{g}|\alpha>$$
$$\Rightarrow \overrightarrow{p}-e\overrightarrow{A}= \mathscr{g}(\overrightarrow{p_{canonical}}-e\overrightarrow{A}-e\nabla\phi)\mathscr{g}$$
