Gauge invariance of a magnetic Schrödinger operator Good morning,
I am studying the properties of the magnetic Schrödinger operator
$$
\mathcal{L}_A = \left( -\mathrm{i} \nabla -A \right)^2 = \left( -\mathrm{i} \nabla -A \right)^\dagger \left( -\mathrm{i} \nabla -A \right),
$$
where $A \colon \mathbb{R}^3 \to \mathbb{R}^3$ is a (smooth) vector potential corresponding to the magnetic field $B = \operatorname{curl}A$. I have a (probably) trivial question: let us assume that $B$ is constant (or homogeneous, as physicists say), i.e. it is a constant field on $\mathbb{R}^3$. May I say that, up to a gauge tranformation $A \mapsto A +\nabla \chi$, I can always assume that $A=0$?
 A: Let us turn this around:


*

*Assume that there is a gauge choice where the magnetic vector potential $\vec{A}=\vec{0}$ vanishes. 

*Then after a possible gauge tranformation, the  magnetic vector potential $\vec{A}=\vec{\nabla}\chi$ is a gradient field.

*But the curl of a gradient vanishes identically $\vec{\nabla}\times\vec{\nabla}\equiv \vec{0} $.

*So the magnetic field $\vec{B}=\vec{\nabla}\times\vec{A}=0$ vanishes.
A: A magnetic field is a gauge-invariant quantity (it's physical) which means that you cannot change it by performing a gauge transformation. The gauge invariance of $B$ follows from
$$A\to A + \nabla\chi \implies \nabla\times A \to \nabla\times A + \nabla\times \nabla \chi \implies B\to B$$
since $\nabla\times \nabla \equiv 0$. Now if we had $A=0$ in one gauge then we would have $B=\nabla\times A = 0$ in all gauges which is a contradiction.
What you can always assume is that $\nabla\cdot A = 0$; this is the so-called Coloumb gauge. Other choices that people sometimes use are: ${\bf n}\cdot A = 0$ where ${\bf n}$ is any fixed vector (axial gauge) and ${\bf r}\cdot A = 0$ where ${\bf r}=(x,y,z)$ (Poincare's gauge).
