Symmetries of a Uniform Magnetic Field Simple question.  A system with a uniform electric field everywhere in space has translational invariance in the directions perpendicular to the electric field but no translational invariance parallel to it.  This system also has rotational invariance in the plane perpendicular to the electric field.
What about a uniform magnetic field?  Judging from the Hamiltonian $(\vec{p}-q\vec{A})^2/2m$, it would seem that the symmetries depend on our choice of gauge.  Choosing a different $\vec{A}$ breaks different symmetries.  Is there a most symmetric choice for $\vec{A}$?
 A: If your system consists of (a) a uniform electric field along direction $\hat{\mathbf{n}}$, or (b) a uniform magnetic field along direction $\hat{\mathbf{n}}$, then its symmetries are


*

*translation orthogonal to $\hat{\mathbf{n}}$,

*translation parallel to $\hat{\mathbf{n}}$, and

*rotations about $\hat{\mathbf{n}}$.


Some of these transformations are equivalent to a gauge transformation. However, gauge transformations are still symmetries of the system, as they respect the physics. More specifically, the motion of a charged particle in those fields is preserved by the symmetries listed above, and it is the motion that matters.
This is somewhat clearer if one spells it out explicitly. Consider a particle of charge $q$ under the action of the electric and magnetic fields $\mathbf E=E_0\hat{\mathbf{z}}$ and $\mathbf B=B_0\hat{\mathbf{z}}$. The hamiltonian for the system can always be written as
$$
H=\frac1{2m}\left(\mathbf p-q\mathbf A\right)^2+\varphi,
$$
where the potentials $\mathbf A=\mathbf A(x,y,t)$ and $\varphi=\varphi(z,t)$ must satisfy
$$
\nabla\times\mathbf A=\mathbf B=B_0\hat{\mathbf{z}}
\quad\text{and}\quad
-\frac{\partial\mathbf A}{\partial t}-\nabla \varphi=\mathbf E=E_0\hat{\mathbf{z}}.
$$
Sample potentials that satisfy this are $\varphi_0=E_0z$ and $\mathbf A_0=\tfrac12 B_0(x\hat{\mathbf{y}}-y\hat{\mathbf{x}})$, and all other acceptable potentials can be obtained from these via
$$
\mathbf A=\mathbf A_0+\nabla\chi
\quad\text{and}\quad
\varphi=\varphi_0-\frac{\partial\chi}{\partial t}
$$
for a suitable function $\chi=\chi(x,y,z,t)$, which has no influence at all on the physics.
Suppose now that you transform the system with the transformation above: You translate by a vector $\mathbf a$ and you rotate about $\hat{\mathbf z}$ with a rotation matrix $R$, so you take $\mathbf r \mapsto \mathbf r'=R(\mathbf r+\mathbf a)$ and $\mathbf p\mapsto \mathbf p'=R\mathbf p$. In terms of the new variables, the old hamiltonian is
\begin{align}
H
&=
\frac1{2m}\left(\mathbf p-q\mathbf A(\mathbf r,t)\right)^2+\varphi(\mathbf r,t)
\\&=
\frac1{2m}\left(R^{-1}\mathbf p'-q\mathbf A(R^{-1}\mathbf r'-\mathbf a,t)\right)^2+\varphi(R^{-1}\mathbf r'-\mathbf a,t)
\\&=
\frac1{2m}\left(\mathbf p'-qR\mathbf A(R^{-1}\mathbf r'-\mathbf a,t)\right)^2+\varphi(R^{-1}\mathbf r'-\mathbf a,t),
\end{align}
because $R$ is orthogonal and preserves norms, so $(R\mathbf u)^2=\mathbf u^2$. This can be written in the same way as before
$$
H\mapsto H'=\frac1{2m}\left(\mathbf p'-q\mathbf A'(\mathbf r',t)\right)^2+\varphi'(\mathbf r',t),
$$
for the transformed potentials
$$
\mathbf A'(\mathbf r',t)=R\mathbf A(R^{-1}\mathbf r'-\mathbf a,t)
\quad\text{and}\quad
\varphi'(\mathbf r',t)=\varphi(R^{-1}\mathbf r'-\mathbf a,t).
$$
We now use the fact that the forms of the old potential are constrained to be gauge transforms of a standard set, so 
\begin{align}
\mathbf A'(\mathbf r',t)&=R\mathbf A_0(R^{-1}\mathbf r'-\mathbf a,t)+R\nabla\chi(R^{-1}\mathbf r'-\mathbf a,t)
\quad\text{and}\quad\\
\varphi'(\mathbf r',t)&=\varphi_0(R^{-1}\mathbf r'-\mathbf a,t)-\frac{\partial\chi}{\partial t}(R^{-1}\mathbf r'-\mathbf a,t)
,\end{align}
and the fact that we know how the standard set transforms: writing the rotation matrix explicitly as
$$
R=\begin{pmatrix}\cos\theta&\sin\theta & 0\\ -\sin(\theta)&\cos \theta & 0\\0&0&1\end{pmatrix}
$$
we can transform the potentials to
\begin{align}
R\mathbf A_0(R^{-1}\mathbf r'-\mathbf a,t)
&=
\frac{B_0}{2}
\begin{pmatrix}\cos\theta&\sin\theta & 0\\ -\sin(\theta)&\cos \theta & 0\\0&0&1\end{pmatrix}
\begin{pmatrix}
-\sin\theta x'-\cos\theta y'+a_y 
\\ 
\cos\theta x'-\sin\theta y'-a_x
 \\ 0
\end{pmatrix}
\\&=
\frac{B_0}{2}
\begin{pmatrix}
-y'+\cos\theta a_y-\sin\theta a_x
\\
x'-\sin\theta a_y-\cos\theta a_x
\\0
\end{pmatrix}
\\&=
\mathbf A_0(\mathbf r',t)+\mathbf b
\end{align}
and 
\begin{align}
\varphi_0(R^{-1}\mathbf r'-\mathbf a,t)=E_0(z'-a_z)=\varphi_0(z',t)-\phi,
\end{align}
where $\mathbf b$ and $\phi$ are constant.
This transformation is now explicitly in the form of a gauge transformation, with $\chi=\mathbf b·\mathbf r-\phi t$. As I explained before, gauge transformations do not change the physics, so this completes the proof.
A: Your hamiltonian is still gauge invariant because the canonical momentum, that is the generator of translations, is $m \dot{\vec{x}} - q \vec A$. In other words, your Hamiltonian is still $H = \frac{1}{2m}\Pi^2 - q \Phi$, where $\vec \Pi = m \vec{ \dot{x}}$ is the mechanical momentum.
