Consider a particle in two dimensions with an external magnetic field in the $z$-direction. The vector potential can be chosen to be $$\mathbf{A}=-By\ \hat{x}$$ so that the Hamiltonian given by $$H=\frac{(p_x+eBy)^2}{2m}+\frac{p_y^2}{2m}.$$ The system has rotational symmetry. This makes sense since you have a constant field in $z$-direction, and the rotating thing will not change anything. To see prove this, We need to prove that $$[H,L_z]=0.$$

It's easy to show with little algebra that $$[H,L_z]\not = 0.$$ The reason is that our chosen gauge breaks the rotational symmetry. If we instead choose, $$\vec{A}=(-By/2,Bx/2,0),$$ we can restore this symmetry.

Why does the symmetry of the system depend on the gauge we choose? Since, Physically, of course, Any consequence following the symmetry should be followed. Furthermore, What is the gauge in which there is a translation symmetry? Can we find the gauge which have both of these symmetries?

Consider a particle in two dimensions with an external magnetic field in the $z$-direction. The vector potential can be chosen to be $$\mathbf{A}=-By\ \hat{x}$$ so that the Hamiltonian given by $$H=\frac{(p_x+eBy)^2}{2m}+\frac{p_y^2}{2m}.$$ The system has rotational symmetry. This makes sense since you have a constant field in $z$-direction, and the rotating thing will not change anything. To see prove this, We need to prove that $$[H,L_z]=0.$$

It's easy to show with little algebra that $$[H,L_z]\not = 0.$$ The reason is that our chosen gauge breaks the rotational symmetry. If we instead choose, $$\vec{A}=(-By/2,Bx/2,0),$$ we can restore this symmetry.

Why does the symmetry of the system depend on the gauge we choose? Since, Physically, of course, Any consequence following the symmetry should be followed. Furthermore, What is the gauge in which there is a translation symmetry? Can we find the gauge which have both of these symmetries?

With a couple of hints and references, It looks like $\Pi$ is the right thing to look at since it gauges invariant but still, I'm not finding $[H,\Pi]=0$.

Consider a particle in two dimensions with an external magnetic field in the z-direction. The vector potential can be chosen to be $\mathbf{A}=-By\ \hat{x}$ so that the Hamiltonian given by $$H=\frac{(p_x+eBy)^2}{2m}+\frac{p_y^2}{2m}$$ The system has rotational symmetry. This makes sense since you have a constant field in $z$-direction, and the rotating thing will not change anything. To see prove this, We need to prove that $$[H,L_z]=0$$

It's easy to show with little algebra that $$[H,L_z]\not = 0$$ The reason is that our chosen gauge breaks the rotational symmetry. If we instead choose, $$\vec{A}=(-By/2,Bx/2,0)$$, we can restore this symmetry.

Why does the symmetry of the system depend on the gauge we choose? Since, Physically, of course, Any consequence following the symmetry should be followed. Furthermore, What is the gauge in which there is a translation symmetry? Can we find the gauge which have both of these symmetries?

Consider a particle in two dimensions with an external magnetic field in the $z$-direction. The vector potential can be chosen to be $$\mathbf{A}=-By\ \hat{x}$$ so that the Hamiltonian given by $$H=\frac{(p_x+eBy)^2}{2m}+\frac{p_y^2}{2m}.$$ The system has rotational symmetry. This makes sense since you have a constant field in $z$-direction, and the rotating thing will not change anything. To see prove this, We need to prove that $$[H,L_z]=0.$$

It's easy to show with little algebra that $$[H,L_z]\not = 0.$$ The reason is that our chosen gauge breaks the rotational symmetry. If we instead choose, $$\vec{A}=(-By/2,Bx/2,0),$$ we can restore this symmetry.

Why does the symmetry of the system depend on the gauge we choose? Since, Physically, of course, Any consequence following the symmetry should be followed. Furthermore, What is the gauge in which there is a translation symmetry? Can we find the gauge which have both of these symmetries?