A naive question on the $U(1)$ gauge transformation of electromagnetic field? For simplicity, in the following we set the electric charge $e=1$ and consider a lattice spinless free electron system in an external static magnetic field $\mathbf{B}=\nabla\times\mathbf{A}$ described by the Hamiltonian $H=\sum_{ij}t_{ij}c_i^\dagger c_j$, where $t_{ij}=\left | t_{ij} \right |e^{iA_{ij}}$ with the corresponding lattice gauge-field $A_{ij}$. As we know the transformation $\mathbf{A}\rightarrow \mathbf{A}+\nabla\theta$ does not change the physical magnetic field $\mathbf{B}$, and the induced transformation in Hamiltonian reads $$H\rightarrow H'=\sum_{ij}t_{ij}'c_i^\dagger c_j$$ with $t_{ij}'=e^{i\theta_i}t_{ij}e^{-i\theta_j}$. Now my confusion point is:
Do these two Hamiltonians $H$ and $H'$ describe the same physics? Or do they describe some same quantum states? Or what common physical properties do they share?
I just know $H$ and $H'$ have the same spectrum, thank you very much.
 A: For constant $t_{ij}$, the transformation may be considered as a simple redifinition of the quantum state basis.
A natural basis for you quantum states are the $|\psi_j \rangle = c_j^+|0\rangle$. In this basis, you have : $H|\psi_j \rangle = t_{ij}|\psi_i \rangle$, so this means that $H_{ij}=t_{ij}$, so we may write $H = \sum H_{ij}~ c^+_i c_j $.
Now, we may decide to change the basis $|\psi' \rangle = U |\psi \rangle$, with $U = Diag (e^{i\theta_1},e^{i\theta_2}, ....e^{i\theta_n})$, so that $|\psi_j \rangle \to |\psi'_j \rangle = e^{i\theta_j} |\psi_j \rangle$. The matrix $U$ is unitary, and it transforms an orthonormal basis into an other orthonormal basis.
In this new basis, the hamiltonian is simply $H' = U H U^{-1}$, or expressing the elements of the operator $H'$, we get  : $H'_{ij} = e^{i\theta_i} H_{ij}e^{-i\theta_j} $
As you know, multiplying a quantum basis state  $|\psi_j \rangle$ by a unit phase $e^{i\theta_j}$ does not change the physical state (which is $|\psi_j \rangle \langle|\psi_j| $), so the physics described by $H$ and $H'$ is the same, the eigenvalues $E_k$ of $H$ and $H'$ are the same, etc... 
