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...

This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Trimok