# 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.

• If you do the transformation $c_i^{(\dagger)}\to e^{-(+)i\theta_i}c_i^{(\dagger)}$, you see that $H'\to H$, and the model is indeed gauge invariant. The physics is thus the same. – Adam Oct 9 '13 at 0:37
• @ Adam Yes,you're right. But why we do this transformation? Can it be deduced from the underlying microscopic model or it's just the requirement for gauge invariant to describe the same physics? – Kai Li Oct 9 '13 at 8:33
• In the underlying model, it corresponds to the transformation $A_i(x)\to A_i(x)+\partial_i \theta(x)$ and $\hat\psi(x)\to e^{i\theta(x)}\hat\psi(x)$ (with maybe some minus signs) in the microscopic Hamiltonian. Or if you take your lattice Hamiltonian as "fundamental" (as in lattice QCD), you can show that it reproduces the usual continuous Hamiltonian in the limit of vanishing lattice spacing. So maybe you are asking what does the U(1) gauge transformation means in quantum mechanics ? – Adam Oct 9 '13 at 14:20
• A system is not only determined by its Hamiltonian, but also by its Hilbert space. U(1) gauge theory can come from constraint of your Hilbert space. – Shenghan Jiang Oct 16 '13 at 21:09
• Remarks: $e^{i\theta _i \hat{n}_i}c_j^\dagger e^{-i\theta _i \hat{n}_i}=\delta _{ij}e^{i\theta _i} c_j^\dagger +(1-\delta _{ij})c_j^\dagger$, and hence $H'=UHU^{-1}$ with $U=\prod _ie^{i\theta _i\hat{n}_i}$. – Kai Li Nov 27 '13 at 13:38

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...

• @ Trimok Thanks for your comment. By the way, I think only the global phase is un physical, while the relative phase(here is the local $U(1)$ transformation) is physical. – Kai Li Oct 9 '13 at 9:43
• To be fair, I consider only global transformations in my answer (with constants $t_{ij}$), so I realize it does not answer the question... The local $U(1)$ invariance is not really "physical". The $U(1)$ invariance of the EM field is a mathematical redundancy, not a true symmetry. This means that we overcount the total number of states, that is that one physical state is represented by many mathematical states, and it is necessary to reduce this number, to keep only one representant for one physical state. – Trimok Oct 9 '13 at 10:07
• Is the notation $|\psi_j \rangle \langle \psi_j|$ in your last paragraph the same as in the expression of density operator in statistical mechanics? I don't know how does it represent the physical state? Thank you very much. – Kai Li Jan 20 '15 at 15:24
• Another naive problem: The local phase may affect the superposition state, e.g., $\psi+e^{i\theta}\phi$. Thus from this point, the local phase seems "physical"? – Kai Li Jan 20 '15 at 15:31