# Is the Hamiltonian some sort of connection/gauge field?

I'm not sure if this is a well-defined question, but I was just looking through some old notes and noticed that the Hamiltonian in usual QM has a similar transformation as gauge fields in QFT: under time-dependent unitary transformations $$|\psi'\rangle = U(t) |\psi\rangle$$ the Hamiltonian transforms as \begin{align*} H' = U H U^\dagger + i (\partial_tU) U^\dagger \end{align*} Similarly under the gauge transformation $$\phi'(x) = U(x) \phi(x)$$ a non-abelian gauge field transforms as \begin{align*} A_\mu' = U A_\mu U^\dagger + \frac{i}{g}(\partial_\mu U) U^\dagger \end{align*} Is this just a coincidence or is there some deeper underlying reasons why these two objects should transform in the same way?

• The key word for this concept is "Berry connection" Commented Mar 3, 2023 at 23:15

the Hamiltonian transforms as \begin{align*} H' = U H U^\dagger + i (\partial_tU) U^\dagger \end{align*} Similarly under the gauge transformation $$\phi'(x) = U(x) \phi(x)$$ a non-abelian gauge field transforms as \begin{align*} A_\mu' = U A_\mu U^\dagger + \frac{i}{g}(\partial_\mu U) U^\dagger \end{align*}

Is this just a coincidence or is there some deeper underlying reasons why these two objects should transform in the same way?

As an initial (somewhat pedantic) matter, I note that these two object do not transform in the same way. They transform in a similar way. That is, the symbols involved in the transformation look similar if you ignore the $$g$$ and the $$\mu$$s.

The vector potential is a spacetime four-vector: $$A^\mu = (A^0, \vec A)\;,$$ where $$A^0\equiv \Phi$$ is the electrostatic potential (or "colorstatic" potential--or whatever--for your non-abelian field). This electrostatic potential transforms, per your equations, as: $$\Phi \to U\Phi U^\dagger + \frac{i}{g}\frac{\partial U}{\partial t}U^\dagger\;.$$

Therefore, assuming your $$g$$ is the charge, the potential energy part transforms as: $$g\Phi \to Ug\Phi U^\dagger + i\frac{\partial U}{\partial t}U^\dagger\;,$$ which is the same transformation as the Hamiltonian.

This is not unexpected since both are the zeroth component of a spacetime vector: $$P^\mu = (H, \vec P)$$ and: $$A^\mu = (A^0, \vec A)\;.$$

Further, in the Coulomb gauge, we know that the Hamiltonian looks like: $$H = \ldots + g\Phi\;,$$ so we might expect $$H$$ and $$\Phi$$ to transform similarly.