# Coupling of electrc field with electric dipole moment in the hamiltonian

I'm studying linear response theory from "Fundamentals of many body physics" by Nolting. We want to study electrical AC conductivity, so what happens when we apply a spatially homogeneous oscillating electric field to a system of charged particles.

Nolting starts with the assumption that the presence of the electric field $$\vec E(t)$$ causes an additional term in the hamiltonian given by $$V(t)=-\vec P\cdot\vec E(t)$$ where $$\vec P$$ is the electric dipole moment operator $$\vec P=\sum_iq_i\hat{\vec r_i}$$ where $$q_i$$ and $$\hat{\vec r_i}$$ are the charge and the position operator of the i-th particle.

I know in classical physics that this is the form of the interaction energy of an electric dipole with an external electric field, but I don't understand how to get this in quantum mechanics.

If the hamiltonian of the i-th particle is $$H_i=\dfrac{1}{2m}\left( \vec p_i-q_i\vec A(\vec r_i,t) \right)^2$$ to have an alternating omogeneous electric field we can choose the potential $$\phi=\text{electrostatic potential energy between the particles}$$ $$\vec A(\vec r_i,t)=\vec A_0\cos(\omega t)$$ if we expand we have, assuming all $$q_i=e$$ $$H_i=\dfrac{p_i^2}{2m}+\dfrac{e^2}{2m}A(\vec r_i)^2-\dfrac{e}{2m}\vec p_i\cdot\vec A(\vec r_i,t)-\dfrac{e}{m}\vec A(\vec r_i,t)\cdot \vec p_i$$

where I considered the fact that momentum and vector potential do not commute. The third term is zero if the vector potential is homogeneous and maybe we could neglect the second term if we are considering a small electric field, but the remaining term does not look like the one Nolting introduces. Here I find that the vector potential couples with the total momentum of the system, and not that the electric field couples with the positions.

How can we arrive to additional term in the hamiltonian given by $$V(t)=-\vec P\cdot\vec E(t)$$??

But the general Hamiltonian for a charged particle in external EM field isn't just kinetic energy, it is kinetic energy plus potential energy:

$$H = \frac{(\mathbf p-q\mathbf A)^2}{2m} + q\phi.$$

You can derive the dipole interaction Hamiltonian by neglecting the effect of $$\mathbf A$$, and approximating electric potential, in a small enough region where $$\mathbf E$$ is constant, by a linear function of coordinates:

$$H = q ( - \mathbf r \cdot \mathbf E ).$$

• Yeah, so it's just using another gauge, because both my 4-potential and yours give the same electric field. I wonder now, since classical QM is not a relativistic theory, if the results are gauge dependent. Because if I start with a different gauge I get a different Hamiltonian, so I'm not sure. Commented Jan 14, 2023 at 8:39
• $|\psi|$ does not depend on choice of $\mathbf A$. Other things such as $\langle \phi_k | \psi \rangle$ may depend on that choice, depending on how $\phi_k$ are defined. Commented Jan 14, 2023 at 15:55