# Magnetic dipole Hamiltonian from current-current interaction

In the Coulomb gauge, we can write the electromagnetic Hamiltonian as

$$$$\label{eq:em-hamiltonian}\tag{1} H_\mathrm{EM} = - \int d^3 x \, \mathbf{j}(\mathbf{x}) \cdot \mathbf{A}(\mathbf{x}) + \frac{1}{2} \int d^3 x \, d^3 x' \, \frac{\rho(\mathbf{x}) \rho(\mathbf{x}')}{4\pi \vert \mathbf{x} - \mathbf{x}' \vert},$$$$ where $$\mathbf{j}(\mathbf{x})$$ is a current density, $$\mathbf{A}(\mathbf{x})$$ is the vector current, and $$\rho(\mathbf{x})$$ is a charge density.

For a magnetic dipole in a uniform magnetic field, we can write the interaction Hamiltonian as $$\label{eq:dipole-hamiltonian}\tag{2} H_\mathrm{dipole} = - \mathbf{\mu}\cdot\mathbf{B}$$ where we can express the dipole moment $$\mathbf{\mu}$$ in terms of the current density via $$\mathbf{\mu} = \frac{1}{2} \int_{\mathbb{R}^3} d^3 x \, \mathbf{x} \times \mathbf{j}(\vec{x}).$$

How can we recover \eqref{eq:dipole-hamiltonian} from \eqref{eq:em-hamiltonian}? We know that $$\mathbf{B}(\mathbf{x}) = \mathbf{\nabla} \times \mathbf{A}(\mathbf{x})$$, and can assume that $$\mathbf{j}(\mathbf{x})$$ is localized in space.

Start with the (particular choice) of $$\mathbf{A}(\mathbf{x})$$ for a uniform field, $$\mathbf{A}(\mathbf{x}) = -\frac{1}{2} \mathbf{x} \times \mathbf{B},$$ and set $$\rho(\mathbf{x}) = 0$$. Then $$H_\mathrm{EM}$$ reduces to $$H_\mathrm{EM} = \frac{1}{2} \int d^3 x \, \mathbf{j}(\mathbf{x}) \cdot \big( \mathbf{x} \times \mathbf{B} \big).$$ Using the scalar triple-product identity $$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = - (\mathbf{b} \times \mathbf{a}) \cdot \mathbf{c}$$, we can rewrite $$H_\mathrm{EM}$$ as $$H_\mathrm{EM} = -\frac{1}{2} \int d^3 x \, \big( \mathbf{x} \times \mathbf{j}(\mathbf{x}) \big) \cdot \mathbf{B}.$$ Because $$\mathbf{B}$$ is independent of $$\mathbf{x}$$, we can pull it out of the integration, writing $$H_\mathrm{EM} = -\left(\frac{1}{2} \int d^3 x \, \mathbf{x} \times \mathbf{j}(\mathbf{x}) \right) \cdot \mathbf{B}.$$ We then recognize the term inside the brackets as $$\mathbf{\mu}$$, and recover $$H_\mathrm{EM} = - \mathbf{\mu} \cdot \mathbf{B} = H_\mathrm{dipole}.$$