How do we derive the minimal coupling Hamiltonian?

Question

Is there a way to rigorously derive the minimal coupling hamiltonian for a system interacting with electromagnetic radiation. How de we arrive at the expression: $$\hat{H} = \frac{1}{2m}(p-\frac{q}{c}A)^2 + q.\phi(r)$$

Answer 1

To derive the minimal coupling Hamiltonian, one starts by calculating the Lagrangian $L$ of a particle in an electromagnetic field. The acting force is the Lorentz force $\boldsymbol{F} = q(\boldsymbol{E} + \boldsymbol{v}/c \times \boldsymbol{B})$. Since $\boldsymbol{F}$ depends on velocity, we have to find a generalized potential $U$ that satisfies the following equation: $$F_j = -\frac{\partial U}{\partial x_j} + \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial U}{\partial \dot{x}_j}\right).$$ From Maxwell's equations $$\boldsymbol{B} = \nabla \times \boldsymbol{A}, \qquad \nabla \times \boldsymbol{E} = - \frac{1}{c}\frac{\partial \boldsymbol{B}}{\partial t},$$ we get $$\boldsymbol{E} = -\nabla \phi - \frac{1}{c} \frac{\partial \boldsymbol{A}}{\partial t}.$$ Now by plugging our results into the Lorentz force equation and doing some vector calculus, we end up with $$\boldsymbol{F} = q\left(-\nabla \phi - \frac{1}{c}\left(\nabla (\boldsymbol{v}\cdot \boldsymbol{A}) - \frac{\mathrm{d}\boldsymbol{A}}{\mathrm{d}t}\right)\right).$$ (Here we make use of the fact that $\boldsymbol{v}\times (\nabla \times\boldsymbol{A}) = \nabla(\boldsymbol{v}.\boldsymbol{A})-(\boldsymbol{v}.\nabla)\boldsymbol{A} $ and $d_{t}\boldsymbol{A} = (\boldsymbol{v}.\nabla)\boldsymbol{A}+\partial_{t}\boldsymbol{A}$.)

To obtain the generalized potential $U$, a final observation is needed, namely $$\frac{\mathrm{d}\boldsymbol{A}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial}{\partial \boldsymbol{v}}(\boldsymbol{v}\cdot \boldsymbol{A} - q \phi)\right),$$ which is true since the electrostatic potential $\phi$ does not depend on the velocity. Comparing $$\boldsymbol{F} = -\nabla \left(q\phi - \frac{q}{c}(\boldsymbol{v}\cdot \boldsymbol{A})\right) + \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial}{\partial \boldsymbol{v}}(\boldsymbol{v}\cdot \boldsymbol{A} - q \phi)\right)$$ with the equation for the generalized potential, we get $U = q\phi - q/c(\boldsymbol{v}\cdot \boldsymbol{A})$, which allows us to write down the Lagrangian $L = T - V = 1 / 2 m \boldsymbol{v}\cdot \boldsymbol{v} - U$.

To derive the minimal coupling Hamiltonian, you have to transform the kinetic momentum (classically this would be $m \boldsymbol{v}$) to the canonical momentum $\boldsymbol{p} = \partial L / \partial \boldsymbol{v} = m \boldsymbol{v} + q / c \boldsymbol{A}$. In quantum mechanics, the kinetic momentum corresponds to the momentum operator $\hat{p}$, so the canonical momentum operator becomes $\hat{p} - q/c A_j$. The Hamiltonian may be obtained by performing the Legendre transformation on $L$: $$H = \boldsymbol{p} \cdot \boldsymbol{v} - L = \frac{1}{2m}\left(m \boldsymbol{v} - \frac{q}{c} \boldsymbol{A}\right)^2 + q \phi \quad \Leftrightarrow \quad \hat{H} = \frac{1}{2m}\left(\hat{p} - \frac{q}{c} A_j\right)^2 + q \phi.$$