# How do we derive the minimal coupling Hamiltonian?

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: $$\hat{H} = \frac{1}{2m}(p-\frac{q}{c}A)^2 + q.\phi(r)$$

• What do you want to derive it from? I can think of a couple ways, but if this is your first exposure to the subject, all of them are going to look far more technical and not any more satisfying. – knzhou Dec 17 '19 at 5:54
• Hi EverydayFoolish. Echoing @knzhou's comment, do you know the derivation of the classical Hamiltonian? – Qmechanic Dec 17 '19 at 8:57
• I am familiar with some of the basics of quantum dynamics from Sakurai, I've seen that minimum coupling hamiltonian is central to the description of matter interacting with E-M field. I was wondering start to read about fundamental aspects about this and also derive the minimal coupling hamiltonian in the current form. – EverydayFoolish Dec 17 '19 at 13:55
• But ... your minimal coupling hamiltonian written is not even complete. Non-minimal coupling Pauli moment terms, over and above this, describe the interactions of neutral neutrons with magnetic fields! – Cosmas Zachos Dec 17 '19 at 22:39

To derive the minimal coupling Hamiltonian, someone 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 $$\boldsymbol{F} = q\left(-\nabla \phi - \frac{1}{c}\left(\nabla (\boldsymbol{v}\cdot \boldsymbol{A}) - \frac{\mathrm{d}\boldsymbol{A}}{\mathrm{d}t}\right)\right).$$ To obain 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 $$L$$: $$H = \boldsymbol{p} \cdot \boldsymbol{v} - L = \frac{1}{2m}\left(m \boldsymbol{v} - \frac{q}{c} \boldsymbol{A}\right)^2 - q \phi \quad \Leftrightarrow \quad \hat{H} = \frac{1}{2m}\left(\hat{p} - \frac{q}{c} A_j\right)^2 - q \phi.$$
• The last part involves performing canonical quantization, that is elevating the conjugate variables $x^i$ and $p_i$ to operators which satisfy the canonical commutator relation $[x^i, p_j] = i \hbar \delta^i_j$. – Tobi7 Apr 17 at 12:43