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

  • 3
    $\begingroup$ 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. $\endgroup$
    – knzhou
    Dec 17, 2019 at 5:54
  • $\begingroup$ Hi EverydayFoolish. Echoing @knzhou's comment, do you know the derivation of the classical Hamiltonian? $\endgroup$
    – Qmechanic
    Dec 17, 2019 at 8:57
  • $\begingroup$ 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. $\endgroup$
    – Fracton
    Dec 17, 2019 at 13:55
  • $\begingroup$ 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! $\endgroup$ Dec 17, 2019 at 22:39

1 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 $L$: $$H = \boldsymbol{p} \cdot \boldsymbol{v} - L = \frac{1}{2m}\left(m \boldsymbol{v} - \frac{q}{c} \boldsymbol{A}\right)^2 + q \phi \quad \Leftrightarrow \quad \hat{H} = \frac{1}{2m}\left(\hat{p} - \frac{q}{c} A_j\right)^2 + q \phi.$$

  • 1
    $\begingroup$ Brilliant answer! :D Helped me decipher page 23 of Goldstein's classical book! $\endgroup$
    – Lopey Tall
    Apr 11, 2020 at 14:23
  • 1
    $\begingroup$ Can you expound ion the final paragraph however? I'm struggling with the introduction of quantum here... why is this justified? $\endgroup$
    – Lopey Tall
    Apr 15, 2020 at 20:20
  • 1
    $\begingroup$ 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 $[x^i, p_j] = i \hbar \delta^i_j$. $\endgroup$
    – Tobi7
    Apr 17, 2020 at 12:43
  • $\begingroup$ @Tobi7 I've rejected this edit as it is too drastic, but you should consider incorporating it. $\endgroup$ Sep 24, 2020 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.