Given external radiation, produced from an electric field travelling from the left and an electric field travelling from the right, producing a standing wave $$E_0e^{i(kx - \omega t)} + E_0e^{-i(kx - \omega t)},$$the Stark energy shift of a two level system under the influence of this electric field is $$\Delta E(x) = \hbar \Delta \omega_0 (x) = \frac{\hbar \Omega^{2}(x)}{4 \delta} = \hbar \bigg( \frac{\vec{d} \cdot \vec{E}_0}{ \hbar} \bigg)^2/ 4 \delta,$$ where $\Omega$ is the Rabi frequency and $\delta$ is the detuning $\omega_0 - \omega$.

How does it follow that the force on the particle depends on the gradient of the electric field at the position of the particle?

  • $\begingroup$ You have both of your fields travelling to the positive $x$; the direction of propagation comes from the relative phase between $kx$ and $\omega t$, i.e. your exponentials should be $~i(kx - \omega t)$ for the + travelling wave and $i(kx+\omega t)$ for then - travelling wave $\endgroup$ Commented Apr 22, 2017 at 19:10
  • 1
    $\begingroup$ @JohnDoe In the context of what Emilio already said in his answer, you may want to consider the Feynman-Hellmann theorem, en.wikipedia.org/wiki/Hellmann–Feynman_theorem, with the position of the system CoM as parameter. It's standard procedure for forces on "slow coordinates" in the Born-Oppenheimer approximation. $\endgroup$
    – udrv
    Commented Apr 23, 2017 at 15:15

1 Answer 1


Simply put, force is always the spatial gradient of energy. That's what $$ \mathrm dU=-\mathbf F\cdot\mathrm d\mathbf r $$ really means: if you have an energy that depends on position, then you have a force that points in the direction of lower energy.

In this specific case it looks a bit mysterious, because you've been speaking about internal degrees of freedom all the while, and then suddenly boom, there's a force acting on the centre of mass, so, say what?

The way you deal with it is with manipulations that are at heart identical to the ones in How does one account for the momentum of an absorbed photon?. First off, you do a Born-Oppenheimer-style separation, where you assume the internal dynamics are much faster than the center-of-mass movement (which is plenty reasonable), and therefore that you can solve for the internal dynamics at some fixed centre-of-mass position $\mathbf r$ and worry about its dynamics later. When you do get around to worrying about those dynamics, though, what you have is a Schrödinger equation with hamiltonian $$ H_\mathrm{COM} = -\frac{\hbar^2}{2m}\nabla_\mathbf{r}^2 \pm \frac12 \Delta E(\mathbf r), $$ where $\Delta E(\mathbf r)$ is the energy shift you state in the question (with things like up for one state and down for another state or whatever) and it is exactly analogous to a Born-Oppenheimer potential-energy surface.

Finally, once you've identified the relevant hamiltonian with the relevant potential, you might decide (but then again you might not) that the quantized motion of the center of mass is not really a good description and that that degree of freedom really lies in the classical limit. That gets rid of all the quantumsy stuff, but you keep the potential (because it's the classical limit of the dynamics with that potential) and therefore you start talking about the force that points along its gradient.


Your Answer

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