Categories: Optics, Physics, Quantum mechanics.

# Electric dipole approximation

Suppose that an electromagnetic wave is travelling through an atom, and affecting the electrons. The general Hamiltonian of an electron in such a wave is given by:

\begin{aligned} \hat{H} &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi \\ &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi \end{aligned}

With charge $$q = - e$$, canonical momentum operator $$\vu{P} = - i \hbar \nabla$$, and magnetic vector potential $$\vb{A}(\vb{x}, t)$$. We reduce this by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$, so that $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$:

\begin{aligned} \comm*{\vb{A}}{\vu{P}} \psi &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi) \\ &= i \hbar (\nabla \cdot \vb{A}) \psi = 0 \end{aligned}

Where $$\psi$$ is an arbitrary test function. Assuming $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible, we split $$\hat{H}$$ as follows, where $$\hat{H}_1$$ can be regarded as a perturbation to $$\hat{H}_0$$:

\begin{aligned} \hat{H} = \hat{H}_0 + \hat{H}_1 \qquad \quad \hat{H}_0 \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi \qquad \quad \hat{H}_1 \equiv - \frac{q}{m} \vu{P} \cdot \vb{A} \end{aligned}

In an electromagnetic wave, $$\vb{A}$$ is oscillating sinusoidally in time and space:

\begin{aligned} \vb{A}(\vb{x}, t) = \vb{A}_0 \sin\!(\vb{k} \cdot \vb{x} - \omega t) \end{aligned}

Mathematically, it is more convenient to represent this with a complex exponential, whose real part should be taken at the end of the calculation:

\begin{aligned} \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp\!(i \vb{k} \cdot \vb{x} - i \omega t) \end{aligned}

The corresponding perturbative electric field $$\vb{E}$$ is then given by:

\begin{aligned} \vb{E}(\vb{x}, t) = - \pdv{\vb{A}}{t} = \vb{E}_0 \exp\!(i \vb{k} \cdot \vb{x} - i \omega t) \end{aligned}

Where $$\vb{E}_0 = \omega \vb{A}_0$$. Let us restrict ourselves to visible light, whose wavelength $$2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$$. Meanwhile, an atomic orbital is several Bohr $$\sim 10^{-10} \:\mathrm{m}$$, so $$\vb{k} \cdot \vb{x}$$ is negligible:

\begin{aligned} \boxed{ \vb{E}(\vb{x}, t) \approx \vb{E}_0 \exp\!(- i \omega t) } \end{aligned}

This is the electric dipole approximation: we ignore all spatial variation of $$\vb{E}$$, and only consider its temporal oscillation. Also, since we have not used the word “photon”, we are implicitly treating the radiation classically, and the electron quantum-mechanically.

Next, we want to rewrite $$\hat{H}_1$$ to use the electric field $$\vb{E}$$ instead of the potential $$\vb{A}$$. To do so, we use that $$\vu{P} = m \: \dv*{\vu{x}}{t}$$ and evaluate this in the interaction picture:

\begin{aligned} \vu{P} = m \dv*{\vu{x}}{t} = m \frac{i}{\hbar} \comm*{\hat{H}_0}{\vu{x}} = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0) \end{aligned}

Taking the off-diagonal inner product with the two-level system’s states $$\ket{1}$$ and $$\ket{2}$$ gives:

\begin{aligned} \matrixel{2}{\vu{P}}{1} = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1} = m i \omega_0 \matrixel{2}{\vu{x}}{1} \end{aligned}

Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$, where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap, close to which we assume that $$\vb{A}$$ and $$\vb{E}$$ are oscillating, i.e. $$\omega \approx \omega_0$$. We thus get:

\begin{aligned} \hat{H}_1(t) &= - \frac{q}{m} \vu{P} \cdot \vb{A} = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp\!(- i \omega t) \\ &\approx - q \vu{x} \cdot \vb{E}_0 \exp\!(- i \omega t) = - \vu{d} \cdot \vb{E}_0 \exp\!(- i \omega t) \end{aligned}

Where $$\vu{d} \equiv q \vu{x} = - e \vu{x}$$ is the transition dipole moment operator of the electron, hence the name electric dipole approximation. Finally, we take the real part, yielding:

\begin{aligned} \boxed{ \hat{H}_1(t) = - \vu{d} \cdot \vb{E}(t) = - q \vu{x} \cdot \vb{E}_0 \cos\!(\omega t) } \end{aligned}

If this approximation is too rough, $$\vb{E}$$ can always be Taylor-expanded in $$(i \vb{k} \cdot \vb{x})$$:

\begin{aligned} \vb{E}(\vb{x}, t) = \vb{E}_0 \Big( 1 + (i \vb{k} \cdot \vb{x}) + \frac{1}{2} (i \vb{k} \cdot \vb{x})^2 + \: ... \Big) \exp\!(- i \omega t) \end{aligned}

Taking the real part then yields the following series of higher-order correction terms:

\begin{aligned} \vb{E}(\vb{x}, t) = \vb{E}_0 \Big( \cos\!(\omega t) + (\vb{k} \cdot \vb{x}) \sin\!(\omega t) - \frac{1}{2} (\vb{k} \cdot \vb{x})^2 \cos\!(\omega t) + \: ... \Big) \end{aligned}

## References

1. M. Fox, Optical properties of solids, 2nd edition, Oxford.
2. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.