Categories: Electromagnetism, Optics, Perturbation, 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 an electromagnetic field is:

H^=(P^qA)22m+qΦ=P^22mq2m(AP^+P^A)+q2A22m+qΦ\begin{aligned} \hat{H} &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \Phi \\ &= \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 \Phi \end{aligned}

Where q<0q < 0 is the electron’s charge, P^=i\vu{P} = - i \hbar \nabla is the canonical momentum operator, A\vb{A} is the magnetic vector potential, and Φ\Phi is the electric scalar potential. We start by fixing the Coulomb gauge A=0\nabla \cdot \vb{A} = 0 such that P^\vu{P} and A\vb{A} commute; let ψ\psi be an arbitrary test function:

[A,P^]ψ=(AP^P^A)ψ=iA(ψ)+i(Aψ)=i(A)ψ=0\begin{aligned} \comm{\vb{A}}{\vu{P}} \psi &= (\vb{A} \cdot \vu{P} - \vu{P} \cdot \vb{A}) \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}

Meaning AP^=P^A\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}. Furthermore, we assume that A\vb{A} is so small that A2\vb{A}{}^2 is negligible, so the Hamiltonian is reduced to:

H^P^22mqmP^A+qΦ\begin{aligned} \hat{H} &\approx \frac{\vu{P}{}^2}{2 m} - \frac{q}{m} \vu{P} \cdot \vb{A} + q \Phi \end{aligned}

We now split H^\hat{H} like so, where H^1\hat{H}_1 can be regarded as a perturbation to the “base” H^0\hat{H}_0:

H^=H^0+H^1H^0P^22m+qΦH^1qmP^A\begin{aligned} \hat{H} = \hat{H}_0 + \hat{H}_1 \qquad\qquad \hat{H}_0 \equiv \frac{\vu{P}{}^2}{2 m} + q \Phi \qquad\qquad \hat{H}_1 \equiv - \frac{q}{m} \vu{P} \cdot \vb{A} \end{aligned}

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

A(x,t)=A0sin(kxωt)\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:

A(x,t)=iA0exp(ikxiωt)\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 E\vb{E} is then given by:

E(x,t)=At=E0exp(ikxiωt)\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 E0=ωA0\vb{E}_0 = \omega \vb{A}_0. Light in and around the visible spectrum has a wavelength 2π/k107m2 \pi / |\vb{k}| \sim 10^{-7} \:\mathrm{m}, while an atomic orbital is several Bohr radii 1010m\sim 10^{-10} \:\mathrm{m}, so kx\vb{k} \cdot \vb{x} is very small. Therefore:

E(x,t)E0exp(iωt)\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 E\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 H^1\hat{H}_1 to use the electric field E\vb{E} instead of the potential A\vb{A}. To do so, we use that momentum P^mdx^/dt\vu{P} \equiv m \: \idv{\vu{x}}{t} and evaluate this in the interaction picture:

P^=mdx^dt=mi[H^0,x^]\begin{aligned} \vu{P} &= m \dv{\vu{x}}{t} = m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}} \end{aligned}

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

2P^1=mi2H^0x^x^H^01=miω02x^1\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}

Where ω0(E2 ⁣ ⁣E1)/\omega_0 \equiv (E_2 \!-\! E_1) / \hbar is the resonance of the energy gap, close to which we assume that A\vb{A} and E\vb{E} are oscillating, i.e. ωω0\omega \approx \omega_0. Therefore, P^/m=iω0x^\vu{P} / m = i \omega_0 \vu{x}, so we get:

H^1(t)=qmP^A=(ii)qω0x^A0exp(iωt)d^E0exp(iωt)\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 - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t) \end{aligned}

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

H^1(t)=d^E(t)=qx^E0cos(ωt)\begin{aligned} \boxed{ \begin{aligned} \hat{H}_1(t) &= - \vu{d} \cdot \vb{E}(t) \\ &= - q \vu{x} \cdot \vb{E}_0 \cos(\omega t) \end{aligned} } \end{aligned}

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

E(x,t)=E0(1+(ikx)+12(ikx)2+...)exp(iωt)\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:

E(x,t)=E0(cos(ωt)+(kx)sin(ωt)12(kx)2cos(ωt)+...)\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}


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