Beyond the Electric Dipole Approximation

Some atomic states have no lower energy state that satisfies the E1 selection rules to decay to. Then, higher order processes must be considered. The next order term in the expansion of \bgroup\color{black}$e^{-i\vec{k}\cdot\vec{r}}=1-i\vec{k}\cdot\vec{r}+...$\egroup will allow other transitions to take place but at lower rates. We will attempt to understand the selection rules when we include the \bgroup\color{black}$i\vec{k}\cdot\vec{r}$\egroup term.

The matrix element is proportional to \bgroup\color{black}$-i\langle\phi_n\vert(\vec{k}\cdot\vec{r})(\hat{\epsilon}^{(\lambda)}\cdot\vec{p}_e)\vert\phi_i\rangle$\egroup which we will split up into two terms. You might ask why split it. The reason is that we will essentially be computing matrix elements of at tensor and dotting it into two vectors that do not depend on the atomic state.

\begin{displaymath}\bgroup\color{black} \vec{k}\cdot\langle\phi_n\vert\vec{r}\vec{p}_e)\vert\phi_i\rangle\cdot\hat{\epsilon}^{(\lambda)} \egroup\end{displaymath}

Putting these two vectors together is like adding to \bgroup\color{black}$\ell=1$\egroup states. We can get total angular momentum quantum numbers 2, 1, and 0. Each vector has three components. The direct product tensor has 9. Its another case of

\begin{displaymath}\bgroup\color{black} 3\otimes 3=5_S \oplus 3_A \oplus 1_S. \egroup\end{displaymath}

The tensor we make when we just multiply two vectors together can be reduced into three irreducible (spherical) tensors. These are the ones for which we can use the Wigner-Eckart theorem to derive selection rules. Under rotations of the coordinate axes, the rotation matrix for the 9 component Cartesian tensor will be block diagonal. It can be reduced into three spherical tensors. Under rotations the 5 component (traceless) symmetric tensor will always rotate into another 5 component symmetric tensor. The 3 component anti symmetric tensor will rotate into another antisymmetric tensor and the part proportional to the identity will rotate into the identity.


The first term is symmetric and the second anti-symmetric by construction.

The first term can be rewritten.

{1\over 2}\langle\phi_n\vert[(\vec{k}\cdot\vec{r})(\hat{\epsil...
...c{r}\vec{r}\vert\phi_i\rangle\cdot\hat{\epsilon}^{(\lambda)} \\

This makes the symmetry clear. Its normal to remove the trace of the tensor: \bgroup\color{black}$\vec{r}\vec{r}\rightarrow\vec{r}\vec{r}-{\delta_{ij}\over 3}r^2$\egroup. The term proportional to \bgroup\color{black}$\delta_{ij}$\egroup gives zero because \bgroup\color{black}$\vec{k}\cdot\hat{\epsilon}=0$\egroup. The traceless symmetric tensor has 5 components like an \bgroup\color{black}$\ell=2$\egroup operator; The anti-symmetric tensor has 3 components; and the trace term has one. This is the separation of the Cartesian tensor into irreducible spherical tensors. The five components of the traceless symmetric tensor can be written as a linear combination of the \bgroup\color{black}$Y_{2m}$\egroup.

Similarly, the second (anti-symmetric) term can be rewritten slightly.

\begin{displaymath}\bgroup\color{black} {1\over 2}[(\vec{k}\cdot\vec{r})(\hat{\e...
...s\hat{\epsilon}^{(\lambda)})\cdot(\vec{r}\times\vec{p}) \egroup\end{displaymath}

The atomic state dependent part of this, $\vec{r}\times\vec{p}$, is an axial vector and therefore has three components. (Remember and axial vector is the same thing as an anti-symmetric tensor.) So this is clearly an \bgroup\color{black}$\ell=1$\egroup operator and can be expanded in terms of the \bgroup\color{black}$Y_{1m}$\egroup. Note that it is actually a constant times the orbital angular momentum operator \bgroup\color{black}$\vec{L}$\egroup.

So the first term is reasonably named the Electric Quadrupole term because it depends on the quadrupole moment of the state. It does not change parity and gives us the selection rule.

\begin{displaymath}\bgroup\color{black} \vert\ell_n-\ell_i\vert\leq 2\leq \ell_n+\ell_i \egroup\end{displaymath}

The second term dots the radiation magnetic field into the angular momentum of the atomic state, so it is reasonably called the magnetic dipole interaction. The interaction of the electron spin with the magnetic field is of the same order and should be included together with the E2 and M1 terms.

\begin{displaymath}\bgroup\color{black} {e\hbar\over 2mc} (\vec{k}\times\hat{\epsilon}^{(\lambda)})\cdot\vec{\sigma} \egroup\end{displaymath}

Higher order terms can be computed but its not recommended.

Some atomic states, such as the 2s state of Hydrogen, cannot decay by any of these terms basically because the 2s to 1s is a 0 to 0 transition and there is no way to conserve angular momentum and parity. This state can only decay by the emission of two photons.

While E1 transitions in hydrogen have lifetimes as small as \bgroup\color{black}$10^{-9}$\egroup seconds, the E2 and M1 transitions have lifetimes of the order of $10^{-3}$ seconds, and the 2s state has a lifetime of about ${1\over 7}$ of a second.

Jim Branson 2013-04-22