I am currently reading a book "Time-Dependent Density Functional Theory: Concepts and Applications". This book says as follows:

In its ground state, the valence electrons of $\rm Na_9^+$ form a closed shell with a doubly occupied $s$ orbital and three doubly occupied $p$ orbitals. We assume that the laser to be polarized along the $z$-axis, so that the initial rotational symmetry (and thus the $m$ quantum number of each TDKS orbital) is conserved.

I have a question on the conservation of $m$. In this case the vector potential can be written as $$\mathbf{A}=A_0 e^{i \mathbf k \cdot \mathbf r -wt} \hat z$$ and the perturbing Hamiltonian becomes $$H'=\frac{e}{mc}\mathbf p \cdot \mathbf A = \frac{eA_0}{mc} \hat{p}_z e^{i \mathbf k \cdot \mathbf r -wt}$$. The conservation of the quantum number $m$ means that $$[H', L_z]=0.$$ However, $$[p_z e^{i\mathbf k \cdot \mathbf r}, L_z] = p_z[e^{i\mathbf k \cdot \mathbf r}, xp_y-yp_x]=i\hbar(xk_y-yk_x)e^{i\mathbf k \cdot \mathbf r} p_z$$ is generally nonzero. (Note that $k_z=0$ and $[p_z, L_z]=0$.) How can I resolve this situation?


Atomic transitions involve states that are localized to length scales of the order of the Bohr radius, $a_0 \sim 0.05 \:\mathrm{nm}$, interacting with light in the visible range of wavelength of the order of $\lambda \sim 500 \:\mathrm{nm}$, which means that $r\sim a_0 \ll \lambda$, by a factor of about $10^3$ or more, which therefore means that $kr\ll 1$. The implication is that we can safely disregard the variation of $\mathbf A$ with respect to space, and substitute it with a homogeneous field that takes the value at the nuclear position. This is known as the dipole approximation.

If you want to be a bit more rigorous about it, then the standard formalism works by taking a multipolar series expansion of $\hat z e^{i\mathbf k\cdot \mathbf r}$, which then gives rise to a rising ladder of levels of interaction: electric dipole (E1), magnetic dipole (M1), electric quadrupole (E2), and so on. Each of those layers has its own set of selection rules (this Wikipedia page does a good job of explaining the framework), and each layer is weaker than the ones immediately below by a factor of $\sim a_0/\lambda$.

This means that, unless you're doing extremely high-precision work, you keep the lowest level available and disregard the rest. For some lines, the electric dipole transition is forbidden (say, transitions between $S$ and $D$ states), so you use an E2 transition or higher. But if there is an E1 transition permitted at that energy, and you're not doing something very specific, then it's completely safe to work within the dipole approximation.

  • $\begingroup$ So the point is that under the dipole approximation, $H'=A_0e^{-iwt} p_z$ and this commutes with $L_z$, right? $\endgroup$ – eigenvalue May 1 at 13:21
  • 1
    $\begingroup$ Yes, precisely. (Note that it's $\omega$, typeset $\omega$, and not $w$, though.) $\endgroup$ – Emilio Pisanty May 1 at 13:23

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