Systematic expansion of $e^{i\vec{k}\cdot\vec{r}}$ in atomic physics in terms of Legendre polynomials and identifying different $l$ terms

Question

In the context of light-matter interaction one often makes the approximation $e^{i\vec{k}\cdot\vec{r}}\approx 1$. Keeping higher order terms in $e^{i\vec{k}\cdot\vec{r}}$ give magnetic dipole, electric quadrupole etc transitions.

This makes me believe that $e^{i\vec{k}\cdot\vec{r}}$ must have similar multipole expansion like $|\vec{r}-\vec{r}^\prime|^{-1}$ in electrostatics where the latter can be systematically expanded in terms of Legendre polynomials $P_l(\cos\gamma)$. Different $l$ values lead to different multipole terms. In particular, $l=0,1,2,...$ respectively represent monopole, dipole, qudrupole etc terms.

Now, there exists an expansion of $e^{i\vec{k}\cdot\vec{r}}$ in atomic physics (c.f. Sakurai) $$e^{i\vec{k}\cdot\vec{r}}=\sum\limits_{l=0}^{\infty}(2l+1)i^l j_l(kr)P_l(\cos\gamma)$$ where $\gamma$ is the angle between $\vec{k}$ and $\vec{r}$, $j_l(x)$ are spherical Bessel functions, and in particular, $j_0(x)=\frac{\sin x}{x}$.

The problem is that unlike my expectation from electrostatics, $l=0$ gives the electric dipole approximation $e^{i\vec{k}\cdot\vec{r}}= 1$ and not $l=1$. Where am I wrong? Further, can we associate this $l$ value to the angular momentum of the emitted/absorbed radiation?

Answer 1

The expression you found in Sakurai's textbook is the correct one and can be further expanded in spherical harmonics, if needed.

The difference with electrostatics is due to the different position the "dipole" emerges in the case of quantum treatment of light-matter interaction.

In electrostatic is the term $l=1$ of the Legendre polynomial expansion of the charge density which gives rise to the dipole term in the multipoles expansion of the charge density and then to the dipole term in the electrostatic potential.

When dealing with interaction between EM radiation and an atomic electron, one starts with the perturbation Hamiltonian which contains a term linear in the momentum of the electron $\bf p$ and then the relevant matrix elements of the perturbations are of the kind $$ \left< i | {\bf {\boldsymbol \epsilon} \cdot p}~ e^{i {\bf k \cdot r}}|f \right> $$ ($|f>$ and $<i|$ being the ket and the bra of the final and initial state and $\boldsymbol \epsilon$ the polarization vector). If the exponential is expanded for small values of its argument, retaining only the zeroth-order $1$, and using the possibility of re-expressing $\bf p$ in terms of the commutator of the position $\bf r$ with the unperturbed Hamiltonian to find that the matrix elements can be written as proportional to $ \left< i | {\bf {\boldsymbol \epsilon} \cdot r} |f \right>$.

Therefore the operator whose matrix elements have to be evaluated has the same symmetry of a dipole with all the known consequences on selection rules etc. In conclusion, the reason in the case of radiation the term $l=0$ does correspond to the dipole approximation, while in electrostatic it is the term $l=1$, is that the functions which are expanded in Legendre Polynomials in the two cases are different and play a different role in the theory, although at some point a dipolar-like term appears.

Edit

I see that the previous answer was not considered satisfactory. I will add reference to a well known textbook (although I thought it was not necessary). Many textbooks of QM touch this topic. On Leonard Schiff's Quantum Mechanics Third edition, the calculation can be found in sections 44 and 45.

As I have already indicated, one has to start with the expression for the rate of transition between two states, evaluated using time dependent perturbation theory. The transitions corresponding to these matrix elements are called electric dipole transitions, since only the matrix element of the electric dipole moment of the particle ($e\bf r$) are involved.

Answer 2

The approximation $e^{i\vec k \cdot \vec r}=1$, in other words $\vec k= 0$ or $\lambda = \infty$, is made for atomic transitions because for such transitions the wavelength is much larger than the size of the atom. The dipole character of the transition is caused by the fact that the perturbing hamiltonian is proportional to $\vec A \cdot \vec p$, where $\vec A$ is the electromagnetic vector potential.

It is the presence of the $\vec p$ operator that is responsible for the $\Delta \cal l =\pm 1$ selection rule for dipole transitions, although the very small deviations from the $\vec k= 0$ approximation in principle also contribute to dipole forbidden transitions.

Answer 3

Yes, it's called a spherical wave expansion. It's given by

$$ e^{i\mathbf{k\cdot r}} = 4\pi \sum_{l=0}^\infty i^l j_l(kr) \sum_{m=-l}^l Y^*_{lm}(\theta, \phi) Y_{lm}(\theta', \phi') $$

where the $j_l$ are spherical Bessel functions, and the primed coordinates are the the angles that specify those of the $\mathbf{k}$ vector, and the unprimed are those of $\mathbf{r}$. Recall that the spherical harmonics are a function of the generalized Legendre polynomials (i.e. when you don't have azimuthal symmetry)