The selection rules for electric quadrupole radiation in a Hydrogen-like atom are: $$ \begin{aligned} \Delta l &= 0,\pm2 \hspace{1cm}(l=0\leftrightarrow l'=0 \textrm{ is forbidden}) \\ \Delta m &= 0,\pm1,\pm2 \end{aligned} $$ My first question is why is the transition $l=0\leftrightarrow l'=0$ forbidden?

I know that the rule $\Delta m = \pm 1$ comes from the matrix element $\langle nlm|xz|n'l'm'\rangle$ and $\langle nlm|yz|n'l'm'\rangle$, while the rule $\Delta m = \pm 2$ follows from the matrix element $\langle nlm|xy|n'l'm'\rangle$. My second question is that what about $\Delta m = 0$ where does it come from?


There are five relevant quadrupole moment operators, and when labeled by the change $\Delta m$ in angular momentum projection they read $$ \begin{array}{c|ccccc} \Delta m & -2 & -1 & 0 & 1 & 2\\ \hat Q_{2m}& (x-iy)^2 & (x-iy)z & x^2+y^2-2z^2 & (x+iy)z & (x+iy)^2 \end{array} $$ These operators arise as (a basis for) all the homogeneous second-degree polynomials that are 'traceless' in the sense that they integrate to zero over the unit sphere.

As such, quadrupole transitions with $\Delta m=0$ arise from the transition quadrupole moment operator $\hat Q_{20}=x^2+y^2-2z^2$, for which it is easy to check that $$⟨lm|\hat Q_{20}|l'm'⟩=0 \quad\text{whenever}\ m≠m',$$ since then the longitudinal integral over $\phi$ vanishes.

A bit more physically, perhaps, transitions caused by $x^2+y^2-2z^2$ represent absorption of photons that are propagating in the $z$ direction, driven not by the electric field of the wave (as is the case for dipole transitions) but by its (generally small, but nonzero) spatial change across the atom in the direction of propagation.

Similarly, if you try to do a quadrupole transition between $l=0$ and $l'=0$, then both $m$ and $m'$ are forced to be zero, so transition amplitudes are proportional to $⟨00|\hat Q_{20}|00⟩$, and when integrated explicitly over the unit sphere this gives \begin{align} ⟨00|\hat Q_{20}|00⟩ & = \frac{1}{4\pi}\int x^2+y^2-2z^2 \:\mathrm d\Omega \quad\text{(since the states have a flat wavefunction)} \\ & = \frac{1}{4\pi}\int_0^{2\pi}\int_0^\pi (\sin^2(\theta)-2\cos^2(\theta))\sin(\theta)\mathrm d\theta\mathrm d\phi \\ & = \frac{1}{2}\int_0^\pi (1-3\cos^2(\theta))\sin(\theta)\mathrm d\theta \\ & = \frac{1}{2}\int_{-1}^1 (1-3u^2)\mathrm du =\frac12 \left[u-u^3\right]_{-1}^1 \\ & = 0. \end{align} Similarly, the transition $l=0\leftrightarrow l'=0$ is always forbidden from dipole radiation upwards, because it leads essentially to the inner product between the relevant transition operator $Y_{lm}$ and the flat $Y_{00}$ that arises from the wavefunctions, and those spherical harmonics are always orthogonal.

  • $\begingroup$ I think I get it now why the transition between $l=0$ states is forbidden. As for my second question, what concerns me actually is about the physical arrangement between the atom and light polarization which leads to each possible change $\Delta m$. For example, if the light propagates in the z direction and polarized in x or y direction then it will lead to the matrix element $\langle n'l'm'|xz|nlm\rangle$ or $\langle n'l'm'|yz|nlm\rangle$ and hence leads to the change $\Delta m = \pm 1$ (because $xz$ and $yz$ are each proportional to a linear combination between $Y_2^1$ and $Y_2^{-1}$). $\endgroup$
    – nougako
    Jul 2 '16 at 9:00
  • $\begingroup$ If the light propagates in the x direction and polarized in y direction then it will lead to the matrix element $\langle n′l′m′|xy|nlm \rangle$ and hence leads to the change $\Delta m=\pm 2$ (because xy is proportional to a linear combination between $Y_2^2$ and $Y_2^{-2}$). But what about the arrangement of the radiation which leads to $\Delta m=0$, in which directions should the photon propagate and be polarized? $\endgroup$
    – nougako
    Jul 2 '16 at 9:02
  • $\begingroup$ There’s a slightly faster route, saving one integration, if one writes $\hat Q_{20}= r^2-3z^2$. The $r^2$ comes out of the integral and one is left with the same $3z^2=3\cos^2\theta$ with slightly less trig manipulations. $\endgroup$ Dec 17 '18 at 16:04
  • $\begingroup$ @ZeroTheHero sure, but it's not really worth changing this now. $\endgroup$ Dec 17 '18 at 17:41
  • $\begingroup$ @EmilioPisanty I was thinking that my comment would be more of an addendum to your answer than a suggestion for change. $\endgroup$ Dec 17 '18 at 18:08

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