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.

• 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 $Y_2^{-1}$). Commented Jul 2, 2016 at 9:00
• 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? Commented Jul 2, 2016 at 9:02
• 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. Commented Dec 17, 2018 at 16:04
• @ZeroTheHero sure, but it's not really worth changing this now. Commented Dec 17, 2018 at 17:41
• @EmilioPisanty I was thinking that my comment would be more of an addendum to your answer than a suggestion for change. Commented Dec 17, 2018 at 18:08

I am writing this supplement @EmilioPisanty, which I feel gives an incorrect answer to the second question, "My second question is that what about $$\Delta m =0$$ where does it come from?".

To start, the physics of quadrupole transitions comes from combining the polarization vector $$\mathbf{\epsilon}$$ and momentum vector $$\mathbf{k}$$ of light, which is to say the matrix element looks like $$M\sim \langle f \vert (\mathbf{\epsilon} \cdot \mathbf{r}) (\mathbf{k} \cdot \mathbf{r}) \vert i \rangle$$.

So, for example, to get a $$\Delta m =0$$ transition you would need to have the $$z$$-components of the angular momentum for the two vectors be zero or cancel out. Likewise, for $$\Delta m = 2$$ you would need the $$z$$-components to add together.

Question 1: As @EmilioPisanty states, your first question about $$l=0 \rightarrow l'=0$$ being forbidden is true to all multipolar orders. This physically arises from the conservation of angular momentum, which is not possible if a single photon is absorbed and the atom does not change angular momentum state at all.

Question 2: There is no geometry for a plane wave photon which allows only $$\Delta m=0$$ while also forbidding $$\Delta m = \pm1, \pm2$$. The underlaying reason is that the photon's polarization vector $$\epsilon$$ and momentum $$k$$ must be perpendicular, so the two vectors cannot strictly cancel each other along all $$x,y,z$$ axes.

Nonetheless, $$\Delta m=0$$ transitions are possible because the $$z$$ components, but not $$x,y$$ components, of the two vectors can be canceled.

One such geometry which allows for $$\Delta m=0$$ is shown below. Here, $$\epsilon$$ and $$k$$ are 45 and 135 degrees from the $$+z$$ direction. In this geometry $$\epsilon \sim x-z$$ and $$k \sim x+z$$, giving $$(x-z)(x+z) = x^2 -z^2$$ which allows for both $$\Delta m =0$$ and $$\Delta m = \pm2$$, but not allowing for any $$\Delta m = \pm1$$.

Note: One can get pure $$\Delta m = 0$$ if one goes beyond plane waves and uses vortex beams which carry orbital angular momentum. There, you can arrange the orbital part to be $$+1$$ along $$z$$ and the polarization to circularly polarized $$-1$$ to give pure $$\Delta m = 0$$. This can be understood as a coherent sum of plane waves which cancel out the $$\Delta m = \pm2$$ contribution. See this paper for more detailed math.

• There is huge literature on quadrupole transitions in deformed nuclei. See this article co-written by a Nobel laureate as a good survey of the physics of quadrupolar objects. Commented Dec 11, 2023 at 4:23
• @ZeroTheHero thank you. Are you thinking there is something missing in the answer with respect to that? Quadrupolar transitions are indeed found in many interesting contexts. Commented Dec 11, 2023 at 6:03