How to derive the allowed beta decay selection rule $\Delta L=0$? Starting from the beta decay matrix element and Fermi's golden rule how can one show that the matrix element at the leading order vanishes unless $\Delta L=0$?
 A: I could be wrong, but I think this is basic kinematics and has nothing to do with fancy quantum mechanics. For a particle with mass $m$ and kinetic energy $E$, the nonrelativistic momentum is $p=\sqrt{2mE}$, whereas in the ultrarelativisic limit we have $p=E/c$. The nucleons are normally going to be nonrelativistic and the leptons ultrarelativistic. Assuming the energies to be of the same order of magnitude, say 1 MeV, we expect $p_\text{lepton}/p_\text{nucleon}\sim (1/c)\sqrt{E/2m}\sim 0.03$. So to a pretty good approximation, we're just replacing, say, a neutron with a proton in the same state of motion.
A: Let the matrix element for $\beta$ decay from state $X$ to state $X^{\prime}$ be
$\cal{M}_{if}=g \langle (e^{i \vec{p}\cdot \vec{x}})(e^{i \vec{q}\cdot \vec{x}}) \psi_{x^{\prime}} \mid \cal{O_\beta} \mid \psi_x \rangle  $,
where $p$ and $q$ are momenta of electron and neutrino.
Electron and neutrino wavelengths ($\lambda$) are many times ($n$) size ($x$) of the nucleus  (  $\lambda \sim nx \sim 1/p$ so from this $px \sim x/(nx) \sim 1/n$ is small). So one can expand both plane waves in Taylor series $1 + \frac{i \vec{p} \cdot \vec{x}}{\hbar} + ... $ or better in Legendre polynomials with Bessel functions (similar for neutrino with $q$)
$ e^{i \vec{p} \vec{x} } = e^{ipx \cos(\theta)} = \Sigma_{l=0}^{\infty} i^l (2l+1)j_l(px) P_l( \cos(\theta) ) $
When only $L=0$ is present, the two $l=0$ terms of electron and neutrino naturally remain from the expansions, which are  $j_0(px) = \frac{ \sin(px)}{px}$, which for small $px$ goes to 1 (the leading term).
The other pairs of $l_p=l_q>0$ that form together $L=0$ are of the next next order and together go to zero (see the Bessel function systematics, but carefully about simple formulae).

The matrix element then turns into
$\cal{M}_{if} \sim g \langle  \psi_{x^{\prime}} \mid \cal{O_\beta} \mid \psi_x \rangle + g \langle  \psi_{x^{\prime}} \mid \cal{O_\beta} \mid \psi_x \rangle  {(px)(qx)} + ... $
which is the desired leading order part (for the Fermi superallowed decay $\psi_{x^{\prime}}$ and $\psi_{x}$ are virtually the same configurations, so the matrix element is like 1.)
Now to the question : if $\Delta L \ne 0$, the both plane waves with $l=0$ cannot be present at once and at least one term containing $px$ and/or $qx$ takes place. As seen in the picture, the Bessel functions other than $l=0$ go to zero. No term with ${kx}^0$ is thus present anymore.
