# Why does a magnetic transition $\mathrm{M}(\ell)$ have approximately the same probability as an electric $\mathrm{E}(\ell+1)$ transition?

When discussing electromagnetic decays and multipolarity, B. Povh, et al.$$^1$$ state that the magnetic transition $$\mathrm{M}(\ell)$$ have approximately the same probability as an electric $$\mathrm{E}(\ell+1)$$ transition. Why?

Furthermore, why does a photon of multipolarity $$\mathrm{M}(\ell)$$ have parity $$(-1)^{\ell+1}$$? Where does the extra $$(-1)$$ factor come from compared to an electric transition?

$$^1$$ B. Povh, et al. Particles and Nuclei, Springer, Berlin, Heidelberg p. 37, (2015).

This is not really a universally true fact, just a rough numerical coincidence for typical sizes of nuclei and typical gamma-ray energies. The ratio of the $$\mathrm{M}\ell$$ to $$\mathrm{E}(\ell+1)$$ is expected according to the Weisskopf estimate to be something like $$R\lambda_C/\lambda^2$$, where $$R$$ is the size of the nucleus, $$\lambda_C$$ is the Compton wavelength of a nucleon, and $$\lambda$$ is the wavelength of the gamma rays. This just happens to be somewhere on the order of 1 for typical cases.
• I'm not familiar with the nuclear physics case of this problem, but in electronic transitions there is a good reason for these two to be comparable- they come at the same order in the expansion of exp$(-i\vec{k}\cdot\vec{x})$ (see physics.stackexchange.com/questions/498039/…). Is there a similar reasoning underlying this coincidence? Dec 3 '20 at 1:43