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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).

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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.

Under a parity transformation, an electric dipole moment flips, but a magnetic dipole moment doesn't.

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  • $\begingroup$ Thanks for the additional details! This makes sense. $\endgroup$
    – Jack G
    Dec 2 '20 at 21:34
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    $\begingroup$ 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? $\endgroup$
    – Rococo
    Dec 3 '20 at 1:43

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