we reorganized the various contributions listed in Table II as follows: “Elastic” denotes $\Delta\bar E^{el}+\Delta E^{Th}$, and “Nuclear” is sum all nuclear contributions, $\Delta E^{PWBA}+\Delta E^{FSI}+\Delta E^\perp+\Delta E^\beta$.

Ref. [@rosenfelder] quotes 1.500 meV $2P-2S$ correction due to the deuteron nuclear electric dipole polarizability; in Ref. [@pachuckiD] a result of 1.680(16) meV is obtained by considering the electric polarizability (and various corrections thereto), elastic and hadronic contributions, and magnetic polarizability. Ref. [@pachuckiD] furthermore obtains the sum of the proton and neutron intrinsic polarizabilities to the Lamb shift in muonic deuterium by rescaling the [*total*]{} Lamb shift for muonic hydrogen, $\Delta E_{\mu H}=36.9\,\mu$eV obtained in Ref. [@Carlson:2011zd] with the ratio $(\mu_r^D/\mu_r^H)^3$ with the result $\Delta E_{\mu D}^{hadr.}=43(3)\,\mu$eV. This estimate is not correct because the main contribution to $\Delta
E_{\mu H}$ is due to the elastic contribution, and only about a third of it, $13.5\,\mu$eV comes from polarizabilities. Since proton and neutron electric polarizabilities are very close, $\alpha_p\approx\alpha_n$, one should expect that the result for the deuteron should be roughly equal to their sum, $\Delta E_{\mu D}^{hadr}\sim 2\Delta E_{\mu
  H}=27\,\mu$eV. Indeed, our result (third entry in Table \[tab2\]) is consistent with this simple estimate, $\Delta E_{\mu D}^{hadr}=28(2)\,\mu$eV. This suggests that after correction the 