# state transition in the hydrogen atoms

Weinberg, in Lectures on quantum mechanics (p. 46) mentions that the transition rate for the hydrogen atom to decay from a state $\psi$ to a state $\psi'$ is proportional to $\left|\intop \psi'X\psi\right|^2$, and under a change of variables $x \rightarrow-x$ thw expressions is multiplied by $(-1)^{l'+l+1}$.

He than goes on to say that an electron at state 2p can decay to 1s by emitting a single photon while 2s cannot do so and must emit 2 photons.

My questions are:

1. Why do we care about the factor $(-1)^{l'+l+1}$ if we care only about square absolute value? Shoudln't the factoe be $(-1)^{l'+l}$ because $dx\rightarrow-dx$?
2. Why can't 2s decay to 1s by emitting a single photont? how do we 2 are required?
• anyone? there must be an answer :) – proton Nov 28 '17 at 15:28
• Remember that a photon has one full unit of spin. So the angular momentum needs to change by 1 in any one-photon atomic transition, as long as the conservation of angular momentum holds. s states all have 0 angular momentum, so you need two photons to go from s to s. – Gilbert Nov 29 '17 at 2:50
• why to? one with $m=1$ and the other with $m=-1$? and each will have half of the energy? – proton Nov 29 '17 at 7:06
• that's right! But two-photon transitions are generally less likely to happen, and their likelihood increases with light intensity. – Gilbert Nov 29 '17 at 13:55