Photon is spin 1 and electron is spin 1/2, so when a photon is absorbed by an electron it is destroyed and the electron becomes excited by that amount of energy. The next moment the electron will go back to it's ground state and emits a photon with the same energy as the original and everything seems good so far. I wonder what happens to the spin of the photon? Is it similar to the momentum of photon scattered by a free electron that relatively speaking electron is at rest the frequency of photon emitted is shifted or electron moves while frequency of photon remains the same as original?
The spin of the electron was introduced because of the electrons behavior in magnetic fields:
Wolfgang Pauli in 1924 was the first to propose a doubling of electron states due to a two-valued non-classical "hidden rotation". In 1925, George Uhlenbeck and Samuel Goudsmit at Leiden University suggested the simple physical interpretation of a particle spinning around its own axis, in the spirit of the old quantum theory of Bohr and Sommerfeld.
The Lorentz force is related to the particles spin. Lorentz equation describes the phenomenon, that electrons and anti-protons get deflected in one direction while positrons and protons get deflected in the opposite direction.
In retrospect, the first direct experimental evidence of the electron spin was the Stern–Gerlach experiment of 1922.
For the photons all the mentioned experiments do not work. And as you say, the absorbed photon does not change the intrinsic spin of a free moving electron. Hitting the electron not in its center the photons moment could induce a rotation of the electron. And a circular polarized photon could give the electron a rotation too.
The spin of the photon is simply different from the behavior of the spin of charges. But as is the spin of a photon? The best answer I’ve seen was this:
But! Nobody found a correct way to represent it in the sum of two gauge invariant operators for angular orbital momentum and spin (L and S). This concludes the theoretical knowledge - operator of photon's spin is not known.
It is possible to measure the Z-component of the angular momentum - helicity.
The photon must have the exact energy for an allowed state for the electron. Spin-space is orthogonal from n,l,m-space, it may seems energy distribution goes from one space to an orthogonal space. But the other 1/2-spin will account for the n,l and m for the excited state of the electron.
If not coupled into otbital angular momentum, electron can flip its spin to compensate the spin of a photon. In the case require electron to flip spin, at the moment electron and photon interact they must have opposite spin to begin. So they can engage into quantum entanglement to allow the process happen.
As has been indicated above, it is not the electron which will gain some internal angular momentum (spin) from the photon, but it is the atom which absorbs the energy as well as the angular momentum of the photon (1 hbar). This will result in an increase of the orbital angular momentum exectly by 1 hbar (l=l+1) plus an increase of the energy (n=n+1,2,3...) depending on the frquency (=energy) of the photon. Remember: Total energy and total angular momentum are the quantities which both have to be conserved!