Why don't we consider ##E_{com}## while solving a two body problem?

[Kashmir]

While studying two interacting particles such as a Hydrogen atom, I learned how to reduce the problem into two independent parts by using center of mass coordinates and the relative coordinates.

The resulting two independent energy eigenvalue equations give me two eigenvalues for energy as:
$\begin{aligned} H_{C M} \psi_{C M}(\mathbf{R}) &=E_{C M} \psi_{C M}(\mathbf{R}) \\ H_{r e l} \psi_{r e l}(\mathbf{r}) &=E_{r e l} \psi_{r e l}(\mathbf{r}) \end{aligned}$

The total energy of the system is the sum of these two.

However while studying ahead, the energy contribution of centre of mass Hamiltonian was never considered again and all the talk was about the energy eigenvalue we get from the relative Hamiltonian.

Why is this so? Why don't we consider the other part? If we measure the energy in the laboratory, what result will I get, the total energy as a sum of two parts or the relative one?

Thank you.

 

[PeroK]

However while studying ahead, the energy contribution of centre of mass Hamiltonian was never considered again and all the talk was about the energy eigenvalue we get from the relative Hamiltonian.

Why is this so? Why don't we consider the other part? If we measure the energy in the laboratory, what result will I get, the total energy as a sum of two parts or the relative one?

Thank you.

If a hydrogen atom undergoes a transition by absorbing or emitting a photon, then the energy is almost entirely the difference in the electron's energy level.

You could calculate for yourself a small correction by considering the atom recoiling by conservation of momentum.

That said, in astronomy, the redshift due to relative velocity of the source and detector affects the measured wavelengths.

 

[andresB]

In the laboratory, there is also a widening of the spectral lines due to the Doppler effect caused by the translational motion of the atoms.