Is $e^-$ in 1st excited state in the $\rm H$-atom in superposition of possible ($l,m,s$ for $n=2$) substates/in superposition of all possible states? I want to know how we can determine the probability of various states of hydrogen as there are infinite of them, or electron in a hydrogen atom is in the superposition of which states, maybe it is in the superposition of degenerate states? I don't know, i am just guessing, please help.
 A: This depends on how the state was prepared. All of the possible outcomes,

*

*the system is in a common eigenstate of the usual CSCO, with well-defined $l,m_l$ and $m_s$;

*the system is in some nontrivial superposition with respect to that basis;

*the system is in a mixed state that describes a classical probabilistic distribution over the states of that basis;

can occur. Which one you get will depend on how your system got to that first excited state.
For example, if the system was prepared by taking ground-state hydrogen atoms and bringing them up to $n=2$ using collisional excitation with a beam of electrons of roughly the right energy, then the process will likely have been quite messy, and you should expect a classical probabilistic distribution where all the available states are equally likely to be prepared.
On the other hand, it is perfectly possible to prepare a pure state on whatever basis state you choose by exciting the atom using a tightly-controlled laser beam with a very stable and well-defined frequency, using the laser's polarization to control the details of the state via the dipole selection rules of the excitation.
Moreover, once you're at such a basis state, you can use the toolbox of quantum coherent control to reach any linear superposition of states in that basis that you desire.
All of these possibilities are on the table, and the choice of which one to use will depend on the experiment in question. In general, in the absence of any additional information you have to take the first option (the "maximally mixed" state representing an evenly-weighted classical probability distribution over all the available basis states), but "absence of additional information" only describes some, and not all, the possible experiments you might want to run.

And that said: be careful with the phrase "first excited state".
If you only include the Coulomb interaction, as generally done on a first approach, then indeed the full $n=2$ level manifold is degenerate in energy. However, if you go beyond that and include relativistic effects (known as fine structure) then the $s$ and the $p$ states become nondegenerate. (You can confirm this directly using the NIST ADS level listings.) Moreover, if you go beyond that and you include magnetic interactions with the nucleus (known as hyperfine structure) then there's additional splittings within the manifold. And if you're there, then you probably want to include the Zeeman effect of any magnetic fields in the experiment, either intentional or stray.
All of these things will generally break the degeneracy, limiting the number of degenerate states or even bringing it down to one. However, they only do so at tiny energy splittings, and it's important to understand whether those levels of the theory will be relevant to your experiment:

*

*if you're using a state-of-the-art laser system with a frequency stability better than one part in $10^{15}$, then a lot of those things will be relevant,

*but, on the other hand, if you're using a messy electron-beam source with a $0.1\:\rm eV$ bandwidth, then fine and hyperfine structure go completely out of the window and they get swamped by the noise in the source, so you might as well consider a fully-degenerate manifold of states.

