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3

What happens, essentially, is that the S and P wavefunctions get mixed to produce eigenstates that have shifted centres. This means the atom gets an induced electric dipole moment, whose interaction with the external field either lowers or raises the eigenenergy. More specifically, consider the wavefunctions of the states $|200\rangle$ and $|210\rangle$: ...

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To get it in the momentum representation, one has to do the Fourier transform of this function. This reference can be useful: http://forum.sci.ccny.cuny.edu/Members/lombardi/publications/MOMREP-H-atom.pdf/view At the end, separation of variables after transformation to the momentum space is not trivial, and the mixing of quantum number is presented.

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A hydrogen atom ion $H^{+}$, with an atomic mass number of A=1, charge number Z=1, is the same as a proton. A hydrogen ion thus usually just refers to a proton. Depending on context, however, you may also have a hydrogen ion which is (a) an ion of a deuterium atom, in which case it is a bound state of a neutron and a proton, with atomic mass number A=2, ...

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I believe you will find that the expectation value of the momentum is zero, which will sort of mess up your calculation of the wave length. Calculating the wavelength of the ground state of any bound system is folly anyway, since it will not have any nodes. The characteristic you may be looking for is the average radius, or the uncertainty in the position.

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I had a look at the paper, and I think the author means that the energy for the reaction: $$H + H^+ \rightarrow H_2^+$$ is negative i.e. the ground state energy of $H_2^+$ is less than the sum of the ground state energies of $H$ and $H^+$. The reason for this is simply the observation that the $H_2^+$ ion is stable. If the energy of $H_2^+$ were higher ...

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