How do I write the wave function of hydrogen atom taking into consideration of nucleus spin? For example consider $2S_{\frac{1}{2}}$ state with nucleus spin $I$, then wave function $\psi=\langle2S_{\frac{1}{2}},F,F_{3}|$ where $F$ is the total angular momentum of hydrogen atom $F=J+I$ and $F_{3}$ is the projection of it along z axis. Now what will be the explicit form of the wave function? Thanks in advance.
[Physics] Wave function of hydrogen atom including spin of nucleus
hydrogenquantum mechanicswavefunction
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The orbital wavefunctions of the hydrogen atom, which obey the eigenvalue equation $$ \left[-\frac{1}{2\mu}\nabla^2-\frac{e^2}{r^2}\right]\psi_{nlm}=E_{nl}\psi_{nlm}, $$ are functions of the separation vector $\mathbf r$ which points from the proton towards the electron. This is a standard trick in the two-body problem and it is done in both the classical and the quantum versions to factor away the motion of the bigger body (which is close to the centre of mass) and leave an effective one-body problem which is easier to treat.
This means that the orbital angular momentum, with total angular momentum number $l$, is in fact the combined angular momentum of the electron and the proton about their centre of mass. In essence, the proton partakes in part of the orbital motion and takes out some of the angular momentum from the electron. (Note, though, that this is classical language and it explicitly does not hold for the hydrogen atom, where the angular momentum of the motion is essentially indivisible.)
This raises an apparent paradox, which is resolved through the fact that the separation vector obeys dynamics through the reduced mass $\mu=1/\left(\tfrac{1}{m_e}+\tfrac{1}{m_p}\right)\approx\left(1-O\left({m_e\over m_p}\right)\right)m_e\lesssim m_e$ of the system, and this is slightly smaller than the electron mass. This slightly enlarges the orbital radius of the electron (since the Bohr radius is inversely proportional to the mass). The velocity stays constant (at $\alpha c$), which means that the angular momentum $L\sim \mu r v$ stays constant as well.
That said, the proton does have spin angular momentum of its own, but this couples weakly to the electronic motion. This coupling is via the same spin-orbit couplings as the electron, but its much higher moment of inertia means that the relevant energies are much smaller, as are the corresponding hyperfine splittings in the spectrum.
This is one of the mysteries of quantum mechanics. If you could measure the velocity of an electron, you would get a non-zero value. But what you can't do is use that velocity to predict where to find it next. The act of measurement disturbs the electron in an essential way.
One popular interpretation of quantum mechanics is a statistical one. This says that the wave function provides the probability density for finding the result of a measurement on an ensemble of identically prepared systems. That is, I start with an atom and measure the velocity of its electron. I get a value. I then prepare an identical atom and measure the velocity of its electron. I get a different value. This is very different from classical mechanics. It also might make a connection for you to @EmilioPisanty comment.
There is no way in which our traditional notion of "orbit" makes sense. There's no way our traditional notion of angular momentum applies. We notice that atoms behave as if they have angular momentum, and then go about building a mathematical structure that describes it. We find that the idea of electrons moving from here to there just has no place. Our description of nature does not include the idea that electrons in atoms move from here to there the way macroscopic objects do.
Best Answer
The easy way
If we do not take into account the dependence of the electron state on the spin state of the nucleus, the wavefunction is just a product of electron and nucleus wavefunctions: $$ \psi = \psi_e(\mathbf{r} - \mathbf{R}) \psi_n(\mathbf{R}) $$ Both are spinors of rank 1 (columns of functions). The spinor $\psi_e$ consists of two components. The number of the components of $\psi_n$ depends on the total spin of the nucleus $I$ and is equal to $2I+1$.
The hard way
If the spin of the nucleus affects the electron state, then the total wavefunction is a spinor of rank 2 i.e. a table of functions with dimensions $2 \times (2I+1)$.