I learned that wavefunction for a hydrogen atom depends on three quantum numbers, not four. Generally, can I think the wavefunction does not depend on the spin quantum number $m_s$?
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$\begingroup$ For the Schrödinger solution or Dirac solution? $\endgroup$– Jeanbaptiste RouxCommented Nov 26 at 11:56
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$\begingroup$ @JeanbaptisteRoux I was talking about the Schrodinger solution $\endgroup$– user378019Commented Nov 26 at 12:06
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4 Answers
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From a pure mathematical point of view, it depends on your defintion of "wavefunction". The quantum state of an electron subjected to the Hydrogen potential is represented by a state vector which can be spanned in terms of base vectors of the form $$\Psi_{n,\ell, m_\ell, m_s} = \psi_{n,\ell,m_\ell}\otimes |m_s\rangle\:.$$ If you define the "wavefunction" as the orbital part of the above state vector, it is $$\psi_{n,\ell,m_\ell} = \psi_{n,\ell,m_\ell}(\vec{r})$$ and $m_s$ does not take place in it. However the complete basis of states also includes the spin part $ |m_s\rangle$ which can attain two values $|1/2\rangle$ and $|-1/2\rangle$.
Another much more physical issue concerns the explicit form of the state vector of an electron in a Hydrogen atom when it has definite energy. Apparently $m_\ell$ and $m_s$ are free parameters (in absence of other exernal potentials than the Coulomb interaction) and an eigenstate of the Hamiltonian could be represented as a completely unpolarized density matrix (in order to keep the isotropy of the space). Hoewer that is not the case, as there are finer interactions which take place when one consider a relativistic perspective, as the spin-orbit interaction.
Coming back to the title of the post (sorry I focused attention on the main text only), if we consider several electrons around the proton, the spin part of the state vector of an electron of the set plays an explicit role in view of the exclusion principle (the fact that the full state must be completely antisymmetric under interchange of electrons). For instance, referring to the state of a pair of electrons: if they have the same orbital part, then the total spin part must be antisymmetric (the spin-$0$ singlet).
$$\psi_{n,\ell,m_\ell}(\vec{r}_1)\otimes \psi_{n,\ell,m_\ell}(\vec{r}_2) \bigotimes (|1/2\rangle_1\otimes |-1/2\rangle_2 - |-1/2\rangle_1\otimes |1/2\rangle_2)$$
answered Nov 26 at 12:21
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1$\begingroup$ But not taking the whole thing to be your wave function makes the exclusion principle hard to formulate, no? $\endgroup$ Commented Nov 26 at 13:01
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$\begingroup$ Yes you are perfectly right: the exclusion principle refers to the whole form of the state, including the spin part. However it concerns a state with many electrons, above I was considering a unique electron around a proton. I just realized that the title of the question refers to two electrons! I only read the main text... $\endgroup$ Commented Nov 26 at 13:17
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You have probably solved the hydrogen atom in absence of any external field in your physics course. Then two electrons with quantum numbers $m_s=\pm 1/2$ have the same probability density $|\Psi_i(\mathbf{r})|^2$. However, if you apply a magnetic field, the spin of the electron will couple to that field. This results that the two degenerated states split up. See for example the Stern Gerlach experiment (https://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment). The spin is further necessary, because two electrons in the ground state of Helium (or any other 2 electron system) can not have the same wavefunction, but this is a different topic.
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$\begingroup$ Even if there is no external field the probability of spin 1/2 is different than the probability of spin -1/2 due to the magnetic dipole moment of the proton. $\endgroup$ Commented Nov 26 at 13:31
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Generally, not. It is true that when we consider the non-relativistic Hamiltonian with Coloumb interaction as the only interaction, there is no reference to spin, which indicates that all spin states would have the same energies (to change the spin state does not affect the energy), and we don't care about it. However, we could increase the precision of the model by including other interaction terms, like the spin-orbit interaction $H_{SO}$. By doing this, the energy levels (Hamiltonian eigenvalues) changes, and some states with different spin would have different energy levels. You should read about The fine structure of Hydrogen atom.
The quantum numbers are related to the number of observables we should consider in the description of the system. This number is not "given by nature", but it is a choice that we do in order to describe what we want.
We could start a description of the hydrogen atom with one quantum number, for example. But then, we know that the angular momentum is conserved for central potentials, so we want to describe the system with states with well defined total angular momentum. But the degeneracy of the angular momentum operator is high, so we include one of its components, usually considered the $z$-component. For a lot of applications, we don't need the fine structure, so we don't care about spin, but spin is important to understand The Zeeman effect for example, and then we should include it.
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Short answer: no. Non-relativistic solutions of the wavefunction like what is learned in first quantum course doesn't have a variable for electron spin. There is I believe another solution to the non-relativistic wavefunction which includes an electron spin component (I believe in Heisenberg matrix formulation) but usually solutions including the electron spin-component also factor in special relativity (Dirac equation) and this is usually more complicated than a first course on quantum will get to.
