The common explanation of spin contamination is like the following:

The non-relativistic electronic Hamiltonian commutes with $S^2$ and $S_z$. Hence, the exact eigenfunction should be an eigenfunction for $S^2$ and $S_z$ operators.

However, the non-relativistic Hamiltonian has no spin variables. Any wavefunction contains spin, the spin part will commute with the Hamiltonian anyway.

For instance, in addition of angular momentum, http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect15.pdf One can form $\{J_1^2, J_2^2, J_{1z}, J_{2z}\}$ set of variables. They commute with Hamiltonian, but no longer eigenfunction of $S^2$, i.e. $\{J^2, J_z, J_{1}^2, J_{2}^2\}$.

If my argument is correct, what would be the actual justification for spin adaptation?

  • 1
    $\begingroup$ OK. Finally I have some time to continue the discussion. As it appears now, you question has very little to do with spin contamination, rather it is about the thing you called "spin adaptation". It is about why do we want a trial electronic wave function to be eigenfunction of total spin operators, not how do we achieve this goal in the unrestricted calculations. Am I right? $\endgroup$
    – Wildcat
    May 8, 2016 at 12:55
  • $\begingroup$ No. The concept spin-adaption and contamination are connected. If we want to have a spin-adapted wavefunction, it will not have contamination. But, why? I can build a spin wavefunction as uncoupled basis, they also commute with Hamiltonian. From your answer, it is just convention. $\endgroup$
    – Rodriguez
    May 8, 2016 at 19:29
  • $\begingroup$ I don't understand where do you see a contradiction out there. Yes, you can require the electronic wf to be an eigenfunction of operators of the individual sources of spin, rather then of the total spin, i.e. you can use the uncoupled picture instead of the coupled one. But when working in the uncoupled picture you don't know the total spin of the system which is often desirable. So what is your problem with this? $\endgroup$
    – Wildcat
    May 8, 2016 at 20:05
  • $\begingroup$ The problem is, I do not think it is just a convention. $\endgroup$
    – Rodriguez
    May 8, 2016 at 20:10

1 Answer 1


Commuting operators indeed admit a set of simultaneous eigenfunctions, and since the non-relativistic electronic Hamiltonian commutes with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, the exact non-relativistic electronic wave function (which is an eigenfunction of the electronic Hamiltonian) is also an eigenfunction of the total spin operators. And this requirement can be also imposed on an approximate non-relativistic electronic wave function.

This is certainly true that the non-relativistic electronic Hamiltonian commutes not only with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, but with operators of the individual sources of spin $\hat{S}_{iz}$ and $\hat{S}_{i}^{2}$ as well, or to be more precise, with their trivial extensions on the state space of a many particle system. Thus, this is indeed true that one could adopt the uncoupled picture as well, rather then the usual coupled one, but what is the usefulness of such picture? In the absolute majority of cases the most important characteristic of a system is a property which describes it as a whole, rather that its components individually. For this particular case, recall, for instance, that the total spin of a molecular system is a very important characteristic of it. For instance, it is heavily used in electronic spectroscopy: important terms like term symbols, spin-allowed and spin-forbidden transitions, fluorescence, phosphorescence, etc. are all directly formulated in terms of total spin angular momentum of electronic states.

Now if you decide to use the uncoupled picture in which the total spin is unspecified you basically loose all the body of knowledge of electronic spectroscopy. But what do you gain instead? What is the point of using the uncoupled picture? You can specify the spin of each electron, yes, but but so what?

  • $\begingroup$ Thanks. I have revised the second sentence. But you have not answered the core problem: I can write other spin operators than $S^2$, they will commute with the Hamiltonian. I can have other wavefunction which will be the eigenfunction of such spin operator, e.g. $S_1^2$ and $S_2^2$. But no longer eigenfunction of $S^2$. The aspect emphasising $S^2$ will let it has a more privilege position than other choice of spin operators, e.g. $S_1^2$ and $S_2^2$. $\endgroup$
    – Rodriguez
    May 3, 2016 at 21:12
  • $\begingroup$ I don't quite understand what do you mean by "other spin operators"? You mean operators for individual sources of spin (spin of individual electrons)? $\endgroup$
    – Wildcat
    May 3, 2016 at 21:23
  • $\begingroup$ for instance, uncoupled basis in additional of angular momentum. $\endgroup$
    – Rodriguez
    May 3, 2016 at 21:25
  • $\begingroup$ @Rodriguez, then you're wrong. The total angular momentum squared $\hat{J}^{2}$ commutes both with $\hat{J}_{1}^{2}$ and $\hat{J}_{2}^{2}$, thus, an eigenfunction of $\hat{J}^{2}$ is an eigenfunction of $\hat{J}_{1}^{2}$ and $\hat{J}_{2}^{2}$ as well. $\endgroup$
    – Wildcat
    May 3, 2016 at 21:27
  • $\begingroup$ The problem is, the coupled and uncoupled angular momentum are two sets of basis. $|j_1 j_2 m_1 m_2\rangle$ and $|j m j_1 j_2 \rangle$. Once you choose the uncoupled basis, it will be eigenfunction of corresponding spin operator $j_1^2, j_2^2, j_{1z}, j_{2z}$ and Hamiltonian, but not $j^2$. $\endgroup$
    – Rodriguez
    May 3, 2016 at 21:31

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