# Connection between singlet, triplet two-electron states and the Slater determinant

I'm confused about a number of things concerning two-electron systems and spin. Here is (perhaps too much) exposition, skip to "the problem" if you want:

Consider the helium atom in the simplified picture where we ignore electron-electron repulsion, and let $\psi_{nlm}$ be the usual space-wavefunctions, and let $\chi_{\uparrow}$, $\chi_{\downarrow}$ denote the simultaneous eigenfunctions of $\mathbf{S}^2$ and $S_z$. Then write the one-electron wavefunctions as $$\phi(i) = \psi(\mathbf{r}_i) \chi(s_i), \qquad i \in \{1,2\}$$ where $i$ is short for the space and spin coordinates $(\mathbf{r}_i, s_i)$. Given two one-electron wavefunctions $\phi_1$, $\phi_2$ we can always form a corresponding antisymmetric two-electron wavefunction with the Slater determinant: $$\phi(1,2) = \frac{1}{\sqrt{2}} \det \begin{bmatrix} \phi_1(1) & \phi_1(2) \\ \phi_2(1) & \phi_2(2) \end{bmatrix}.$$

Now, I'm having trouble constructing the eigenstates of helium (in this simplified picture). Following Griffiths, we assume one electron is in "the ground state" (i.e. has space-wavefunction $\psi_{100}$) while the other electron is in some "state" $\psi_{nlm}$.

In the case where both electrons have the space-wavefunction $\psi_{100}$ then we let $$\phi_1 = \psi_{100} \chi_{\uparrow}, \qquad \phi_2 = \psi_{100} \chi_{\downarrow}$$ and using the Slater determinant, we get $$\phi(1,2) = \psi_{100}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2) \left[\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) - \chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right].$$ We recognise the spin part as the "singlet", with total spin $0$. We also notice that this is the only possible wavefunction (up to overall phase) where both electrons are in the "$\psi_{100}$ state". I feel like I understand this case but I included it to make sure we're on the same page.

The problem (tl;dr): Consider now the case where one electron is in $\psi_{100}$ and the other is in some other "state", e.g. $\psi_{200}$. Griffiths tells me that we find possible two-electron states by constructing the symmetric and antisymmetric space-wavefunctions $$\psi_{\text{S}}(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}} \left(\psi_{100}(\mathbf{r}_1)\psi_{200}(\mathbf{r}_2) + \psi_{200}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2) \right),$$ $$\psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}} \left(\psi_{100}(\mathbf{r}_1)\psi_{200}(\mathbf{r}_2) - \psi_{200}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2) \right)$$ and pairing them with an antisymmetric and antisymmetric spin-functions respectively: $$\phi(1,2) = \psi_{\text{S}}(\mathbf{r}_1, \mathbf{r}_2)\left[\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) - \chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right], \qquad \text{(singlet)}$$ $$\phi(1,2) = \begin{cases} \psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2)\chi_{\uparrow}(s_1)\chi_{\uparrow}(s_2) & \\ \psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2)\left[\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) + \chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right] & \text{(triplet)} \\ \psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2)\chi_{\downarrow}(s_1)\chi_{\downarrow}(s_2) & \end{cases}$$

I can see why these are valid states, but it seems to me that these are not all the possible states. For example, letting
$$\phi_1 = \psi_{100} \chi_{\uparrow}, \qquad \phi_2 = \psi_{200} \chi_{\downarrow}$$ then using the Slater determinant, we get another antisymmetric wavefunction with total spin $0$: $$\phi(1,2) = \frac{1}{\sqrt{2}} \left[\psi_{100}(\mathbf{r}_1)\psi_{200}(\mathbf{r}_2)\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) - \psi_{200}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2)\chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right] .$$ How do these two approaches fit together? Is it possible to form this last wavefunction by taking linear combinations of the singlet/triplet wavefunctions? Can you form the singlet/triplet wavefunctions with slater determinants? Also, is there reason to prefer one approach to the other?

• Of course you can get every wavefunction from the singlet+triplet! With $\phi_1,\phi_2,\phi_3,\phi_4$ as your four singlet/triplet states in order, you have $\phi(1,2) = \phi_1(1,2) + \phi_3(1,2)$ (up to normalization). I'm not sure what the question about that is. – ACuriousMind Apr 26 '16 at 13:08
• @ACuriousMind That completely escaped me, but I see it now. Then I guess the question remains, can the singlet and triplet states above be written as Slater determinants, and is there any reason why these singlet/triplet states are predominantly used? – Pétur Apr 26 '16 at 13:27
• Well, writing them as Slater determinants is just another exercise in staring at these functions a bit longer until you see how, and "predominantly" is a primarily opinion-based question (as you can certainly find applications where you have superpositions of them), but surely many use them because they correspond to the irreducible representations of the rotation group, and hence a constant value of $S^2$, i.e. the singlet is spin-0 and the triplet is spin-1 (regardless of the "middle" triplet state having spin-z value of 0). – ACuriousMind Apr 26 '16 at 14:09
• @ACuriousMind Then I guess the question is pretty trivial in retrospect. If you would like to turn your comments into an answer, I will accept it. – Pétur Apr 26 '16 at 14:30

## 1 Answer

First, observe that the Slater determinant you have written down is the linear combination of the singlet state and the z-spin-0 state of the triplet. Vice versa, you can produce the singlet and triplet states as linear combinations of Slater determinants.

Whether you prefer the Slater determinant or the singlet/triplet formalism for writing down your two-fermion states depends on your personal preference and on the specific application you have in mind.