Numerics values of Slater determinants with spin orbitals

Question

Consider a Slater determinant of a two-electron system with spin orbitals $\phi$ and $\overline{\phi}$, i.e.

$$ \Psi = \frac{1}{\sqrt{2}}\left| \begin{matrix} \phi_{1s}(x_1) & \overline{\phi_{1s}}(x_1) \\ \phi_{1s}(x_2) & \overline{\phi_{1s}}(x_2) \\ \end{matrix} \right| $$

with an overbar indicating $\beta$ spin, and no overbar $\alpha$, and both electrons occupy a 1s state.

I am interested in simply calculating a value for $\Psi$ at some specific set of electron coordinates $x_1$ and $x_2$. However I must somehow deal with the spin functions contained in these orbitals, which to my understanding aren't really functions after all but rather symbols to indicate orthogonality. (Disclaimer: This question is coming from a background of quantum chemistry where concepts and formalisms may not be taught as stringently as considered necessary from a physicist's perspective.)

Apparently, in Quantum Monte Carlo the wavefunction is split into spin-up and spin-down Slater determinants, $\Psi = \psi^{\alpha}\psi^{\beta}$. (See also this unanswered question.) How exactly this is done? Does $\psi^{\alpha}$ contain only the spin orbitals with $\alpha$ spin, $\psi^{\beta}$ only those with $\beta$ spin? Then our two-electron wavefunction would be $\Psi = \phi_1(x_1) \overline{\phi_1}(x_2)$, which does not obey the Pauli principle.

So, how would one calculate numeric values of wavefunctions containing spin orbitals at specific sets of electron coordinates? Is this possible at all?

Answer 1

The notation here is a bit confusing, so I suggest taking a step back from the Slater determinant. It is only a compact way of writing the fermionic antisymmetry. In particular, the subscript 1 on the wave functions is unnecessary, and I will omit it in my response.

It seems that you are mixing spatial and spin wave functions. We can give each of these separate indices: if we have a set of orthogonal spatial wave functions $\psi_i$, then if we also impart spin, we obtain a second set of orthogonal spatial wave functions. We can distinguish these two sets by adding up and down arrows as $\psi_i^{\uparrow}$, $\psi_i^{\downarrow}$. For different spatial orbitals $i$ and $j$, the wave functions are orthogonal regardless of spin. For identical spatial orbitals with different spin, the wave functions are still orthogonal. We can capture both of these features by writing \begin{equation} \int dx\ \left(\psi^{\sigma}_i(x)\right)^* \psi^{\rho}_j(x) = \delta_{\sigma \rho} \delta_{ij}, \end{equation} where $\sigma$ and $\rho$ indicate spin, and $i$ and $j$ indicate spatial orbital. So, \begin{equation} \int dx\ \left(\psi^{\uparrow}_i(x)\right)^* \psi^{\uparrow}_j(x) = \int dx\ \left(\psi^{\downarrow}_i(x)\right)^* \psi^{\downarrow}_j(x) =\delta_{ij}, \end{equation} and \begin{equation} \int dx\ \left(\psi^{\uparrow}_i(x)\right)^* \psi^{\downarrow}_j(x) = \int dx\ \left(\psi^{\downarrow}_i(x)\right)^* \psi^{\uparrow}_j(x) = 0. \end{equation} This answers one of your questions$-$$\psi^{\alpha}$ only concerns electrons with spin $\alpha$, and $\psi^{\beta}$ only concerns electrons with spin $\beta$. If we know we have $N_{\alpha}$ electrons of spin $\alpha$ and $N_{\beta}$ electrons of spin $\beta$, then we can safely antisymmetrize $\psi^{\alpha}$ and $\psi^{\beta}$ separately.

In quantum Monte Carlo, the wave function you refer to, $\Psi = \phi(x_1) \bar{\phi}(x_2)$, actually does not violate the Pauli principle. This is because these studies consider separately the number of spin-up and spin-down electrons. The Pauli principle applies to fermions with all quantum numbers equal. If you know that one has spin up and one has spin down, though, they have different quantum numbers, and the principle does not apply. It does still apply within each spin sector, so for $\psi^{\alpha} = \psi(x_1,x_2,\ldots,x_n)$, where all $n$ electrons have spin $\alpha$, $\psi^{\alpha}$ must be antisymmetric with respect to all $n$ arguments. $\psi^{\beta}$ must obey a similar condition, but it is independent of $\psi^{\alpha}$.

If the only difference between $\phi(x)$ and $\bar{\phi}(x)$ is spin, then the total wave function can be written as \begin{equation} \Psi(x_1,x_2) = \frac{1}{\sqrt{2}} \left[\phi(x_1) \bar{\phi}(x_2) - \phi(x_2)\bar{\phi}(x_1)\right]. \end{equation}