# Numerics values of Slater determinants with spin orbitals

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?

• Can you clarify your notation? Why do you have subscript 1's on your $\phi$'s? Do all of those states have the same spatial wavefunctions $\phi_1$ but different spin wavefunctions $\alpha$ and $\beta$? – J. Murray Nov 11 '19 at 23:47
• Yes, that's what I meant with it - actually I originally wanted to make it both into "1s" labels and then dropped the ball. My apologies. I've added them in. Of course, it doesn't strictly have to be 1s spatial orbitals, just as long as the spatial part for both electrons is the same so that the wavefunction would become 0 without the different spins. – Antimon Nov 12 '19 at 4:59
• $1s$ is actually not any more clarifying than no subscript at all. – user138962 Nov 12 '19 at 5:01

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 $$$$\int dx\ \left(\psi^{\sigma}_i(x)\right)^* \psi^{\rho}_j(x) = \delta_{\sigma \rho} \delta_{ij},$$$$ where $$\sigma$$ and $$\rho$$ indicate spin, and $$i$$ and $$j$$ indicate spatial orbital. So, $$$$\int dx\ \left(\psi^{\uparrow}_i(x)\right)^* \psi^{\uparrow}_j(x) = \int dx\ \left(\psi^{\downarrow}_i(x)\right)^* \psi^{\downarrow}_j(x) =\delta_{ij},$$$$ and $$$$\int dx\ \left(\psi^{\uparrow}_i(x)\right)^* \psi^{\downarrow}_j(x) = \int dx\ \left(\psi^{\downarrow}_i(x)\right)^* \psi^{\uparrow}_j(x) = 0.$$$$ This answers one of your questions$$-\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 $$$$\Psi(x_1,x_2) = \frac{1}{\sqrt{2}} \left[\phi(x_1) \bar{\phi}(x_2) - \phi(x_2)\bar{\phi}(x_1)\right].$$$$

• Sorry, but this doesn't answer any of the questions I had. I know that the spin orbitals are orthogonal by virtue of their spin part, even if the spatial part is the same. Your last equation is of course nothing else than the Slater determinant I wrote down in the first place. My main issue was if it was possible to calculate an actual numeric value of the given antisymmetrized wavefunction at specific electron coordinates. – Antimon Nov 12 '19 at 4:56
• The given wave function is not any different than the most general possible wave function for two identical electrons. No more information can be given without providing what kind of potential the electrons are subject to. – user138962 Nov 12 '19 at 5:00
• For sake of this argument, it's enough to assume that both of those orbitals are just hydrogenic 1s orbitals; it really doesn't matter as long as I have expressions for those functions. If my Slater determinant has no spin labels in it, I can just put in some electron coordinates and calculate an actual numeric value for the wavefunction, but what do I do if I have those spin labels in there? – Antimon Nov 12 '19 at 5:03
• How many electrons do you have? Because if it's more than two, the hydrogenic wave function isn't even valid. If your Slater determinant doesn't have spin labels, then it must be referring to one spin type. If it has labels, then you have two Slater determinants. – user138962 Nov 12 '19 at 5:08
• It's just those two electrons, one with $\alpha$ spin, one with $\beta$. The spin labels are indicated on the orbitals by overbars, as described in the original post. This Slater determinant is the exact same as e.g. eq. (30) here, if it helps to have another take on it. Isn't that just one Slater determinant? And it's equal to your last equation that you wrote down. – Antimon Nov 12 '19 at 5:16