# Why can't you always write the anti-symmetric state as slater determinant?

For example, consider the following state $$|\Psi\rangle_a = [|r_1 r_2\rangle -|r_2 r_1\rangle ]\otimes \left[|\uparrow \downarrow\rangle +|\downarrow \uparrow \rangle \right]$$ You can't write this as a slater determinant but can write as sum of two slater determinants.

What does this show? From what I understand is that, The anti-symmetric Projector

$$A\equiv \frac{1}{N!}\sum_\alpha \epsilon_\alpha \mathcal{P}_\alpha$$ projector only basic set of Anti-symmetric subspace of vector space $$\mathcal{V}$$ and one need to form a linear combination to get the whole states. I'm not able to make this idea more rigorous. Is that true or always true? What's the better way to say it?

• Well, even for two distinguishable particles, the general wave function cannot be written as a simple tensor. Instead, it is a linear combination of such. Why do you expect that the case for two indistinguishable fermions will be different? Feb 11, 2022 at 11:00

Let us consider an example of two particles: $$H(x_1,x_2)=H_0(x_1) + H_0(x_2) + V(x_1,x_2),\\ H_0(x)\phi_n(x)=\epsilon_n\phi_n(x)$$ A valid state where particles have quantum numbers $$n,m$$ of non-interacting single-particle Hamiltonian can be written only as $$\psi_{n,m}(x_1,x_2)=\frac{1}{\sqrt{2}}\left[\phi_n(x_1)\phi_m(x_2)-\phi_n(x_2)\phi_m(x_1)\right]$$ Any valid two-particle state is anti-symmetric, $$\Psi(x_1,x_2)=-\Psi(x_2,x_1)$$ and can be expanded in terms of the single-particle basis as $$\Psi(x_1,x_2)=\sum_{n,m}c_{n,m}\psi_{n,m}(x_1,x_2)=\sum_{n,m}\frac{c_{n,m}}{\sqrt{2}}\left[\phi_n(x_1)\phi_m(x_2)-\phi_n(x_2)\phi_m(x_1)\right]$$ However, this is obviously not the only possible expansion and not the only possible basis. Moreover, for a non-zero interaction $$V(x_1,x_2)$$ we do not really expect that the eigenfunctions would be the same as those of the non-interacting Hamiltonian $$H_0(x_1) + H_0(x_2)$$.