5

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?

Himanshu
  • 12,211

2 Answers2

3

Single Slater determinant states only make up a small fraction of the possible antisymmetric states in a many-body Hilbert space. Indeed, you've written down one such state that is not expressible as a single Slater determinant. However, Slater determinant states do form a basis for the antisymmetric subspace of a many-body Hilbert space, meaning that any antisymmetric state is expressible as a linear combination of Slater determinant states. (For more on this, see here.) This is also what you're seeing when you observe that your state is expressible as a sum of single Slater determinant states.

Note that this is equivalent to the statement that fermionic Fock states are a basis for a fermionic many-body Hilbert space, as fermionic Fock states are single Slater determinant states.

As far as the projector you've written down, this acts trivially on antisymmetric states, acts as a Slater determinant on product states, and annihilates symmetric states. The full many-body Hilbert space includes orthogonal subspaces of states which are symmetric and anti-symmetric, and all bosonic and fermionic many-body states lie in these subspaces, respectively. There is more discussion of this here.

Chris
  • 444
1

Slater determinant is a prescription for forming a correctly anti-symmetrized state from single-particle states. The inverse is not true - that is not every anti-symmetric many-particle state can be represented as a sum of products of single-particle states.

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 $H_0(x_1) + H_0(x_2)$.

Remark: This answer is essentially an attempt to present in pedestrian way what has been already succinctly formulated in the answer by @Chris.

Roger V.
  • 68,984