3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ Commented Feb 11, 2022 at 11:00

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.