# Single-particle wavefunction in Slater determinant

The ground state of $$N$$ non-interacting fermions can be written using a Slater determinant as:

$$\Phi_{GS}(\textbf{r}_{1}, ..., \textbf{r}_{N}) = \frac{1}{\sqrt{N!}} \begin{vmatrix} \phi_{\mu_{1}}(\textbf{r}_{1}) & \cdots & \phi_{\mu_{N}}(\textbf{r}_{1})\\ \vdots & \ddots & \vdots \\ \phi_{\mu_{1}}(\textbf{r}_{N}) & \cdots & \phi_{\mu_{N}}(\textbf{r}_{N}) \end{vmatrix},$$

where $$\phi_{\mu} (\textbf{r}_i)$$ corresponds to single-particle wave functions, with $$\mu$$ orbital and $$\textbf{r}_{i}$$ coordinate of the $$i$$th particle.

What (basis) functions can you choose to describe $$\phi_{k}$$ ? Do they have to be orthogonal? For example, would $$\phi_{\mu} = e^{ik_i r_i}$$ , $$k_i$$ being the coordinate of the $$i$$th particle in momentum space, be a good choice?

• It is not necessarily (or even typically) true that the ground state is always a Slater determinant. There are other antisymmetric configurations that may sometimes have lower energy expectations.
– Buzz
Commented May 20, 2022 at 0:58

In the case of a ground state, for instance, you would want to pick as part of your set the lowest $$N/2$$ energy states of some relevant non-interacting Hamiltonian, and maybe a few states above the $$N/2$$th, especially if some states are closely spaced near this energy. If you choose a poor set, you will need lots of determinants to get reasonable results.
• Thank you for your answer. But would then $\phi_{\mu} = e^{ik_i r_i}$ be also a valid choice for single particle wave-function? Commented Feb 20, 2022 at 16:07
• It's a terrible choice because it is an eigenstate of a free particle (not a bound state), it cannot be normalized in the standard way and the state $\phi_\mu$ does not satisfy any reasonable boundary conditions for your system. I'm not an expert at using "bad" choices: maybe it can be done but it would certainly be an overly laborious choice in practice. Commented Feb 20, 2022 at 16:12