quantum mechanics - What would be the Slater's determinant representation for an excited state?

Setup

Introducing this spinorbital notation:

\begin{align} \Psi_1=\chi_{(r1)}\alpha_{(\omega1)} = 1 \\ \Psi_1=\chi_{(r1)}\beta_{(\omega1)} = \bar{1} \end{align}

and the Slater's determinant, for a system of two electrons, is:

\begin{align} |\Psi\rangle = \frac{1}{\sqrt{2}} \Big| \begin{array}{cc} 1 & \bar{1} \\ 2 & \bar{2} \\ \end{array} \Big| = \frac{|1\bar{2}\rangle-|\bar{1}2\rangle}{\sqrt2} \end{align}

Question

Based on the example, what would be the Slater's determinant representation for an excited state like $\Psi_a^r$ (i.e. with parallel spins like: $|\bar{1}\bar{2}\rangle$) and $\Psi_{ab}^{rs}$ (feel free to use any number) ?

Answer

The slater determinant is only a "trick" to get a total antisymmetric wave function. This is required by the Pauli principle.

For understand this you need to think in indistinguishability of particles. So a any allowed state of a particle need to be assigned equally at each indistinguishable particle in your system. So if you have to state allowed: $|1\rangle$ and $|2\rangle$, then your state of the system is:

$$ |\psi\rangle=\frac{|1\rangle|2\rangle - |2\rangle|1\rangle}{\sqrt2} $$

For a $N$ allowed states is something like:

$$ |\psi \rangle= \frac{1}{\sqrt{N}}\sum_{\sigma \in S_n} p(\sigma) \bigotimes_{i=1}^N |\sigma_i\rangle $$

where $\sigma$ is a permutation of $(1,2,...,N)$, $p(\sigma)$ is the parity of the permutation $\sigma$, and $\sigma_i=\sigma(i)$ is the i-th number of the permutation $\sigma$. The big "O" with an "x" inside is:

$$ \bigotimes_{i=1}^n |\sigma _i\rangle= |\sigma _1\rangle \otimes |\sigma _2\rangle \otimes ...\otimes |\sigma _n\rangle = |\sigma _1,\sigma _2,...,\sigma _n \rangle $$ where $\otimes$ is a tensor product.

Note that the tensor product is not commutative. Then, if you fixing some basis as the position ones, you get:

$$ \psi (x_1,x_2,...x_n)= \frac{1}{\sqrt{N}}\sum_{\sigma \in S_n} p(\sigma) \prod_{i=1}^n \psi_i(x_{\sigma_i}) $$

The non-commutativity goes to $i$ labels, that maintains the order of the permutation $\sigma$. This is precise the formula of the determinant as you can see.

If you want to deal with extra state, you need to do the enumeration $(n,l,m_l,s)\rightarrow i$. e.g. the states allowed is $n=p_1,p_2$ and $s=\pm \frac{1}{2}$. We can enumerate: $(p_1,+\frac{1}{2})\rightarrow 1$, $(p_2,+\frac{1}{2})\rightarrow 2$, $(p_1,-\frac{1}{2})\rightarrow 3$ and $(p_2,-\frac{1}{2})\rightarrow 4$. Some properties of determinant can help you to understanding the structure of this antisymmetric state.