A lithium atom has three electrons. These occupy individual particle states corresponding to the sets of four quantum numbers given by .

$\left(n,l,m_l,m_j\right)\mathbf{=}\left(1,0,0,+\frac{1}{2}\right)\mathbf{,}\left(1,0,0,-\frac{1}{2}\right)\mathbf{}\mathbf{\text{and}}\mathbf{}\left(2,0,0,+\frac{1}{2}\right)$

Using ${\mathit{\psi }}_{\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}\left(r_j\right)\mathbf{\uparrow }\mathbf{,}{\mathit{\psi }}_{\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}\left(r_j\right)\mathbf{\downarrow }\mathbf{,}\mathbf{}\mathbf{\text{and}}\mathbf{}{\mathit{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}\left(r_j\right)\mathbf{\uparrow }$ to represent the individual-particle states when occupied by particle$j$ . Apply the Slater determinant discussed in Exercise 42 to find an expression for an antisymmetric multiparticle state. Your answer should be sums of terms like .

${\mathit{\psi }}_{\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}\left(r_1\right)\mathbf{\uparrow }\mathbf{,}{\mathit{\psi }}_{\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}\left(r_2\right)\mathbf{\downarrow }\mathbf{,}\mathbf{}\mathbf{\text{and}}\mathbf{}{\mathit{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}\left(r_3\right)\mathbf{\uparrow }$

The Slater determinant mainly describes a multi-fermionic system’s wave function. This determinant mainly satisfies the anti-symmetry requirements.

As there are three individual states of particle that are represented by the function ${\psi }_{1,0,0}\left(r_j\right)\uparrow {\psi }_{1,0,0}\left(r_j\right)\downarrow ,{\psi }_{2,0,0}\left(r_j\right)\uparrow$. For the j particle, the Slater determinant consists of three columns in which one column is for each state of the individual particle. There are also three types of electrons, hence the Slater determinant will consist of three rows which are $j=3$ to $j=1$ as one row has each electron.

With the help of the previous analysis, the next possible step is implementing Slater determinant and the equation is expressed as:

$\left|\begin{array}{ccc}{\psi }_{1,0,0}\left(r_1\right)\uparrow & {\psi }_{1,0,0}\left(r_1\right)\downarrow & {\psi }_{2,0,0}\left(r_1\right)\uparrow \\ {\psi }_{1,0,0}\left(r_2\right)\uparrow & {\psi }_{1,0,0}\left(r_2\right)\downarrow & {\psi }_{2,0,0}\left(r_2\right)\uparrow \\ {\psi }_{1,0,0}\left(r_3\right)\uparrow & {\psi }_{1,0,0}\left(r_3\right)\downarrow & {\psi }_{2,0,0}\left(r_3\right)\uparrow \end{array}\right|$

The above determinant mainly represents the three-electron antisymmetric state, let it be denoted by $\psi -\left(r_1,r_2,r_3\right)$. Hence, expanding the above determinant with the help of the first row, the above equation can be expressed as:

$$\begin{gathered} \psi -\left(r_1,r_2,r_3\right)={\psi }_{1,0,0}\left(r_1\right)\uparrow \left({\psi }_{1,0,0}\left(r_2\right)\downarrow {\psi }_{2,0,0}\left(r_3\right)\uparrow -{\psi }_{2,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\downarrow \right) \\ -{\psi }_{1,0,0}\left(r_1\right)\downarrow \left({\psi }_{1,0,0}\left(r_2\right)\uparrow {\psi }_{2,0,0}\left(r_3\right)\uparrow -{\psi }_{2,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\uparrow \right) \\ +{\psi }_{2,0,0}\left(r_1\right)\uparrow \left({\psi }_{1,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\downarrow -{\psi }_{1,0,0}\left(r_2\right)\downarrow {\psi }_{1,0,0}\left(r_3\right)\uparrow \right) \end{gathered}$$

The previous result can be expanded for writing the final answer.

$$\begin{gathered} \psi -\left(r_1,r_2,r_3\right)={\psi }_{1,0,0}\left(r_1\right)\uparrow {\psi }_{1,0,0}\left(r_2\right)\downarrow {\psi }_{1,0,0}\left(r_3\right)\uparrow -{\psi }_{1,0,0}\left(r_1\right)\uparrow {\psi }_{2,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\downarrow \\ -{\psi }_{1,0,0}\left(r_1\right)\downarrow {\psi }_{1,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\uparrow +{\psi }_{1,0,0}\left(r_1\right)\downarrow {\psi }_{2,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\uparrow \\ +{\psi }_{2,0,0}\left(r_1\right)\downarrow {\psi }_{1,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\uparrow -{\psi }_{2,0,0}\left(r_1\right)\uparrow {\psi }_{2,0,0}\left(r_2\right)\downarrow {\psi }_{1,0,0}\left(r_3\right)\uparrow \end{gathered}$$

Thus, the expression for an antisymmetric multiparticle state is .

$$\begin{gathered} {\psi }_{1,0,0}\left(r_1\right)\uparrow {\psi }_{1,0,0}\left(r_2\right)\downarrow {\psi }_{1,0,0}\left(r_3\right)\uparrow -{\psi }_{1,0,0}\left(r_1\right)\uparrow {\psi }_{2,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\downarrow \\ -{\psi }_{1,0,0}\left(r_1\right)\downarrow {\psi }_{1,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\uparrow +{\psi }_{1,0,0}\left(r_1\right)\downarrow {\psi }_{2,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\uparrow \\ +{\psi }_{2,0,0}\left(r_1\right)\downarrow {\psi }_{1,0,0}\left(r_2\right)\uparrow {\psi }_{1,0,0}\left(r_3\right)\uparrow -{\psi }_{2,0,0}\left(r_1\right)\uparrow {\psi }_{2,0,0}\left(r_2\right)\downarrow {\psi }_{1,0,0}\left(r_3\right)\uparrow \end{gathered}$$