Question

The Slater determinant is introduced in Exercise 42. Show that if states $\mathit{n}$and $\mathbf{n}\mathbf{\text{'}}$of the infinite well are occupied and both spins are up, the Slater determinant yields the antisymmetric multiparticle state:${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\uparrow }{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\uparrow }\mathbf{-}{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\uparrow }{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\uparrow }$

Short answer

The resultant answer is proved.

Step-by-step solution

Given data

The given data is ${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow$.

Concept of Slater determinant

A determinant is an expression that describes the wave function of a multi-fermionic system.

It satisfies anti-symmetry requirements, and consequently the Pauli principle, by changing sign upon exchange of two electrons (or other fermions).

Simplify the expression

The multi particle for fermions of anti-symmetric character can be expressed with the help of Slater determinant.

The two wave functions$\mathrm{\psi }\left({\mathrm{x}}_1\right)$ and $\mathrm{\psi }\left({\mathrm{x}}_2\right)$ have states $\mathrm{n}$ and ${\mathrm{n}}^{\text{'}}$ that are occupied for both spins up.

The Slater determinant of these states can be expressed by the determinant of $2\times 2$matrix as shown below.

$\begin{array}{l}\mathrm{\psi }=\left|\begin{array}{ll}{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow & {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow \\ {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow & {\mathrm{\psi }}_{\text{m}}\left({\mathrm{x}}_2\right)\uparrow \end{array}\right|\\ \mathrm{\psi }={\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{m}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow \end{array}$