Question

The Slater determinant is introduced in Exercise 42. Show that if states $\mathbf{n}$ and $\mathbf{n}\mathbf{\text{'}}$of the infinite well are occupied. with the particle in state $\mathbf{n}$ being spin up and the one in being spin down. then the Slater determinant yields the antisymmetric multiparticle state: ${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\uparrow }{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\downarrow }\mathbf{-}{\mathbf{\psi }}_{{\mathbf{m}}^{\mathbf{2}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\uparrow }$.

Short answer

The resultant answer is proved.

Step-by-step solution

Given data 

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

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

Slater Determinant is used to express multi-particle states for fermions of anti-symmetric character.

The two wave functions $\mathrm{\psi }\left({\mathrm{x}}_1\right)$ and $\mathrm{\psi }\left({\mathrm{x}}_2\right)$have states $\mathrm{n}$ and $n\text{'}$that are occupied b spin up for state$n$, and spin down for state $n\text{'}$.

The Slater determinant of these states can be expressed by the determinant of $2\times 2$matrix as follows:

$\begin{array}{l}\mathrm{\Psi }=\left|\begin{array}{l}{\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow \\ {\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_2\right)\uparrow {\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow \end{array}\right|\\ \mathrm{\Psi }={\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow -{\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_2\right)\uparrow {\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow \end{array}$

Therefore, the required result is proved.