Question

Suppose you could find a solution$\mathbf{\psi }\left(r_1,r_2,...,r_z\right)$to the Schrödinger equation (Equation 5.25), for the Hamiltonian in Equation 5.24. Describe how you would construct from it a completely symmetric function, and a completely anti symmetric function, which also satisfy the Schrödinger equation, with the same energy.

role="math" localid="1658219144812" $\hat{H}=\sum _{j=1}^Z\left\{-\frac{{ħ}^2}{2m}{\nabla }_j^2-\left(\frac{1}{4\pi {{\displaystyle \stackrel{,}{o}}}_0}\right)\frac{Z{e}^2}{r_j}\right\}+\frac{1}{2}\left(\frac{1}{4\pi {{\displaystyle \stackrel{,}{o}}}_0}\right)\sum _{j\ne 1}^Z\frac{e^2}{\left|r_j-r_k\right|}$ (5.24).

role="math" localid="1658219153183" $\hat{H}\psi =E$ (5.25).

Short answer

${\mathrm{\psi }}_{-}=\mathrm{\psi }\left({\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\right)-\mathrm{\psi }\left({\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\right)-\mathrm{\psi }\left({\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\right)+\mathrm{\psi }\left({\mathrm{r}}_2,{\mathrm{r}}_3,{\mathrm{r}}_1\right)-\mathrm{\psi }\left({\mathrm{r}}_3,{\mathrm{r}}_2,{\mathrm{r}}_1\right)+\left({\mathrm{r}}_3,{\mathrm{r}}_1,{\mathrm{r}}_2\right)-\mathrm{\psi }\left({\mathrm{r}}_1,{\mathrm{r}}_3,{\mathrm{r}}_2\right)=0$

Step-by-step solution

Definition of Schrodinger equation

The fundamental equation for characterizing quantum mechanical phenomena is the Schrodinger equation. It is a partial differential equation that illustrates how a physical system's wave function changes over time. The electron is a wave in the three-dimensional space surrounding the nucleus.

Finding a solution to Schrodinger equation

${\mathrm{\psi }}_{\pm }=\mathrm{A}\left[\mathrm{\psi }\left({\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3,..{\mathrm{r}}_{\mathrm{z}}\right)\pm \mathrm{\psi }\left({\mathrm{r}}_2,{\mathrm{r}}_1,{\mathrm{r}}_3,..,{\mathrm{r}}_{\mathrm{z}}\right)+\mathrm{\psi }\left({\mathrm{r}}_2,{\mathrm{r}}_3,{\mathrm{r}}_1,..,{\mathrm{r}}_{\mathrm{z}}\right)+\mathrm{etc}.\right],$

Where etc. runs over all permutations of the arguments,$\left({\mathrm{r}}_1,{\mathrm{r}}_2,..,{\mathrm{r}}_{\mathrm{z}}\right)$ with a + sign for all even permutations (even number of transpositions $r_i\leftrightarrow r_j$starting from $\left({\mathrm{r}}_1,{\mathrm{r}}_2,..,{\mathrm{r}}_{\mathrm{z}}\right)$and ± for all odd permutations (+ for bosons, – for fermions).

At the end of the process, normalize the result to determine A. (Typically $A=1/\sqrt{Z!}$, but this may not be right if the starting function is already symmetric under some interchanges.)

If ψ is symmetric in the first two arguments (or any other pair), the anti symmetric combination is zero.

For example, if Z = 3,

${\mathrm{\psi }}_{-}=\psi \left[\mathrm{\psi }\left({\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\right)-\mathrm{\psi }\left({\mathrm{r}}_2,{\mathrm{r}}_1,{\mathrm{r}}_3\right)+\mathrm{\psi }\left({\mathrm{r}}_2,{\mathrm{r}}_3,{\mathrm{r}}_1\right)-\left({\mathrm{r}}_3,{\mathrm{r}}_1,{\mathrm{r}}_2\right)+\mathrm{\psi }\left({\mathrm{r}}_3,{\mathrm{r}}_1,{\mathrm{r}}_2\right)-\mathrm{\psi }\left({\mathrm{r}}_1,{\mathrm{r}}_3,{\mathrm{r}}_2\right)\right]=0$

(They cancel in pairs).