91Ó°ÊÓ

In essence, normalizing the wave function entails figuring out the precise shape that guarantees the probability that the particle will be discovered somewhere in space is equal to 1 (i.e., it will be discovered somewhere); this typically entails solving for a constant while keeping in mind the restriction above that the probability is equal to 1.

(a)

Find the function's normalisation constant A.

For give equation-

${\mathrm{ψ}}_{âŠ\yen }(\mathrm{r}{⃗}_1,\mathrm{r}{⃗}_2)=\mathrm{A}\left[{\mathrm{ψ}}_{\mathrm{a}}\right(\mathrm{r}{⃗}_1\left){\mathrm{ψ}}_{\mathrm{b}}\right(\mathrm{r}{⃗}_2)Â\pm {\mathrm{ψ}}_{\mathrm{b}}(\mathrm{r}{⃗}_1\left){\mathrm{ψ}}_{\mathrm{a}}\right(\mathrm{r}{⃗}_2\left)\right]$

It must be valid when it normalise the function:

$$\begin{gathered} 1={\mathrm{ψ}}^{*}{\mathrm{ψ»å}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ ={\left|\mathrm{A}\right|}^2{\left[{\mathrm{ψ}}_{\mathrm{a}}\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{b}},\left(\mathrm{r}{⃗}_2\right)Â\pm {\mathrm{ψ}}_{\mathrm{b}}\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{a}}\left(\mathrm{r}{⃗}_2\right)\right]}^{*}\left[{\mathrm{ψ}}_{\mathrm{a}}\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{b}},\left(\mathrm{r}{⃗}_2\right)Â\pm {\mathrm{ψ}}_{\mathrm{b}}\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{a}}\left(\mathrm{r}{⃗}_2\right)\right]{\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ \frac{1}{{\left|\mathrm{A}\right|}^2}={\left|{\mathrm{ψ}}_{\mathrm{a}}\left(\mathrm{r}{⃗}_1\right)\right|}^2{\left|{\mathrm{ψ}}_{\mathrm{b}}\left(\mathrm{r}{⃗}_2\right)\right|}^2{\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2Â\pm {\mathrm{ψ}}_{\mathrm{a}}^{*}\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{b}},\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{b}}^{*}\left(\mathrm{r}{⃗}_2\right){\mathrm{ψ}}_{\mathrm{b}}\left(\mathrm{r}{⃗}_2\right){\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ Â\pm {\mathrm{ψ}}_{\mathrm{b}}^{*}\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{a}}\left(\mathrm{r}{⃗}_1\right){\mathrm{ψ}}_{\mathrm{a}}^{*}\left(\mathrm{r}{⃗}_2\right){\mathrm{ψ}}_{\mathrm{b}},\left(\mathrm{r}{⃗}_2\right){\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2+{\left|{\mathrm{ψ}}_{\mathrm{a}}\left(\mathrm{r}{⃗}_1\right)\right|}^2{\left|{\mathrm{ψ}}_{\mathrm{b}}\left(\mathrm{r}{⃗}_2\right)\right|}^2{\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ =\left(1Â\pm 0Â\pm 0+1\right) \\ \frac{1}{{\left|\mathrm{A}\right|}^2}=2 \\ \mathrm{A}=\frac{1}{\sqrt{2}} \end{gathered}$$

Hence the value of A is, $\frac{1}{\sqrt{2}}$.

(b)

if${\mathrm{ψ}}_a={\mathrm{ψ}}_b$then:

$$\begin{gathered} 1={\left|\mathrm{A}\right|}^2∫{\left[2{\mathrm{ψ}}_{\mathrm{a}}\left({\mathrm{r}}_1\right){\mathrm{ψ}}_{\mathrm{a}}\left({\mathrm{r}}_2\right)\right]}^{*}\left[2{\mathrm{ψ}}_{\mathrm{a}}\left({\mathrm{r}}_1\right){\mathrm{ψ}}_{\mathrm{a}}\left({\mathrm{r}}_2\right)\right]{\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ =4{\mathrm{A}}^2∫{\left|{\mathrm{ψ}}_{\mathrm{a}}\left({\mathrm{r}}_1\right)\right|}^2{\mathrm{d}}^3{\mathrm{r}}_1∫{\left|{\mathrm{ψ}}_{\mathrm{a}}\left({\mathrm{r}}_2\right)\right|}^2{\mathrm{d}}^3{\mathrm{r}}_2 \\ =4{\mathrm{A}}^2 \\ \mathrm{A}=\frac{1}{2} \end{gathered}$$

Hence the value of A is,$\frac{1}{2}$ .