91影视

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-

${\mathrm{蠄}}_{鈯�}(\mathrm{r}{鈨�}_1,\mathrm{r}{鈨�}_2)=\mathrm{A}\left[{\mathrm{蠄}}_{\mathrm{a}}\right(\mathrm{r}{鈨�}_1\left){\mathrm{蠄}}_{\mathrm{b}}\right(\mathrm{r}{鈨�}_2)卤{\mathrm{蠄}}_{\mathrm{b}}(\mathrm{r}{鈨�}_1\left){\mathrm{蠄}}_{\mathrm{a}}\right(\mathrm{r}{鈨�}_2\left)\right]$

It must be valid when it normalise the function:

$$\begin{gathered} 1={\mathrm{蠄}}^{*}{\mathrm{蠄诲}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ ={\left|\mathrm{A}\right|}^2{\left[{\mathrm{蠄}}_{\mathrm{a}}\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{b}},\left(\mathrm{r}{鈨�}_2\right)卤{\mathrm{蠄}}_{\mathrm{b}}\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{a}}\left(\mathrm{r}{鈨�}_2\right)\right]}^{*}\left[{\mathrm{蠄}}_{\mathrm{a}}\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{b}},\left(\mathrm{r}{鈨�}_2\right)卤{\mathrm{蠄}}_{\mathrm{b}}\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{a}}\left(\mathrm{r}{鈨�}_2\right)\right]{\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ \frac{1}{{\left|\mathrm{A}\right|}^2}={\left|{\mathrm{蠄}}_{\mathrm{a}}\left(\mathrm{r}{鈨�}_1\right)\right|}^2{\left|{\mathrm{蠄}}_{\mathrm{b}}\left(\mathrm{r}{鈨�}_2\right)\right|}^2{\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2卤{\mathrm{蠄}}_{\mathrm{a}}^{*}\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{b}},\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{b}}^{*}\left(\mathrm{r}{鈨�}_2\right){\mathrm{蠄}}_{\mathrm{b}}\left(\mathrm{r}{鈨�}_2\right){\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ 卤{\mathrm{蠄}}_{\mathrm{b}}^{*}\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{a}}\left(\mathrm{r}{鈨�}_1\right){\mathrm{蠄}}_{\mathrm{a}}^{*}\left(\mathrm{r}{鈨�}_2\right){\mathrm{蠄}}_{\mathrm{b}},\left(\mathrm{r}{鈨�}_2\right){\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2+{\left|{\mathrm{蠄}}_{\mathrm{a}}\left(\mathrm{r}{鈨�}_1\right)\right|}^2{\left|{\mathrm{蠄}}_{\mathrm{b}}\left(\mathrm{r}{鈨�}_2\right)\right|}^2{\mathrm{d}}^3{\mathrm{r}}_1{\mathrm{d}}^3{\mathrm{r}}_2 \\ =\left(1卤0卤0+1\right) \\ \frac{1}{{\left|\mathrm{A}\right|}^2}=2 \\ \mathrm{A}=\frac{1}{\sqrt{2}} \end{gathered}$$

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

(b)

if${蠄}_a={蠄}_b$then:

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

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