Question

(a) If ${\mathit{\psi }}_{\mathbf{a}}$and${\mathit{\psi }}_{\mathbf{b}}$ are orthogonal, and both normalized, what is the constant A in Equation 5.10?

(b) Ifrole="math" localid="1658225858808" ${\mathit{\psi }}_{\mathbf{a}}\mathbf{=}{\mathit{\psi }}_{\mathbf{b}}$ (and it is normalized), what is A ? (This case, of course, occurs only for bosons.)

Short answer

(a) The required constant A is $\frac{1}{\sqrt{2}}$.

(b) The ${\psi }_a={\psi }_b$, A is $\frac{1}{2}$.

Step-by-step solution

Definition of normalisation

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.

Determine the normalization constant  A

(a)

Find the function's normalisation constant A.

For give equation-

${\mathrm{\psi }}_{\perp }(\mathrm{r}{⃗}_1,\mathrm{r}{⃗}_2)=\mathrm{A}\left[{\mathrm{\psi }}_{\mathrm{a}}\right(\mathrm{r}{⃗}_1\left){\mathrm{\psi }}_{\mathrm{b}}\right(\mathrm{r}{⃗}_2)\pm {\mathrm{\psi }}_{\mathrm{b}}(\mathrm{r}{⃗}_1\left){\mathrm{\psi }}_{\mathrm{a}}\right(\mathrm{r}{⃗}_2\left)\right]$

It must be valid when it normalise the function:

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

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

Determination of the A

(b)

if${\psi }_a={\psi }_b$then:

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

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