Two particles in a box occupy the $\mathbf{n}\mathbf{=}\mathbf{1}$and${\mathbf{n}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{2}$individual-particle states. Given that the normalization constant is the same as in Example$\mathbf{8}\mathbf{.}\mathbf{2}$(see Exercise 36), calculate for both the symmetric and antisymmetric states the probability that both particles would be found in the left side of the box (i.e., between 0 and$\frac{\mathbf{1}}{\mathbf{3}}\mathbf{L}$)?

Two particles in a box occupy the$\mathrm{n}=1$ and${\mathrm{n}}^{\text{'}}=2$ individual-particle states.

The symmetric function is

${\mathbf{\psi }}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\mathbf{\left(}\mathbf{(}\mathbf{sin}\frac{{\mathbf{\pi x}}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{\mathbf{2}{\mathbf{\pi x}}_{\mathbf{2}}}{\mathbf{L}}\mathbf{+}\mathbf{sin}\frac{\mathbf{2}{\mathbf{\pi x}}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{{\mathbf{\pi x}}_{\mathbf{2}}}{\mathbf{L}}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

${\mathbf{\psi }}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{\pi x}}_1}{L}\sin \frac{2{\mathrm{\pi x}}_2}{L}-\sin \frac{2{\mathrm{\pi x}}_1}{L}\sin \frac{{\mathrm{\pi x}}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{..}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Equations (1) and (2) can be written as

$\mathbf{\psi }\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{\pi x}}_1}{L}\sin \frac{2{\mathrm{\pi x}}_2}{L}\pm \sin \frac{2{\mathrm{\pi x}}_1}{L}\sin \frac{{\mathrm{\pi x}}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{..}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Where "+" sign represents symmetric wave function and “ $\mathbf{-}$” sign represents anti-symmetric wave function. Probability to find one particle in the two particle wave function in rangerole="math" localid="1659950017144" ${\mathbf{x}}_{\mathbf{1}}$to${\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathit{d}{\mathit{x}}_{\mathbf{1}}$and the another in rangerole="math" localid="1659950029523" ${\mathbf{x}}_{\mathbf{2}}$to${\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{dx}}_{\mathbf{2}}$isrole="math" localid="1659950072860" ${\mathbf{\left|}\mathbf{\psi }\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{\right|}}^{\mathbf{2}}{\mathbf{dx}}_{\mathbf{1}}{\mathbf{dx}}_{\mathbf{2}}$,${\mathbf{\left|}\mathbf{\psi }\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{\right|}}^{\mathbf{2}}{\mathbf{dx}}_{\mathbf{1}}{\mathbf{dx}}_{\mathbf{2}}$

The probability of finding both particles in the right half of the box, that is for both$\mathrm{x}$and x z in the range from 0 to$\text{L}/2$is therefore,

${\mathrm{P}}_{\text{righl}}={\int }_0^{\mathrm{I}/2}{\int }_0^{\mathrm{I}/2}\left|\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)\right|{\mathrm{dx}}_1{\mathrm{dx}}_2\dots \dots \mathrm{..}\left(4\right)$

Substitute the value of $\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)$from equation (3) into Equation (4) we get

${\mathrm{P}}_{\text{rightr}}$ localid="1659950570417" $=\frac{2}{{\mathrm{L}}^2}{\int }_0^{\mathrm{L}/2}{\int }_0^{\mathrm{I}/2}{\left(\sin \frac{\mathrm{\pi x}]}{\mathrm{L}}\sin \frac{2\mathrm{\pi xz}}{\mathrm{L}}\pm \sin \frac{2{\mathrm{\pi x}}_1}{\mathrm{L}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{L}}\right)}^2\mathrm{dx}1\mathrm{dxz}\dots \dots .\left(5\right)$

$P_{\text{rights}}=\frac{4}{L^2}\left({\int }_0^{I/2}{\sin }^2\frac{\pi x}{L}dx\right)\left({\int }_0^{1/2}{\sin }^2\frac{2\pi x}{L}dx\right)$localid="1659950915322" $\pm \frac{4}{{\mathrm{l}}^2}{\left({\int }_0^{\mathrm{L}/2}\sin \frac{2\mathrm{\pi x}}{\mathrm{L}}\sin \frac{\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx}\right)}^2\dots \dots \dots$(6)

The integral${\mathrm{x}}_1$in equation (6).

${\int }_0^{1/\mathrm{L}}\sin \frac{2\mathrm{\pi x}}{\mathrm{L}}\sin \frac{\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx}={\int }_0^{\mathrm{IIL}}\left(2\sin \frac{\mathrm{\pi x}}{\mathrm{L}}\cos \frac{\mathrm{\pi x}}{\mathrm{L}}\right)\sin \frac{\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx}=\frac{2\mathrm{L}}{\mathrm{\pi }}{\int }_0^{1/2}{\sin }^2\frac{\mathrm{\pi x}}{\mathrm{L}}\mathrm{d}\left(\sin \frac{\mathrm{\pi x}}{\mathrm{L}}\right)=\frac{\mathrm{LL}}{3\mathrm{\pi }}\dots .\left(7\right)$

The integrals ${\int }_0^{1/2}{\sin }^2\frac{\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx}$in equation (6) have the form

$\int {\sin }^2\mathrm{axdx}=\frac{1}{2}\mathrm{x}-\frac{1}{4\mathrm{a}}\sin 2\mathrm{ax}\dots \dots \text{(8)}$

And can be evaluated as

${\int }_0^{\mathrm{I}/2}{\sin }^2\frac{\mathrm{n\pi x}}{\mathrm{L}}\mathrm{dx}={\left[\frac{1}{2}\mathrm{x}-\frac{\mathrm{L}}{4\mathrm{n\epsilon }}\sin \frac{2\mathrm{n\pi x}}{\mathrm{L}}\right]}_0^{\mathrm{I}/2}=\frac{\mathrm{L}}{4}\dots \dots \mathrm{..}\text{(9)}$

Use Equation$\left(7\right)$ and$\left(9\right)$ to evaluate Equation (6)

$\mathrm{P}=\frac{4}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{4}\right)}^2\pm \frac{4}{{\mathrm{L}}^2}{\left(\frac{2\mathrm{L}}{3\mathrm{x}}\right)}^2=0.25\pm \frac{16}{9{\mathrm{x}}^2}=0.25\pm 0.1803=0.430\text{or}0.070$