91影视

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

The symmetric function is

${\mathbf{蠄}}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\mathbf{\left(}\mathbf{(}\mathbf{sin}\frac{{\mathbf{蟺虫}}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{\mathbf{2}{\mathbf{蟺虫}}_{\mathbf{2}}}{\mathbf{L}}\mathbf{+}\mathbf{sin}\frac{\mathbf{2}{\mathbf{蟺虫}}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{{\mathbf{蟺虫}}_{\mathbf{2}}}{\mathbf{L}}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

${\mathbf{蠄}}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{蟺虫}}_1}{L}\sin \frac{2{\mathrm{蟺虫}}_2}{L}鈭�\sin \frac{2{\mathrm{蟺虫}}_1}{L}\sin \frac{{\mathrm{蟺虫}}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}\mathbf{..}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

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

$\mathbf{蠄}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{蟺虫}}_1}{L}\sin \frac{2{\mathrm{蟺虫}}_2}{L}卤\sin \frac{2{\mathrm{蟺虫}}_1}{L}\sin \frac{{\mathrm{蟺虫}}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}\mathbf{..}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Where "+" sign represents symmetric wave function and 鈥� $\mathbf{-}$鈥� 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" ${\mathbf{\left|}\mathbf{蠄}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{\right|}}^{\mathbf{2}}{\mathbf{dx}}_{\mathbf{1}}{\mathbf{dx}}_{\mathbf{2}}$,${\mathbf{\left|}\mathbf{蠄}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{\right|}}^{\mathbf{2}}{\mathbf{dx}}_{\mathbf{1}}{\mathbf{dx}}_{\mathbf{2}}$

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,

${\mathrm{P}}_{\text{righl}}={鈭�}_0^{\mathrm{I}/2}{鈭�}_0^{\mathrm{I}/2}\left|\mathrm{蠄}({\mathrm{x}}_1,{\mathrm{x}}_2)\right|{\mathrm{dx}}_1{\mathrm{dx}}_2鈥�鈥�\mathrm{..}\left(4\right)$

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

${\mathrm{P}}_{\text{rightr}}$ localid="1659950570417" $=\frac{2}{{\mathrm{L}}^2}{鈭�}_0^{\mathrm{L}/2}{鈭�}_0^{\mathrm{I}/2}{\left(\sin \frac{\mathrm{蟺虫}]}{\mathrm{L}}\sin \frac{2\mathrm{蟺虫z}}{\mathrm{L}}卤\sin \frac{2{\mathrm{蟺虫}}_1}{\mathrm{L}}\sin \frac{{\mathrm{蟺虫}}_2}{\mathrm{L}}\right)}^2\mathrm{dx}1\mathrm{dxz}鈥�鈥�.\left(5\right)$

$P_{\text{rights}}=\frac{4}{L^2}\left({鈭�}_0^{I/2}{\sin }^2\frac{蟺x}{L}dx\right)\left({鈭�}_0^{1/2}{\sin }^2\frac{2蟺x}{L}dx\right)$localid="1659950915322" $卤\frac{4}{{\mathrm{l}}^2}{\left({鈭�}_0^{\mathrm{L}/2}\sin \frac{2\mathrm{蟺虫}}{\mathrm{L}}\sin \frac{\mathrm{蟺虫}}{\mathrm{L}}\mathrm{dx}\right)}^2鈥�鈥�鈥�$(6)

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

${鈭�}_0^{1/\mathrm{L}}\sin \frac{2\mathrm{蟺虫}}{\mathrm{L}}\sin \frac{\mathrm{蟺虫}}{\mathrm{L}}\mathrm{dx}={鈭�}_0^{\mathrm{IIL}}\left(2\sin \frac{\mathrm{蟺虫}}{\mathrm{L}}\cos \frac{\mathrm{蟺虫}}{\mathrm{L}}\right)\sin \frac{\mathrm{蟺虫}}{\mathrm{L}}\mathrm{dx}=\frac{2\mathrm{L}}{\mathrm{蟺}}{鈭�}_0^{1/2}{\sin }^2\frac{\mathrm{蟺虫}}{\mathrm{L}}\mathrm{d}\left(\sin \frac{\mathrm{蟺虫}}{\mathrm{L}}\right)=\frac{\mathrm{LL}}{3\mathrm{蟺}}鈥�.\left(7\right)$

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

$鈭�{\sin }^2\mathrm{axdx}=\frac{1}{2}\mathrm{x}鈭�\frac{1}{4\mathrm{a}}\sin 2\mathrm{ax}鈥�鈥�\text{(8)}$

And can be evaluated as

${鈭�}_0^{\mathrm{I}/2}{\sin }^2\frac{\mathrm{n蟺虫}}{\mathrm{L}}\mathrm{dx}={\left[\frac{1}{2}\mathrm{x}鈭�\frac{\mathrm{L}}{4\mathrm{苍蔚}}\sin \frac{2\mathrm{n蟺虫}}{\mathrm{L}}\right]}_0^{\mathrm{I}/2}=\frac{\mathrm{L}}{4}鈥�鈥�\mathrm{..}\text{(9)}$

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

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