Question

Two particles in a box have a total energy $\mathbf{5}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\hslash }}^{\mathbf{2}}\mathbf{/}\mathbf{2}{\mathbf{mL}}^{\mathbf{2}}$.

(a) Which states are occupied?

(b) Make a sketch of${\mathbf{P}}_{\mathbf{5}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}$versus${\mathbf{x}}_{\mathbf{1}}$for points along the line${\mathbf{x}}_{\mathbf{2}}\mathbf{=}{\mathbf{x}}_{\mathbf{1}}$.

(c) Make a similar sketch of${\mathbf{P}}_{\mathbf{\text{A}}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}$.

(d) Repeat parts (b) and (c) but for points on the line${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{L}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}$

Short answer

(a) The state of one particle is ${\mathrm{n}}_1=1$ and other particles is${\mathrm{n}}_2=2$

(b) The sketch is shown in figure 1

(c) The sketch is shown in figure 2

(d) The sketch is shown in figure 1 with curve B for symmetric function and curve A for asymmetric function.

Step-by-step solution

Given information:

The total energy of particle is $5{\mathrm{\pi }}^2{\mathrm{\hslash }}^2/2{\mathrm{mL}}^2$.

Concept of probability density and energy

(a) The expression for energy of two particles in one dimension is given by,

$\mathbf{E}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mL}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mL}}^{\mathbf{2}}}$

(b) The expression for probability density is given by,role="math" localid="1659954573251" $\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{\left[}\frac{\mathbf{1}}{\mathbf{L}}\left(\sin \left(\frac{\mathrm{\pi x}}{L}\right)\left(\sin \frac{2\mathrm{\pi x}}{L}\right)\pm \sin \left(\frac{2\mathrm{\pi x}}{L}\right)\left(\sin \frac{\mathrm{\pi x}}{L}\right)\right)\mathbf{\right]}}^{\mathbf{2}}$

(c) The expression for probability density is given by,

role="math" localid="1659954661753" $\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{\left[}\frac{\mathbf{1}}{\mathbf{L}}\left(\sin \left(\frac{\mathrm{\pi x}}{L}\right)\left(\sin \frac{2\mathrm{\pi x}}{L}\right)\pm \sin \left(\frac{2\mathrm{\pi x}}{L}\right)\left(\sin \frac{\mathrm{\pi x}}{L}\right)\right)\mathbf{\right]}}^{\mathbf{2}}$

(d) The expression for probability density is given by,

$\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\left[\frac{1}{L}\left(\sin \left(\frac{{\mathrm{\pi x}}_1}{L}\right)\left(\sin \frac{2{\mathrm{\pi x}}_2}{L}\right)\pm \sin \left(\frac{2{\mathrm{\pi x}}_1}{L}\right)\left(\sin \frac{{\mathrm{\pi x}}_2}{L}\right)\right)\right]}^{\mathbf{2}}$

Evaluate total energy 

(a)

Suppose ${\mathrm{n}}_1=1$and ${\mathrm{n}}_2=2$.

The total energy is calculated as,

$\mathrm{E}=\frac{{\mathrm{n}}_1^2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}+\frac{{\mathrm{n}}_2^2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}=\frac{{\left(1\right)}^2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}+\frac{{\left(2\right)}^2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}$

Evaluate probability density of (b)

(b)

The probability density for symmetric function is calculated as,

$\mathrm{P}\left(\mathrm{x}\right)={\left[\frac{1}{\mathrm{L}}\left(\sin \left(\frac{\mathrm{\pi x}}{\mathrm{L}}\right)\left(\sin \frac{2\mathrm{\pi x}}{\mathrm{L}}\right)+\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{L}}\right)\left(\sin \frac{\mathrm{\pi x}}{\mathrm{L}}\right)\right)\right]}^2=\frac{4}{{\mathrm{L}}^2}{\sin }^2\left(\frac{\mathrm{\pi x}}{\mathrm{L}}\right){\sin }^2\left(\frac{2\mathrm{\pi x}}{\mathrm{L}}\right)$

The graph between probability density and $\mathrm{x}$for symmetric and asymmetric function is shown below,

Figure 1: Curve 1 shows the symmetric function and curve 2 shown asymmetric functions.

Evaluate probability density of (c)

(c)

The probability density for anti-symmetric function is calculated as,

$\mathrm{P}\left(\mathrm{x}\right)={\left[\frac{1}{\mathrm{L}}\left(\sin \left(\frac{\mathrm{\pi x}}{\mathrm{L}}\right)\left(\sin \frac{2\mathrm{\pi x}}{\mathrm{L}}\right)-\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{L}}\right)\left(\sin \frac{\mathrm{\pi x}}{\mathrm{L}}\right)\right)\right]}^2=0$

The graph between probability density and $\mathrm{x}$for symmetric and asymmetric function is shown below,

Figure 2: Curve 1 shows the symmetric function and curve 2 shown asymmetric function

Evaluate probability density of (d)

(d)

The probability density is calculated as,

Because of the extra minus sign in the above equation comparison between part (b) and (c) shows P(x) for the symmetric wave function along the path is equal to$\mathrm{P}\left(\mathrm{x}\right)$for anti-symmetric case in case of part (b) and (c), Corresponding to curve B in figure 1 .

Similarly for anti-symmetric wave function along the path is equal to $\mathrm{P}\left(\mathrm{x}\right)$for symmetric case in case of part (b) and (c), corresponding to curve in figure 1.