Question

Two identical spin-zero bosons are placed in an infinite square well (Equation 2.19). They interact weakly with one another, via the potential

$\mathit{V}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}\mathbf{,}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathit{a}{\mathbf{V}}_{\mathbf{0}}\mathbf{\delta }\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{.}$ (2.19).

(where ${\mathit{V}}_{\mathbf{0}}$is a constant with the dimensions of energy, and a is the width of the well).

(a)First, ignoring the interaction between the particles, find the ground state and the first excited state—both the wave functions and the associated energies.

(b) Use first-order perturbation theory to estimate the effect of the particle– particle interaction on the energies of the ground state and the first excited state.

Short answer

(a) Ground state:

${\mathbf{\Psi }}_{\mathbf{1}}^{\mathbf{0}}\left(x_1,x_2\right)\mathbf{=}{\mathbf{\Psi }}_{\mathbf{1}}\left(x_1\right){\mathbf{\Psi }}_{\mathbf{1}}\left(x_2\right)\mathbf{=}\frac{\mathbf{2}}{\mathbf{a}}\mathit{s}\mathit{i}\mathit{n}\left(\frac{\pi x_1}{a}\right)\mathit{s}\mathit{i}\mathit{n}\left(\frac{\pi x_2}{a}\right)\mathbf{;}{\mathbf{E}}_{\mathbf{1}}^{\mathbf{0}}\mathbf{=}\mathbf{2}{\mathbf{E}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{ħ}}^{\mathbf{2}}}{\mathbf{m}{\mathbf{a}}^{\mathbf{2}}}$

First excited state:

${\mathbf{\Psi }}_{\mathbf{2}}^{\mathbf{0}}\left(x_1,x_2\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left[{\Psi }_1\left(x_1\right){\Psi }_2\left(x_2\right)+{\Psi }_2\left(x_1\right){\Psi }_1\left(x_2\right)\right]\mathbf{.}\mathbf{=}{\mathit{E}}_{\mathbf{2}}^{\mathbf{0}}\mathbf{=}{\mathit{E}}_{\mathbf{1}}\mathbf{+}{\mathit{E}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{5}}{\mathbf{2}}\frac{{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{ħ}}^{\mathbf{2}}}{\mathbf{m}{\mathbf{a}}^{\mathbf{2}}}$

(b)$\mathbf{-}\frac{\mathbf{32}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\pi }}\left(\frac{3\pi }{8}-\frac{5\pi }{16}\right)\mathbf{=}\mathbf{-}\mathbf{2}{\mathit{V}}_{\mathbf{0}}$

Step-by-step solution

(a) Finding the ground state and the first excited state

In terms of the one-particle states (Eq. 2.28) and energies (Eq. 2.27):

${\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{a}}}\sin \left(\frac{\mathrm{n\pi }}{\mathrm{a}}\mathrm{x}\right)$ (2.28).

$E_n=\frac{{ħ}^2{k}_n^2}{2m}=\frac{n^2{\pi }^2{ħ}^2}{2m{a}^2}$ (2.27).

Ground state:${\mathrm{\Psi }}_1^0\left({\mathrm{x}}_1{\mathrm{x}}_2\right)={\mathrm{\Psi }}_1\left({\mathrm{x}}_1\right){\mathrm{\Psi }}_1\left({\mathrm{x}}_2\right)=\frac{2}{\mathrm{a}}\sin \left(\frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\right)\sin \left(\frac{{\mathrm{\pi x}}_2}{\mathrm{a}}\right);E_1^0=2E_1=\frac{{\pi }^2{ħ}^2}{m{a}^2}$

First excited state:${\mathrm{\Psi }}_2^0\left({\mathrm{x}}_1{\mathrm{x}}_2\right)=\frac{1}{\sqrt{2}}\left[{\mathrm{\Psi }}_1\left({\mathrm{x}}_1\right){\mathrm{\Psi }}_2\left({\mathrm{x}}_2\right)+{\mathrm{\Psi }}_2\left({\mathrm{x}}_1\right){\mathrm{\Psi }}_1\left({\mathrm{x}}_2\right)\right]$

 Step2: (b) estimating the effect of the particle– particle interaction on the energies of the ground state and the first excited state

$$\begin{gathered} =\left(-a{V}_0\right)\left(\frac{2}{a^2}\right)\int {\int }_0^a{\left[\sin \left(\frac{\pi x_1}{a}\right)\sin \left(\frac{2\pi x_2}{a}\right)+\sin \left(\frac{2\pi x_1}{a}\right)\sin \left(\frac{\pi x_2}{a}\right)\right]}^2\delta \left(x_1-x_2\right)d{x}_{1}d{x}_{2.} \\ =-\frac{2V_0}{a}{\int }_0^a{\left[\sin \left(\frac{\pi x}{a}\right)\sin \left(\frac{2\pi x}{a}\right)+\sin \left(\frac{2\pi x}{a}\right)+\sin \left(\frac{2\pi x}{a}\right)\sin \left(\frac{\pi x}{a}\right)\right]}^{2}dx=-\frac{8V_0}{a}{\int }_0^a\sin \left(\frac{\pi x}{a}\right)\sin \left(\frac{2\pi x}{a}\right)dx \\ =-\frac{8V_0}{a}.\frac{a}{\pi }{\int }_0^{\pi }{\sin }^{2}y{\sin }^2\left(2y\right)dy \\ =-\frac{8V_0}{a}.4{\int }_0^{\pi }{\sin }^{2}y{\sin }^{2}y{\cos }^{2}ydy=-\frac{32V_0}{\pi }{\int }_0^{\pi }({\sin }^{4}y-{\sin }^{6}y)dy \\ -\frac{32V_0}{\pi }\left(\frac{3\pi }{8}-\frac{5\pi }{16}\right)=-2V_{0.} \end{gathered}$$