(a) Write down the Hamiltonian for two noninteracting identical particles in the infinite square well. Verify that the fermion ground state given in Example 5.1 is an eigenfunction of H, with the appropriate eigenvalue.

(b) Find the next two excited states (beyond the ones in Example 5.1) - wave functions and energies - for each of the three cases (distinguishable, identical bosons, identical fermions).

• The Hamiltonian of a system expresses its total energy that is, the sum of its kinetic (motion) and potential (position) energy in terms of the Lagrangian function developed from prior studies of dynamics and the position and momentum of individual particles.
• "Particles come in two types: the particles that make up matter, known as 'fermions,' and the particles that transport forces, known as 'bosons,' according to Carroll.
• Fermions take up space, whereas bosons can be stacked on top of one another.

(a)

It must demonstrate that the following function exists $\psi (x_1,x_2)$ For two non-interacting particles in an infinite square well, is the eigenfunction of a Hamiltonian:

$$\begin{gathered} \mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{\sqrt{2}}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{2{\mathrm{\pi x}}_2}{\mathrm{a}}-\sin \frac{2{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{a}} \\ \hat{\mathrm{H}}\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)=\mathrm{E\psi }({\mathrm{x}}_1,{\mathrm{x}}_2) \\ \hat{\mathrm{H}}\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{-{\hslash }^2}{2\mathrm{m}}{\nabla }_1^2+{\nabla }_2^2\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2) \\ \frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}_1^2}=\frac{\sqrt{2}}{\mathrm{a}}\frac{-{\mathrm{\pi }}^2}{{\mathrm{a}}^2}\sin \frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{2{\mathrm{\pi x}}_2}{\mathrm{a}}+\frac{4{\mathrm{\pi }}^2}{{\mathrm{a}}^2}\sin \frac{2{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{a}} \\ \frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}_2^2}=\frac{\sqrt{2}}{\mathrm{a}}\frac{-4{\mathrm{\pi }}^2}{{\mathrm{a}}^2}\sin \frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{2{\mathrm{\pi x}}_2}{\mathrm{a}}+\frac{1{\mathrm{\pi }}^2}{{\mathrm{a}}^2}\sin \frac{2{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{a}} \\ \hat{\mathrm{H}}\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{-{\hslash }^2}{2\mathrm{m}}\frac{-5{\mathrm{\pi }}^2}{{\mathrm{a}}^2}\hspace{0.33em}\frac{\sqrt{2}}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{2{\mathrm{\pi x}}_2}{\mathrm{a}}-\sin \frac{2{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{a}} \\ \hat{\mathrm{H}}\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{5{\hslash }^2{\mathrm{\pi }}^2}{2{\mathrm{ma}}^2}\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2) \\ \mathrm{E}=\frac{5{\hslash }^2{\mathrm{\pi }}^2}{2{\mathrm{ma}}^2}=5\mathrm{K} \end{gathered}$$

(b)

Particles that can be distinguished

We write total for identifiable particles. As a product, the wave function:

$$\begin{gathered} {\mathrm{\psi }}_{{\mathrm{n}}_1,{\mathrm{n}}_2}({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{2}{\mathrm{a}}\sin \frac{{\mathrm{n}}_1{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{n}}_2{\mathrm{\pi x}}_2}{\mathrm{a}} \\ \mathrm{E}=({\mathrm{n}}_1^2+{\mathrm{n}}_2^2)\mathrm{K} \end{gathered}$$

Second level of elation$n_1-n_2-2:$

isn't a degenerate Degenerate is the third excited state $\begin{array}{c}{n}_1=1\\ n_2=3\end{array}$ or

$$\begin{gathered} {\mathrm{n}}_1=3, \\ {\mathrm{n}}_2=1 \\ {\mathrm{\psi }}_{13}=\frac{2}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{3{\mathrm{\pi x}}_2}{\mathrm{a}}. \\ {\mathrm{\psi }}_{31}=\frac{2}{\mathrm{a}}\sin \frac{3{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{a}} \\ {\mathrm{E}}_{13}-{\mathrm{E}}_{31}=10\mathrm{K} \end{gathered}$$

Bosons that are identical

The overall wave function for identical bosons is:

${\mathrm{\psi }}_{{\mathrm{n}}_1,{\mathrm{n}}_2}({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{\sqrt{2}}{\mathrm{a}}\sin \frac{{\mathrm{n}}_1{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{n}}_2{\mathrm{\pi x}}_2}{\mathrm{a}}+\sin \frac{{\mathrm{n}}_2{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{n}}_1{\mathrm{\pi x}}_2)}{\mathrm{a}}$

The situation$n_1=n_2=2$ is the same as in the preceding example:

$$\begin{gathered} {\mathrm{\psi }}_{22}({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{2}{\mathrm{a}}\sin \frac{2{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{2{\mathrm{\pi x}}_2}{\mathrm{a}} \\ {\mathrm{E}}_{22}=8\mathrm{K} \end{gathered}$$

$\begin{array}{c}{n}_1=1\\ n_2=3\end{array}$is a non-degenerate state:

$$\begin{gathered} {\mathrm{\psi }}_{1,3}({\mathrm{x}}_1,{\mathrm{x}}_2)=-\frac{\sqrt{2}}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{3{\mathrm{\pi x}}_2}{\mathrm{a}}+\sin \frac{3{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{a}} \\ {\mathrm{E}}_{13}=10\mathrm{K} \end{gathered}$$

fermions that are identical

The overall wave function for identical fermions is:

${\mathrm{\psi }}_{{\mathrm{n}}_1,{\mathrm{n}}_2}({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{\sqrt{2}}{\mathrm{a}}\sin \frac{{\mathrm{n}}_1{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{n}}_2{\mathrm{\pi x}}_2}{\mathrm{a}}-\sin \frac{{\mathrm{n}}_2{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{n}}_1{\mathrm{\pi x}}_2}{\mathrm{a}}$

There is no such state as$n_1=n_2=2$

role="math" localid="1658232959725" $\begin{array}{c}{n}_1=1\\ n_2=3\end{array}$is a non-degenerate state:

$$\begin{gathered} {\mathrm{\psi }}_{1,3}({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{\sqrt{2}}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{3{\mathrm{\pi x}}_2}{\mathrm{a}}-\sin \frac{3{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{{\mathrm{\pi x}}_2}{\mathrm{a}} \\ {\mathrm{E}}_{13}=10\mathrm{K} \end{gathered}$$

The following state is$n_1=2,n_2=3$, which is also non-degenerate:

$$\begin{gathered} {\mathrm{\psi }}_{23}({\mathrm{x}}_1,{\mathrm{x}}_2)=\frac{\sqrt{2}}{\mathrm{a}}\sin 2\frac{{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{3{\mathrm{\pi x}}_2}{\mathrm{a}}-\sin \frac{3{\mathrm{\pi x}}_1}{\mathrm{a}}\sin \frac{2{\mathrm{\pi x}}_2}{\mathrm{a}} \\ {\mathrm{E}}_{23}=13\mathrm{K} \end{gathered}$$