Question

(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).

Short answer

(a) For two non-interacting particles in a square box, we established that a given wave function is an eigenfunction of a Hamiltonian in the first task.

The energies of two distinct particles are: $\begin{array}{c}{\mathrm{E}}_{22}\mathrm{is}8\mathrm{K}\\ {\mathrm{E}}_{13}\mathrm{is}10\mathrm{K}\end{array}$

(b) The energies of two identical bosons are as follows: $\begin{array}{c}{\mathrm{E}}_{22}\mathrm{is}8\mathrm{K}\\ {\mathrm{E}}_{13}\mathrm{is}10\mathrm{K}\end{array}$

The energies of two identical fermions are as follows: $$\begin{gathered} {\mathrm{E}}_{13}\mathrm{is}10\mathrm{K} \\ {\mathrm{E}}_{23}\mathrm{is}13\mathrm{K}. \end{gathered}$$

Step-by-step solution

Definition of Hamiltonian and Bosons, 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.

Determine the Hamiltonian for two noninteracting identical particles

(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:

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

Determine the functions and energies for each of the three cases (distinguishable, identical bosons, identical fermions).

(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:

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