Question

Suppose you had three particles, one in state${\mathbf{\psi }}_{\mathbf{a}}\left(x\right)$, one in state${\mathbf{\psi }}_b\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, and one in state${\mathbf{\psi }}_c\mathbf{\left(}\mathbf{x}\mathbf{\right)}$. Assuming ${\mathbf{\psi }}_{\mathbf{a}}\mathbf{,}{\mathbf{\psi }}_{\mathbf{b}}$, and${\mathbf{\psi }}_c$ are orthonormal, construct the three-particle states (analogous to Equations 5.15,5.16, and 5.17) representing

(a) distinguishable particles,

(b) identical bosons, and

(c) identical fermions.

Keep in mind that (b) must be completely symmetric, under interchange of any pair of particles, and (c) must be completely antisymmetric, in the same sense. Comment: There's a cute trick for constructing completely antisymmetric wave functions: Form the Slater determinant, whose first row is${\mathbf{\psi }}_{\mathbf{a}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{,}{\mathbf{\psi }}_{\mathbf{b}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{,}{\mathbf{\psi }}_{\mathbf{c}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}$ , etc., whese second row is${\mathbf{\psi }}_{\mathbf{a}}\mathbf{\left(}{\mathbf{x}}_2\mathbf{\right)}\mathbf{,}{\mathbf{\psi }}_{\mathbf{b}}\mathbf{\left(}{\mathbf{x}}_2\mathbf{\right)}\mathbf{,}{\mathbf{\psi }}_{\mathbf{c}}\mathbf{\left(}{\mathbf{x}}_2\mathbf{\right)}$ , etc., and so on (this device works for any number of particles).

Short answer

a) The distinguishable particles: $\mathrm{\psi }\left({\mathrm{x}}_{1,}{\mathrm{x}}_{2,}{\mathrm{x}}_3\right){\mathrm{is\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)$

b) The identical bosons

$$\begin{gathered} \left({\mathrm{\psi }}_{1,}{\mathrm{\psi }}_{2,}{\mathrm{\psi }}_{3,}\right)-\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \\ +\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \end{gathered}$$

c) The identical fermions

$$\begin{gathered} \left({\mathrm{\psi }}_{1,}{\mathrm{\psi }}_{2,}{\mathrm{\psi }}_{3,}\right)-\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \\ +\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \end{gathered}$$

Step-by-step solution

Definition of identical bosons ,identical ferminos and slater determinant

• "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 Slater determinant is a formula in quantum mechanics that describes the wave function of a multi-fermionic system.

• It satisfies anti-symmetry criteria, and thus the Pauli principle, by changing sign when two electrons are exchanged (or other fermions)

Determine the Slater determinant

$$\begin{gathered} \mathrm{When}\mathrm{N}\mathrm{fermions}\mathrm{are}\mathrm{present},\mathrm{the}\mathrm{whole}\mathrm{wave}\mathrm{function}\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{as}: \\ {\mathrm{x}}_2,...,{\mathrm{x}}_{\mathrm{N}})=\frac{1}{\sqrt{\mathrm{N}!}}\left|\begin{array}{cccc}{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right)& {\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_1\right)& ...& {\mathrm{\psi }}_{\mathrm{N}}\left({\mathrm{x}}_1\right)\\ {\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right)& {\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_1\right)& ...& {\mathrm{\psi }}_{\mathrm{N}}\left({\mathrm{x}}_2\right)\\ ...& & & \\ {\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_{\mathrm{N}}\right)& {\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_{\mathrm{N}}\right)& ...& {\mathrm{\psi }}_{\mathrm{N}}\left({\mathrm{x}}_{\mathrm{N}}\right)\end{array}\right| \\ \mathrm{The}\mathrm{Slater}\mathrm{determinant}\mathrm{is}\mathrm{the}\mathrm{name}\mathrm{for}\mathrm{this}\mathrm{formula}. \end{gathered}$$

Determine the distinguishable particles

(a)

The total wave function for identifiable particles is simply the combination of three wave functions:

$\mathrm{\psi }\left({\mathrm{x}}_{1,}{\mathrm{x}}_{2,}{\mathrm{x}}_3\right)={\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)$

Determine the  identical bosons

(b)

When we permute any two particles, the entire wave function for identical bosons must be symmetric:

$$\begin{gathered} \mathrm{\psi }\left({\mathrm{x}}_{1,}{\mathrm{x}}_{2,}{\mathrm{x}}_{3,}\right)-\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \\ +\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \end{gathered}$$

Determine the  identical fermions

(c)

When we permute any two fermions, the whole wave function must be antisymmetric for identical fermions:

$$\begin{gathered} \mathrm{\psi }\left({\mathrm{x}}_{1,}{\mathrm{x}}_{2,}{\mathrm{x}}_{3,}\right)-\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \\ +\frac{1}{\sqrt{6}}\left[{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)+{\mathrm{\psi }}_{\mathrm{a}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{b}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\mathrm{c}}\left({\mathrm{x}}_3\right)\right] \end{gathered}$$