Question

Suppose you have three particles, and three distinct one-particle state$\left({\Psi }_{\mathrm{a}}\left(X\right),{\Psi }_{\mathrm{b}}\left(X\right),\mathrm{and}{\mathrm{\Psi }}_{\mathrm{c}}\left(\mathrm{x}\right)\right)$are available. How many different three-particle states can be constructed (a) if they are distinguishable particles, (b) if they are identical bosons, (c) if they are identical fermions? (The particles need not be in different states $-{\Psi }_{\mathrm{a}}\left(X_1\right),{\Psi }_a\left(X_2\right){\mathrm{\Psi }}_a\left({\mathrm{x}}_3\right)$would be one possibility, if the particles are distinguishable.)

Short answer

(a)27

(b)10

(c)1

Step-by-step solution

(a)if they are distinguishable

Each particle has 3 possible states: 3 × 3 × 3 = 27.

(b) if they are identical bosons

All in same state: aaa, bbb, ccc⇒3.

2 in one state: aab, aac, bba, bbc, cca, ccb⇒6 (each symmetrized).

3 different states: abc (symmetrized)⇒1.

Total: 10.

If they are identical fermions

Only abc (antisymmetrized) ⟹ 1.