In Example 5.1 and Problem 5.5(b) we ignored spin (or, if you prefer, we assumed the particles are in the same spin state).
(a) Do it now for particles of spin 1/2. Construct the four lowest-energy configurations, and specify their energies and degeneracies. Use the notation Ψ_n1n2 |sm), where Ψ_n1n2 is defined in Example 5.1 and |sm) in Section 4.4.3.
(b) Do the same for spin 1.

Problem 5.5 (b)

Find the next two excited states (beyond the ones given in the example)—wave functions, energies, and degeneracies—for each of the three cases (distinguishable, identical bosons, identical fermions).

Example 5.1

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Example 5.1 Suppose we have two noninteracting (they pass right through one another...never mind how you would set this up in practice!) particles, both of mass m, in the infinite square well (Section 2.2). The one-particle states are Vn(x) = 2 =√√² sin (1x). En = n² K a (where K = ²²/2ma²). If the particles are distinguishable, with number 1 in state 71 and number 2 in state 12, the composite wave function is a simple product: Yn 12 (x1, x2) = Yn1 (x1 )Vn₂ (x2), _En\n² = (n² + n²) K. For example, the ground state is 2 V₁1 = sin (72¹) sin (22). E₁1 = 2K: the first excited state is doubly degenerate: 2 V/12 = sin a V/21 = E12 = 5K, sin (271) sin (X²), E21 = 5K; and so on. If the two particles are identical bosons, the ground state is unchanged, but the first excited state is nondegenerate: a 210 (TXL) S a a 2x2 a sin [sin() sin (22)+sin(2x) sin()] (still with energy 5K). And if the particles are identical fermions, there is no state with energy 2K; the ground state is a ЛХ [sin (¹) sin (22)- sin (2x) sin (²)]. a a a a and its energy is 5 K.