For identical bosons the full wave function is symmetric under the particle exchange operator:

ψ (x_{2}, x_{1}) = ψ (x_{1}, x_{2}) .

In the given position-only model this is the spatial parity under the particle exchange operator. If additional spin variables are present, the symmetry applies to the complete state including them.

In particular it vanishes when the two complete one-particle arguments coincide. Applying this to the position-only model assumes that no independent spin symmetry changes the required spatial parity.

Solution

 0  0

The orthogonal coordinate change preserves the kinetic energy and takes the particle exchange operator to

(x_{+}, x_{-}) \mapsto (x_{+}, - x_{-})

. Thus the spatial wave function is even in

x_{-}

for bosons and odd for fermions.

The total potential energy and Hamiltonian separate as

V = \frac{m ω ^{2}}{2} x_{+}^{2} + \frac{m ω ^{2} ( 1 + 2 κ )}{2} x_{-}^{2}, H = H_{+} (ω) + H_{-} (Ω), Ω = ω 1 + 2 κ .

(1)

Assume

1 + 2 κ > 0

so the relative oscillator is confining. Its spectrum is

E_{n_{+}, n_{-}} = ℏ ω (n_{+} + 1/2) + ℏΩ (n_{-} + 1/2)

, and its parity in the two-particle relative coordinate is

(- 1)^{n_{-}}

The lowest boson state uses

n_{+} = n_{-} = 0

, whereas the lowest fermion state uses

n_{+} = 0

n_{-} = 1

. Therefore

E_{B} = \frac{1}{2} ℏ ω [1 + 1 + 2 κ], E_{F} = \frac{1}{2} ℏ ω [1 + 3 1 + 2 κ] .

(2)

For an oscillator of frequency

Ω

⟨ x_{-}^{2} ⟩_{n} = (n + 1/2) ℏ/ (m Ω)

. Since

(x_{1} - x_{2})^{2} = 2 x_{-}^{2}

⟨(x_{1} - x_{2})^{2} ⟩_{B} = \frac{ℏ}{m Ω}, ⟨(x_{1} - x_{2})^{2} ⟩_{F} = \frac{3ℏ}{m Ω} .

(3)

The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied oscillator wave functions correspond to kinetic energy

P^{2} / (2 m)

; the extra factor

1/2

printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional kinetic energy dictated by particle mass