Section 13.2: Two Particles in a One-dimensional Infinite Well

Please wait for the animation to completely load.

What if we allowed two identical particles into a single infinite well?  If they are non-interacting, the particles would move independently of each other.  (They would not even collide; they would pass directly through each other!)  For one particle we find that for an infinite well from 0 < x < L, ψ_n(x) = (2/L)^1/2 sin(nπx/L) and E_n = n²π²ħ²/(2mL²).  Now add another particle.  The Hamiltonian has changed with the addition of terms that involve the second particle.  The total wave function must now be a function of both positions x₁ and x₂:

[−(ħ²/2m)(∂²/∂x₁²) − (ħ²/2m)(∂²/∂x₂²)] ψ(x₁,x₂) = E ψ(x₁,x₂) .                (13.8)

Since each particle (one at x₁ and one at x₂) is independent of each other, solving this problem proceeds like the two-dimensional case in Section 13.1 and we find that ψ_{n1 n2}(x₁,x₂) = (2/L)sin(n₁πx₁/L)sin(n₂πx₂/L) and

E_{n1 n2} =  n₁²π²ħ²/(8mL²) + n₂²π²ħ²/(8mL²) = π²ħ²/(8mL²) [n₁² + n₂²] ,               (13.9)

which yields the same energy spectrum as in Section 13.1.  This degeneracy, however, is due to the exchange of particle one (n₁) at position x₁ with particle two (n₂) at position x₂.  This is called an exchange degeneracy.  Any time we have two solutions that give the same energy, we can consider a linear combination of these solutions as another solution (since this linear solution will have the same energy as its composite states).  For the case of two particles in a one-dimensional well, we combine the ψ₁₂(x₁,x₂) with the ψ₁₂(x₂,x₁)  solution as

ψ_S(x₁,x₂) = 2^−1/2 [ψ₁₂(x₁,x₂) + ψ₁₂ (x₂,x₁)],                (13.10)

 and

ψ_A(x₁,x₂) = 2^−1/2 [ψ₁₂(x₁,x₂) − ψ₁₂ (x₂,x₁)],                (13.11)

as a symmetric and an antisymmetric solution, respectively.  Ψ_S(x₁,x₂) is symmetric under interchange because ψ_S(x₁,x₂) = ψ_S(x₂,x₁).  Likewise, ψ_A(x₁,x₂) is antisymmetric under interchange because ψ_A(x₁,x₂) = −ψ_A(x₂,x₁).  We can take the square |ψ|² of each wave function and we find that these expressions are invariant to the exchange of the particles

 |ψ_S(x₁,x₂)|² = |ψ_S(x₂,x₁)|²        and         |ψ_A(x₁,x₂)|² = |ψ_A(x₂,x₁)|²,                 (13.12)

respectively, where |ψ_S(x₁,x₂)|² ≠ |ψ_A(x₁,x₂)|².  Shown in the animation are the probability densities for two identical particles in an infinite well with L = 2.  The x axis corresponds to the position of particle 1 and the y axis corresponds to the position of particle 2.  The probability densities depicted are for one of the particles in the ground state and the other particle in the first excited state.  Animation 1 shows the symmetric solution, while Animation 2 shows the antisymmetric solution.  Note that the diagonal line across the animation (x = y) corresponds to x₁ = x₂, which is where both particles are at the same position.  For the antisymmetric case, one never finds both particles at the same position (probability density is zero along this line).  For the symmetric case, one is likely to find both particles at the same position (probability density is non-zero, and is actually the largest at certain points along this line).

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