Java Security Update: Oracle has updated the security settings needed to run Physlets.
Click here for help on updating Java and setting Java security.
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, ψn(x) = (2/L)1/2 sin(nπx/L) and En = 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) = EΨ(x1,x2) .
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 Ψn1 n2(x1,x2) = (2/L)sin(n1πx1/L)sin(n2πx2/L) and
En1 n2 = n12π2ħ2/(8mL2) + n22π2ħ2/(8mL2) = π2ħ2/(8mL2) [n12 + n22] ,
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
ΨS(x1,x2) = 2−1/2 [Ψ12(x1,x2) + Ψ12 (x2,x1)], (13.5)
ΨA(x1,x2) = 2−1/2 [Ψ12(x1,x2) −Ψ12 (x2,x1)], (13.6)
as a symmetric and an antisymmetric solution, respectively. ΨS(x1,x2) is symmetric under interchange because ΨS(x1,x2) = ΨS(x2,x1). Likewise, ΨA(x1,x2) is antisymmetric under interchange because ΨA(x1,x2) = −ΨA(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
|ΨS(x1,x2)|2 = |ΨS(x2,x1)|2 (13.7)
|ΨA(x1,x2)|2 = |ΨA(x2,x1)|2, (13.8)
respectively, where |ΨS(x1,x2)|2 ≠ |ΨA(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).