Triplet Square Well

Consider a two-electron system in one dimension, where both electrons have spins aligned in the same direction (say, up). They interact only through the attractive square well in relative coordinates

(i)

What is the lowest energy of the two-electron state? Assume the total momentum is zero.

SOLUTION

Since the two spins are parallel, they are in a spin triplet state with S = 1 and M = ±1. The spin eigenfunction has even parity. The two-electron wave function is written as an orbital part ѱ(x₁, x₂) times the spin part. The total wave function must have odd parity. Since the spin has even parity, the orbital part must have odd parity: ѱ(x₁, x₂) = - ѱ(x₁, x₂). Since the interaction potential acts only between the electrons, it is natural to write the orbital part in center-of-mass coordinates, where X = (x₁ + x₂)/2 and x = x₁ – x₂.

(1)

The problem stated that the total momentum was zero, so set k = 0. We must now determine the form for the relative eigenfunction ѱ(x). It obeys the Schrodinger equation with the reduced mass μ = m/2, where m is the electron mass:

(2)

We have reduced the problem to solving the bound state of a “particle” in a box. Here the “particle” is the relative motion of two electrons. However, since the orbital part of the wave function must have odd parity, we need to find the lowest energy state which is antisymmetric, ѱ(-x) = -ѱ(x).

Bound states have E = -E_B where the binding energy E_B > 0. Define two wave vectors: α² = 2μ E­­_B/h² for outside the box, |x| > a, and k = 2μ (V₀ - E­­_B)/h² when the particle is in the box, |x| < a. The lowest antisymmetric wave function is

(3)

We match the wave function and its derivative at one edge, say x = a, which gives two equations:

(4)

(5)

We divide these two equations, which eliminates the constants A and B. The remaining equation is the eigenvalue equation for α:

(6)

(7)

(8)

Since α and k are both positive, the cotangent of ka must be negative, which requires that ka > π/2. This imposes a constraint for the existence of any antisymmetric bound state:

(9)

Any attractive square well has a bound state which is symmetric, but the above condition is required for the antisymmetric bound state.