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Section 13.1: The Two-dimensional Infinite Square Well
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Recall that in one-dimension, the infinite square well confines a particle to be between 0 to L in the x direction. In this case, the time-independent Schrödinger equation is
−(ħ2/2m)(d2/dx2) ψ(x) = Eψ(x) ,
which for the application of the boundary conditions gives the solutions ψn(x) = (2/L)1/2 sin(nπx/L) and En = n2π2ħ2/(2mL2) where n = 1, 2, 3,….
We now seek to extend this to two dimensions for a symmetric two-dimensional infinite well, where V(x,y) = 0 for −a < x < a and −b < y < b and V(x,y) = ∞ elsewhere. The Hamiltonian has changed from the one-dimensional case with the addition of a term that involves the derivative in the y direction. The energy eigenfunction must therefore also be a function of both x and y and satisfy the two-dimensional time-independent Schrödinger equation:
[−(ħ2/2m)(∂2/∂x2) − (ħ2/2m)(∂2/∂y2)] ψ(x,y) = Eψ(x,y) . (13.1)
How do we solve this time-independent Schrödinger equation? We begin by using the technique of separation of variables to write the energy eigenfunction as ψ(x,y) = ψ(x)ψ(y). In doing so, we see that only the part of the energy eigenfunction that is of one variable or the other gets differentiated by the differential terms. Thus we expect
[Enx + Eny]ψ(x)ψ(y) = Eψ(x)ψ(y) ,
which we can separate into
−(ħ2/2m)(d2/dx2) ψ(x) = Exψ(x) and −(ħ2/2m)(d2/dy2) ψ(y) = Eyψ(y) .
For the individual boundary conditions yields the solutions ψnx(x) = (1/a)1/2 sin(nxπ(x + 1)/2a) and Enx = nx2π2ħ2/(8ma2) and ψny(y) = (1/b)1/2 sin(nxπ(y + 1)/2b) and Eny = ny2π2ħ2/(8mb2). We have for the entire solution, therefore,
Enx ny = nx2π2ħ2/(8ma2) + ny2π2ħ2/(8mb2) = π2ħ2/(8m2) [nx2/a2 + ny2/b2] , (13.2)
and ψnx ny(x,y) = (1/ab)1/2 sin(nxπ(x + 1)/a)sin(nyπ(y + 1)/b).
Consider the case where a = b = L which is shown in the animation (ħ = 2m = 1 and L = 1). You can view the energy eigenfunction in either a three-dimensional plot or as a contour plot or view the probability density as a three-dimensional plot or as a contour plot. For such a square well, the energy eigenfunction and energy are simplification yields
ψnx ny (x,y) = (1/L) sin(nxπ(x + 1)/L)sin(nyπ(y + 1)/L) (13.3)
Enx ny = = π2ħ2/(8mL2) [nx2 + ny2] , (13.4)
respectively. We note that this equation yields the energies
E11 = 2E1 E21 = E12 = 5E1 E22 = 8E1 E31 = E13 = 10E1 ,
and so forth. We therefore find degeneracies of the energy eigenvalues which is due to the geometrical symmetry of the problem. The result that some energies are degenerate is called a symmetry degeneracy.