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) ,                     (13.1)

which for the application of the boundary conditions gives the solutions ψ_n(x) = (2/L)^1/2 sin(nπx/L) and E_n = 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 kinetic energy associated with the y direction.  The wave function 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) .                     (13.2)

How do we solve this time-independent Schrödinger equation?  We begin by using the technique of separation of variables to write the wave function as ψ(x,y) = ψ(x)ψ(y).  In doing so, we see that only the part of the wave function that is of one variable or the other gets differentiated by the kinetic energy terms.  Thus we expect

[E_nx + E_ny]ψ(x)ψ(y) = Eψ(x)ψ(y) ,                       (13.3)

which we can separate into

−(ħ²/2m)(d²/dx²) ψ(x) = E_xψ(x)       and             −(ħ²/2m)(d²/dy²) ψ(y) = E_xψ(y) .                 (13.4)

For the individual boundary conditions yields the solutions ψ_nx(x) = (1/a)^1/2 sin(n_xπ(x + 1)/2a) and E_nx = n_x²π²ħ²/(8ma²) and ψ_ny(y) = (1/b)^1/2 sin(n_xπ(y + 1)/2b) and E_ny = n_y²π²ħ²/(8mb²).  We have for the entire solution, therefore,

E_{nx ny} =  n_x²π²ħ²/(8ma²) + n_y²π²ħ²/(8mb²) = π²ħ²/(8m²) [n_x²/a² + n_y²/b²] ,                     (13.5)

 and ψ_{nx ny}(x,y) = (1/ab)^1/2 sin(n_xπ(x + 1)/a)sin(n_yπ(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 wave function 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 wave function and energy are simplification yields ψ_{nx ny} (x,y) = (1/L) sin(n_xπ(x + 1)/L)sin(n_yπ(y + 1)/L) and

E_{nx ny} =   = π²ħ²/(8mL²) [n_x² + n_y²] ,                     (13.6)

respectively. We note that this equation yields the energies

E₁₁ = 2E₁        E₂₁ = E₁₂ = 5E₁          E₂₂ = 8E₁         E₃₁ = E₁₃ = 10E₁ ,                      (13.7)

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.

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