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Section 10.2: The Quantum-mechanical Infinite Square Well

In the infinite square well potential, a particle is confined to a box of length L by two infinitely high potential energy barriers:

V = ∞ x ≤ 0 , V = 0    0 < x < L , V = ∞ xL .

We begin with the time-independent Schrödinger equation in one dimension for 0 < x < L:

−(ħ2/2m)(d2/dx2) ψ(x) = E ψ(x) .

The solution to this differential equation is a combination of sines and cosines (or complex exponentials; here, because of the boundary conditions, we choose sines and cosines). The general solution is

ψ(x) = Asin(kx) + Bcos(kx) ,

where k2 = 2mE/ħ2.

We still need to satisfy the boundary conditions, and thereby determine A, B, and k (and therefore E). For the wall on the left we need ψ(x) = 0, for x ≤ 0 and for the wall on the right we need ψ(x) = 0 for xL, which means that

ψ(0) = 0    and    ψ(L) = 0 .

The cosine part of the general solution does not vanish at the x = 0 boundary since, cos(0) ≠ 0, and therefore we must have B = 0. We are left with ψ(x) = Asin(kx) and determining both A and k. This situation is shown in the animation. The boundary condition at x = 0 is already solved and you can vary the energy to see the effect on the energy eigenfunction (this is the shooting method).  In the animation, ħ = 2m = 1.  Restart.  For what values of the energy (and therefore k) is the boundary condition at x = L satisfied?

At the boundary x = L we must have a k such that ψ(L) = Asin(kL) = 0. Satisfying this boundary condition can be accomplished by requiring that kL = nπ, where n = 0, ±1, ±2, ±3,…. We can eliminate the negative values of n as these energy eigenfunctions are different from the positive n values by just an overall phase factor of −1. The n = 0 value must be considered more carefully. If we allow n = 0, this yields k = 0 and amounts to a zero-curvature solution to the time-independent Schrödinger equation. A zero-curvature solution is of the form Ax + B and cannot satisfy the boundary conditions and be non-zero in the well. Therefore, k = 0 is not a valid possibility2 and we have that:

ψn(x) = Asin(nπx/L) for 0 < x < L   with n = 1, 2, 3,… , (10.2)

and zero otherwise. Since k = (2mE/ħ2)1/2, the energy becomes:

En = n2π2ħ2/2mL2. (10.3)

But what about A? The energy eigenfunction must be normalized  to satisfy Born's probabilistic interpretation, so

∫ ψn*(xn(x) dx =  1 . [integral from −∞ to +∞] (10.7)

Since ψn(x) is a real function and most of the spatial integral vanishes because the energy eigenfunction is zero everywhere except between 0 and L, we are simply left with calculating

A2 ∫ sin2(nπx/L) dx     →    A2 ∫ sin2(nπx/L) dx = A2L/2 = 1 ,       [integral from −∞ to +∞] (10.8)

which tells us that A = (2/L)1/2. Therefore the energy eigenfunction

ψn(x) = (2/L)1/2sin(nπx/L) for 0 < x < L (10.4)

satisfies the normalization condition. We also can consider the integral

∫ ψm*(x)ψn(x) dx = Ln cos(nπ)sin(mπ)/((m2 n2)π) + Lmcos(mπ)sin(nπ)/((n2 m2)π) = 0  [integral from 0 to L]

for mn. Therefore we can represent these two equations together as

∫ψm*(xn(x) dx = δmn        [integral from 0 to L]

where δmn is the Kronecker delta which is defined such that δnn = 1 and δm ≠ n = 0. Hence the solutions to the infinite square well are orthogonal and normalized, or orthonormal.