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Section 10.2: The Quantum-mechanical Infinite Square Well

Please wait for the animation to completely load.

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.

Please wait for the animation to completely load.

In the second animation, the first 10 normalized energy eigenfunctions are shown for a box with L = 1, along with the energy spectrum. In the animation ħ = 2m = 1.  You can click-drag in the energy spectrum on the left to change the energy state. As you do so, the displayed energy turns from green to red.

Note that these energy eigenfunctions, ψn(x), are only non-zero in the spatial region 0 < x < L and are zero everywhere else.3  These energy eigenfunctions are not so simple after all. In fact, these energy eigenfunctions have a kink (a discontinuous first derivative) at x = 0 and x = L.  Normally this is not acceptable for a energy eigenfunction, but here the potential energy function for the infinite square well is so badly behaved, these energy eigenfunctions are actually acceptable.

We can now also calculate expectation values of position and momentum. We find that: <x> = L/2 and <p> = 0, as expected. We also find that: <x2> = L2 (1/3 − 1/(2n2π2)), <p2> = n2π2ħ2/L2, and hence <H> =  n2π2ħ2/(2mL2), again as expected.

2Such an argument is also made by M. A. Morrison in Understanding Quantum Physics: A User's Manual. The correct derivation of why the k = 0 case cannot exist in the infinite square well was originally stated by M. Bowen and J. Coster, "Infinite Square Well: A Common Mistake," Am. J. Phys. 49, 80-81 (1980) with follow-up discussion in R. C. Sapp, "Ground State of the Particle in a Box," Am. J. Phys. 50, 1152-1153 (1982) and L. Yinji and H. Xianhuai, "A Particle Ground State in the Infinite Square Well," Am. J. Phys. 54, 738 (1986).
3These energy eigenfunctions can also be written as ψn(x) = (2/L)1/2 sin(nπx/L) Θ(x) Θ(L − x).  This representation uses two Heaviside step functions, Θ(ξ), to explicitly show the region in which the energy eigenfunction is valid. This can be accomplished because the step function, Θ(ξ), is zero for ξ < 0 and is 1 for ξ > 0.

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