Confinement restricts the wavefunctions that can satisfy the boundaries. In a one-dimensional infinite well, only certain standing shapes and corresponding energies are allowed.
Define the well and its boundaries
The potential energy V is zero inside 0 < x < L and is an ideal infinite barrier outside. Here V means potential energy, not voltage. The particle cannot penetrate the barrier, so its wavefunction vanishes outside and at x = 0 and x = L.
A nonzero standing shape must fit n half-wavelengths between the walls:
L = nλn/2 λn = 2L/n ψn = √(2/L) sin(nπx/L) n = 1, 2, 3, ...
The coefficient normalises each state. An n = 0 sine profile would be zero everywhere and could not have total probability 1. Each state has n - 1 interior nodes, in addition to its two boundary nodes.
n = 1: no interior node
The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.
n = 2: 1 interior node
The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.
n = 3: 2 interior nodes
The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.
All three states use X = x/L, with the same walls at 0 and 1. The upper axis is the scaled amplitude √L ψ; the lower is scaled density L|ψ|2. Their separate vertical labels matter. Each density has unit area against X, and all amplitude nodes are also density zeros.
The second state has an interior node at x/L = 1/2; the third has nodes at 1/3 and 2/3. Negative amplitude lobes still give positive density. These stationary states have time-independent position densities. Their full wavefunctions can change phase, but a time factor must not make total probability oscillate between zero and one.
Connect the allowed wavelength to energy
The spatial wavelength sets a momentum scale h/λn = hn/(2L). Combining this scale with the non-relativistic energy relationship gives the allowed well energies:
En = (hn/2L)2/(2m) = h2n2/(8mL2)
This is an energy calculation for a confined standing state. It does not give one definite measured classical momentum, or restrict all momentum measurements to just two values. Confinement requires a momentum spread.
An infinite-well potential is an ideal energy model
Inside: V = 0. Outside: the ideal potential energy is infinite. The wavefunction vanishes at both boundaries and outside. The electron example uses L = 1.00 nm, independently of the earlier 4.00 nm normalization profiles.
Allowed energies have the ratio 1 : 4 : 9
For the supplied 1.00 nm electron box, E1 is about 0.377 eV. The first three energies are about 0.377, 1.51 and 3.39 eV. The increasing gaps are genuine; the levels have not been evenly spaced for convenience.
The infinite barriers are symbolic potential-energy walls, with no tunnelling tail. The separate energy plot has a linear scale: allowed levels are E/E1 = 1, 4 and 9. Zero is the energy reference, not an allowed n = 0 state.
Worked electron well
Keep the width distinct from the earlier profiles
An electron confined across about 1 nm has a ground-energy scale of order 10-19 J, below a few eV. Doubling the width should reduce that energy scale by four.
Use L = 1.00 nm = 1.00 × 10-9 m, me = 9.11 × 10-31 kg and h = 6.63 × 10-34 J s. This is a new width, distinct from the 4.00 nm normalisation examples.
Allowed wavelengths and energies in the supplied 1.00 nm electron well
n
λn / nm
En / eV
1
2.00
0.377
2
1.00
1.51
3
0.667
3.39
The corresponding unrounded energy calculations for n = 2 and 3 give 2.41256 × 10-19 J and 5.42827 × 10-19 J. The ratio 1:4:9 is exact in the model; displayed energies are rounded.
E2 - E1 = 3E1 ≈ 1.13 eV E3 - E2 = 5E1
The increasing gaps are not equally spaced. At fixed n and mass, doubling L quarters En. At fixed n and L, doubling m halves it.
The nonzero ground energy is consistent with the momentum spread needed for confinement. It does not mean that a stationary density graph depicts a ball travelling back and forth at a definite speed. The infinite well is also not the exact potential or energy sequence of an atom.
Optional check For the same particle mass, compare n = 2 in an infinite well of width 2L with n = 1 in a well of width L. How do their energies compare?