9478 / 2027
Quantum Physics overview

Lesson 6 of 7

A particle in a box

  • I can apply superposition to position wavefunctions to explain single-particle interference and standing solutions in a box.Syllabus

    Syllabus 9478, 19(g). Apply superposition to position wavefunctions to explain single-particle interference and standing solutions in a box.

  • I can explain standing wavefunctions for a particle in a one-dimensional infinite square well.Syllabus

    Syllabus 9478, 19(i). Explain standing wavefunctions for a particle in a one-dimensional infinite square well.

  • I can use E_n = h^2 n^2/(8mL^2) for allowed infinite-well energies.Syllabus

    Syllabus 9478, 19(j). Use E_n = h^2 n^2/(8mL^2) for allowed infinite-well energies.

Make a guess. You are not marked.

An electron in a one-dimensional box moves from its n = 1 energy level to its n = 3 level. By what factor does its energy increase?

  1. Two, because the electron has climbed up two levels in the box.
  2. Three, because the energy is proportional to n.
  3. Nine, because the energy is proportional to n squared.
Show the answer

Nine, because the energy is proportional to n squared.

In a box, E = n2 h2 / (8 m L2). Going from n = 1 to n = 3 multiplies n2 from 1 to 9, so the energy becomes nine times as large.

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

n = 1: no interior nodeTwo aligned plots use the same dimensionless position X equals x over L, with walls at zero and one. The upper dimensionless amplitude is square root L times psi-n, equal to square root two sine of 1 pi X. Its full-scale extrema are plus and minus square root two. The lower dimensionless density is L times modulus psi-n squared, equal to two sine squared of 1 pi X; its maxima are two and it is never negative. Both vanish at the boundaries and outside. There is no interior node. Filled markers at interior nodes mark graph zeros, not particle detections. Each lower curve has area one against X.√L ψ1 (dimensionless)-√20+√201/21X = x/LL|ψ1|2 (dimensionless)01201/21X = x/L

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

n = 2: 1 interior nodeTwo aligned plots use the same dimensionless position X equals x over L, with walls at zero and one. The upper dimensionless amplitude is square root L times psi-n, equal to square root two sine of 2 pi X. Its full-scale extrema are plus and minus square root two. The lower dimensionless density is L times modulus psi-n squared, equal to two sine squared of 2 pi X; its maxima are two and it is never negative. Both vanish at the boundaries and outside. An interior node at X one half is zero in both plots. Filled markers at interior nodes mark graph zeros, not particle detections. Each lower curve has area one against X.√L ψ2 (dimensionless)-√20+√201/21X = x/LL|ψ2|2 (dimensionless)01201/21X = x/L

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

n = 3: 2 interior nodesTwo aligned plots use the same dimensionless position X equals x over L, with walls at zero and one. The upper dimensionless amplitude is square root L times psi-n, equal to square root two sine of 3 pi X. Its full-scale extrema are plus and minus square root two. The lower dimensionless density is L times modulus psi-n squared, equal to two sine squared of 3 pi X; its maxima are two and it is never negative. Both vanish at the boundaries and outside. Interior nodes at X one third and two thirds are zero in both plots. Filled markers at interior nodes mark graph zeros, not particle detections. Each lower curve has area one against X.√L ψ3 (dimensionless)-√20+√201/32/31X = x/LL|ψ3|2 (dimensionless)01201/32/31X = x/L

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

An infinite-well potential is an ideal energy modelThe horizontal coordinate is X equals x over L. The impenetrable boundaries are X zero at drawing coordinate eighty and X one at three hundred, exactly the positions used in all subsequent state plots. Potential energy V is zero inside. Vertical wall arrows point upwards towards infinity; their drawn height is symbolic, not a finite barrier value. Shaded outside regions are inaccessible in this ideal model. No wavefunction penetration or tunnelling tail is shown. V denotes potential energy in joules, not electric potential in volts.Potential energy V / J∞∞0V = 001Scaled position X = x/LInfinite arrows are symbolic.

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

Allowed energies have the ratio 1 : 4 : 9The vertical axis is dimensionless E over E1 and uses one linear scale: y equals three hundred and sixty minus twenty-eight times E over E1. The n equals one, two and three levels are at relative energies one, four and nine, so their y coordinates are three hundred and thirty-two, two hundred and forty-eight, and one hundred and eight. The zero dashed baseline is a reference only, not an allowed n equals zero state. The n equals two to one gap is three E1 and the n equals three to two gap is five E1. Horizontal level lengths are not position or energy magnitudes.1n = 14n = 29n = 30E / E1 (dimensionless)Zero is a reference,not an allowed n = 0 state.

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.

E1 = (6.63 × 10-34)2
/ [8(9.11 × 10-31)(1.00 × 10-9)2]
= 6.03141 × 10-20 J
= 0.376963 eV ≈ 0.377 eV
Allowed wavelengths and energies in the supplied 1.00 nm electron well
nλn / nmEn / eV
12.000.377
21.001.51
30.6673.39

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.

Check your understanding

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?

  1. They are equal.
  2. The n = 2 energy is one quarter as large.
  3. The n = 2 energy is four times larger.
Show the answer

They are equal.

E_n is proportional to n2/L2 for fixed mass. Here 22/(2L)2 = 1/L2, the same ratio as the original ground state.