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Standing states in an infinite square well

Key idea: A particle in an infinite well has a wavefunction that vanishes at both walls. Only whole half-wavelengths fit, giving energies proportional to n²/L² and a nonzero ground-state energy.

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  • GCE A-Level H2 Physics 2027
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A guitar string fixed at both ends can only vibrate at certain frequencies, because a whole number of half-wavelengths must fit between its ends. A particle trapped in a box behaves the same way. Its wavefunction can only take certain shapes, so the particle can only have certain energies. This is the simplest example of quantised energy.

Standing waves in a well

In a one-dimensional infinite square well, a particle can move freely between two walls a distance L apart, but the walls are infinitely high, so it can never be found outside. The wavefunction must therefore be zero at both walls.

Waves reflected back and forth between the walls superpose to form standing waves. To have a node at each wall, a whole number of half-wavelengths must fit into the well:

L = nλₙ/2 ⇒ λₙ = 2L/n, n = 1,2,3,…

The allowed wavefunctions are standing waves:

ψₙ = square root of (2/L) sin((nπ x)/L)

inside the well, and zero outside. State n has n half-waves, so it has n-1 nodes between the walls, where the probability density is zero.

The first three infinite-well states

Normalised amplitudes sqrt(L) ψₙ = sqrt(2) sin(n pi x/L) for n = 1, 2 and 3, on the same position scale. Each has unit integrated probability.

Scroll across the figure to read all labels.

Normalised amplitudes sqrt(L) ψₙ = sqrt(2) sin(n pi x/L) for n = 1, 2 and 3, on the same position scale. Each has unit integrated probability.Normalised amplitudes sqrt(L) ψₙ = sqrt(2) sin(n pi x/L) for n = 1, 2 and 3, on the same position scale. Each has unit integrated probability.
All states vanish at the two walls. The n = 1, 2 and 3 states have zero, one and two internal nodes.
Open full-size graph
What to noticeCount half-waves and internal nodes. These are wavefunction amplitudes at a chosen phase, not trajectories or probability-density curves.
View figure data
Figure data: The first three infinite-well states
Position ratio, x/L (1)n = 1n = 2n = 3
0.0000000.000000.000000.0000
0.0083330.037020.074010.1110
0.0166670.074010.147830.2212
0.0250000.110960.221230.3301
0.0333330.147830.294030.4370
0.0416670.184590.366030.5412
0.0500000.221230.437020.6420
0.0583330.257720.506810.7389
0.0666670.294030.575210.8313
0.0750000.330140.642040.9185
0.0833330.366030.707111.0000
0.0916670.401660.770241.0754
0.1000000.437020.831251.1441
0.1083330.472070.889991.2058
0.1166670.506810.946291.2601
0.1250000.541201.000001.3066
0.1333330.575211.050971.3450
0.1416670.608831.099051.3751
0.1500000.642041.144121.3968
0.1583330.674801.186061.4099
0.1666670.707111.224741.4142
0.1750000.738921.260071.4099
0.1833330.770241.291951.3968
0.1916670.801021.320281.3751
0.2000000.831251.345001.3450
0.2083330.860921.366031.3066
0.2166670.889991.383311.2601
0.2250000.918461.396801.2058
0.2333330.946291.406471.1441
0.2416670.973481.412281.0754
0.2500001.000001.414211.0000
0.2583331.025831.412280.9185
0.2666671.050971.406470.8313
0.2750001.075381.396800.7389
0.2833331.099051.383310.6420
0.2916671.121971.366030.5412
0.3000001.144121.345000.4370
0.3083331.165491.320280.3301
0.3166671.186061.291950.2212
0.3250001.205821.260070.1110
0.3333331.224741.224740.0000
0.3416671.242841.18606-0.1110
0.3500001.260071.14412-0.2212
0.3583331.276451.09905-0.3301
0.3666671.291951.05097-0.4370
0.3750001.306561.00000-0.5412
0.3833331.320280.94629-0.6420
0.3916671.333100.88999-0.7389
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0.4083331.355980.77024-0.9185
0.4166671.366030.70711-1.0000
0.4250001.375140.64204-1.0754
0.4333331.383310.57521-1.1441
0.4416671.390530.50681-1.2058
0.4500001.396800.43702-1.2601
0.4583331.402110.36603-1.3066
0.4666671.406470.29403-1.3450
0.4750001.409850.22123-1.3751
0.4833331.412280.14783-1.3968
0.4916671.413730.07401-1.4099
0.5000001.414210.00000-1.4142
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0.6083331.33310-0.88999-0.7389
0.6166671.32028-0.94629-0.6420
0.6250001.30656-1.00000-0.5412
0.6333331.29195-1.05097-0.4370
0.6416671.27645-1.09905-0.3301
0.6500001.26007-1.14412-0.2212
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0.6666671.22474-1.224740.0000
0.6750001.20582-1.260070.1110
0.6833331.18606-1.291950.2212
0.6916671.16549-1.320280.3301
0.7000001.14412-1.345000.4370
0.7083331.12197-1.366030.5412
0.7166671.09905-1.383310.6420
0.7250001.07538-1.396800.7389
0.7333331.05097-1.406470.8313
0.7416671.02583-1.412280.9185
0.7500001.00000-1.414211.0000
0.7583330.97348-1.412281.0754
0.7666670.94629-1.406471.1441
0.7750000.91846-1.396801.2058
0.7833330.88999-1.383311.2601
0.7916670.86092-1.366031.3066
0.8000000.83125-1.345001.3450
0.8083330.80102-1.320281.3751
0.8166670.77024-1.291951.3968
0.8250000.73892-1.260071.4099
0.8333330.70711-1.224741.4142
0.8416670.67480-1.186061.4099
0.8500000.64204-1.144121.3968
0.8583330.60883-1.099051.3751
0.8666670.57521-1.050971.3450
0.8750000.54120-1.000001.3066
0.8833330.50681-0.946291.2601
0.8916670.47207-0.889991.2058
0.9000000.43702-0.831251.1441
0.9083330.40166-0.770241.0754
0.9166670.36603-0.707111.0000
0.9250000.33014-0.642040.9185
0.9333330.29403-0.575210.8313
0.9416670.25772-0.506810.7389
0.9500000.22123-0.437020.6420
0.9583330.18459-0.366030.5412
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1.0000000.000000.000000.0000

Check your understanding 1 Why not n = 0?

Why is there no state with n = 0?

Show answer

With n = 0, sin((0 × π x)/L) = 0 everywhere, so the wavefunction is zero everywhere and the total probability would be zero instead of one. A particle in the well must have n ≥ 1. Compare your explanation for the boundary condition and the total probability: satisfying the wall nodes alone is not enough.

Allowed energies

The standing wave combines travelling components with momenta + pₙ and -pₙ, where pₙ = h/λₙ = nh/2L. The state has zero mean momentum, rather than one definite momentum pₙ. Both components have the same kinetic energy pₙ²/(2m). Taking potential energy as zero inside the well gives

Eₙ = n²h²/8mL²

Only these energies are possible: the energy is quantised. The lowest, the ground state E₁, is not zero, so a trapped particle can never be completely at rest, as the uncertainty principle requires.

The energies are in the ratio 1:4:9:16:…, so the levels get further apart as n increases. A narrower well or a lighter particle gives larger energies: halving L makes every energy four times as large.

Energy levels in a 1D infinite well (scaled)

Discrete stationary energy ratios Eₙ/E₁ = n² for integer n from 1 to 5: 1, 4, 9, 16 and 25.

Scroll across the figure to read all labels.

Discrete stationary energy ratios Eₙ/E₁ = n² for integer n from 1 to 5: 1, 4, 9, 16 and 25.Discrete stationary energy ratios Eₙ/E₁ = n² for integer n from 1 to 5: 1, 4, 9, 16 and 25.
Because Eₙ grows as n², the gaps between successive energy levels get larger as n increases.
Open full-size graph
What to noticeOnly the integer-n points are allowed stationary levels. Successive gaps are 3, 5, 7 and 9 times E₁.
View figure data
Figure data: Energy levels in a 1D infinite well (scaled)
Quantum number, n (1)Eₙ/E₁ = n²
11
24
39
416
525

Worked example 1

An electron in a 1.00 nm well

Problem

An electron is trapped in an infinite square well 1.00 nm wide. Find the energies of its first three states, in electronvolts. The mass of an electron is 9.11 × 10⁻³¹ kg.

Show full solution

  1. Ground state

    Method

    Substitute n = 1 into Eₙ = n²h²/8mL², with L in metres.

    Reason

    The ground-state energy sets the scale for every other level.

    Working

    Answer

    E₁ = ((6.63 × 10⁻³⁴)²)/(8(9.11 × 10⁻³¹)(1.00 × 10⁻⁹)²) = 6.03 × 10⁻²⁰ J = 0.377 eV

  2. Higher states

    Working

    Answer

    Eₙ = n²E₁: E₂ = 4E₁ = 1.51 eV and E₃ = 9E₁ = 3.39 eV

Guided practice 2

A wider well

About 2 min

Problem

The electron is now trapped in a well 2.00 nm wide. Using the worked example, find its new ground-state energy.

Find E₁

Unit: eV

Hints

Hint 1

only L changes

The mass and n are unchanged, and E₁ ∝ 1/L².

Show full solution

  1. Scaling

    Working

    Answer

    Doubling L divides E₁ by 4: 0.377/4 = 0.0942 eV

Try it yourself 3

From the ground state up

Minimal support

Problem

A particle in an infinite square well has a ground-state energy of 0.20 eV. Find the energy of its n = 3 state, and the energy of the photon it would emit in dropping from n = 3 to n = 2.

Find the energies

Unit: eV
Show the steps Optional working
Unit: eV
Unit: eV

Hints

Hint 1

scaling

Use Eₙ = n²E₁. The photon carries away the difference between the two levels.

Show solution step by step

  1. Energy of n = 3

    Reason

    The energies scale as n².

    Working

    Answer

    E₃ = 9(0.20) = 1.8 eV and E₂ = 4(0.20) = 0.80 eV

  2. Photon

    Working

    Answer

    E₃-E₂ = 1.8-0.80 = 1.0 eV

Check your understanding 2 Two wells

Well A traps a particle of mass m and has width L. Well B traps a particle of mass 2m and has width L/2. How do their ground-state energies compare? How many nodes between the walls does the n = 2 state have in each?

Show answer

EB/EA = m/2m × (L/(L/2))² = 1/2 × 4 = 2, so well B’s ground state has twice the energy. Both n = 2 states have one node between the walls, halfway across: the mass and width set the energies, and n sets the shape.

Common mistakes

  • Including a state with n = 0, or saying the ground-state energy is zero.
  • Thinking the energy levels are equally spaced; they go as n².
  • Leaving L in nanometres, or forgetting to square it, in Eₙ = n²h²/8mL².
  • Reading a negative part of ψ as a negative probability; the probability density is |ψ|².

Spotted an error? Report a correction

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