The Infinite Square Well and Energy Quantisation
In a one-dimensional infinite square well, the wall conditions select stationary waves with whole numbers of half wavelengths. Their energies rise as the square of the state number. Solving the boundary-value problem gives both the wavefunctions and their allowed energies.
What a learner can do afterwards
- Explains why the wall conditions select discrete energies and calculates an electron's energy using a supplied infinite-well formula
- Identifies interior nodes and determines the probability density at a wavefunction zero
- Predicts how the energy levels shift when the well is made narrower
1 · Read
For a particle of mass m, set U=0 between x=0 and x=L, with infinite walls outside. The stationary equation is -(ℏ²/2m)ψ''=Eψ, where ℏ=h/(2π). For E>0, write ψ=A cos(kx)+B sin(kx), with E=ℏ²k²/(2m). The wall at 0 gives A=0. The wall at L requires sin(kL)=0, so k=nπ/L for n=1,2,3,... . The n=0 wave is zero everywhere and cannot be normalised.
Substitute k=nπ/L to obtain E_n=n²π²ℏ²/(2mL²)=n²E1. Here m is particle mass and L is well width. The ground state has n=1; n=2 is the first excited state. For example, if E1=0.8 eV, then E2=4×0.8=3.2 eV. The first three energies are E1, 4E1 and 9E1.
A normalised real spatial wave is ψ_n(x)=√(2/L)sin(nπx/L). It has n lobes with alternating signs and n-1 interior nodes. Squaring gives probability density |ψ_n|², with nonnegative lobes and the same zeros. For n=3, the interior nodes are L/3 and 2L/3. For n=2, the centre is a node. The ground-state density peaks at the centre. Integrating density over an interval gives the probability for that interval.
At fixed mass and state number, E_n varies as 1/L². Halving the width makes every level four times higher; doubling it makes every level one quarter as high. Gaps between a fixed pair of levels have the same width factor. At fixed width and state number, energy varies as 1/m.
Energy is shown in units of its value at width L.
A transition needs an interaction that transfers energy. If one emitted photon carries all the energy lost from n=2 to n=1, its energy is E2-E1=3E1. With E1=0.8 eV, this gives 3.2-0.8=2.4 eV. Energy conservation fixes the difference, but does not require every downward transition to emit a photon.
A superposition can combine different stationary states. If it contains only n=1 and n=2, with both present, an energy measurement can return E1 or E2. It cannot return an intermediate energy. The state sets the chances of the two results. Its probability density can change with time.
The walls select k=nπ/L and E_n=n²E1. State n has n-1 interior nodes. At fixed mass, halving the width multiplies every energy by four.
Lesson and questions written for this dataset. OpenStax University Physics Volume 3, sections 7.3 and 7.4, were consulted for physics facts and methods.
Readings and flashcards written for this dataset. OpenStax University Physics Volume 3, sections 7.3 and 7.4, and College Physics 2e, section 29.1, were consulted for physics facts.
More to read
Applying the two wall conditions
Take a well with zero potential between x=0 and x=L and impenetrable walls outside. At positive energy, its interior spatial wavefunction is a sum of cosine and sine terms: ψ=A cos(kx)+B sin(kx).
The left wall sets A=0. At the right wall, B sin(kL) must vanish. Setting B=0 would remove the entire wave, so keep B nonzero and require kL=nπ. Positive integers n label the distinct states. Substitution into E=ℏ²k²/(2m) gives E_n=n²π²ℏ²/(2mL²).
The index n counts half wavelengths across the box. There are n-1 nodes strictly inside it. Thus energy grows as n², not as the square of the interior-node count. The ground state has no interior nodes but has positive energy.
The infinite wall is a special limiting boundary. The wavefunction vanishes there, but its derivative need not match the zero exterior derivative.
Wave signs and measurement results
Normalisation makes the integral of probability density over the box equal to one. A real normalised eigenfunction is ψ_n(x)=√(2/L)sin(nπx/L). All these states have the same amplitude prefactor at a fixed width. Higher n adds oscillations rather than increasing that prefactor.
The wavefunction can have positive and negative lobes. Probability density is its squared magnitude, so density has no negative lobes. Both functions vanish at the same nodes. The ground-state density peaks at the centre; the n=2 density vanishes there and has a peak on each side.
A spatial interval has a probability found by integrating density over it. Density itself is not the probability of one exact position.
A definite-energy state has a density that stays unchanged in time. A superposition of n=1 and n=2 can instead have a changing density. Its energy measurements return E1 or E2, with chances determined by the state.
Comparing widths and energy gaps
For a fixed state number, the infinite-well formula has energy proportional to 1/(mL²). A wider box lowers the energy for the same particle. A heavier particle also has lower energy at the same width and state number.
If width doubles while mass stays fixed, each energy becomes one quarter of its previous value. The difference between any fixed pair of levels also becomes one quarter as large.
The levels are not equally spaced. From E_n=n²E1, the first two adjacent gaps are E2-E1=3E1 and E3-E2=5E1.
An interaction must supply or receive energy when a transition occurs. If one photon receives the entire energy lost in a downward transition, its energy equals the initial energy minus the final energy. Energy conservation does not by itself require photon emission or determine which transitions occur.
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