The finite square well: three regions and two energy sectors
Climbable walls: discrete levels below V₀, a continuum above.
This is where the walls stop doing the quantising for you. Before any root-finding, set the problem up so the rest of the week can stand on it: one constant-coefficient equation per region, the sign of V₀ − E deciding what ψ does outside, and a count of constants that says in advance which energies come out discrete.
Build the model
In every region the equation is easy; the sign of V₀ − E decides which kind of easy.
1The big idea
The finite square well sets V = 0 for |x| < a and V = V₀ for |x| > a: a box whose walls can be climbed. Because V is constant inside each of the three regions, the time-independent Schrödinger equation there reads ψ″ = (2m/ħ²)(V − E)ψ with a constant coefficient, solved by e(λx) with λ² = 2m(V − E)/ħ². So the sign of V − E settles everything locally. Inside, E > 0 and ψ oscillates with k = √(2mE)/ħ. Outside, ψ mixes e(κx) and e(−κx), κ = √(2m(V₀ − E))/ħ, when E < V₀, and oscillates with the smaller q = √(2m(E − V₀))/ħ when E > V₀; below the floor, E < 0, there is no acceptable state at all. A count of constants explains the two sectors before any algebra. Two per region make six, and the four matching conditions at the walls act in both sectors, but normalisability strikes out the two rising exponentials below V₀ and nothing above it. What is left below V₀ is a homogeneous 4 × 4 system with solutions only at special energies; above V₀, two constants stay free at every energy. The weak point of the model is the walls themselves. A real potential climbs over some distance d, and the step can be trusted only while d is short beside both 1/k and 1/κ: true for a semiconductor well grown one atomic layer at a time, crude for a nucleus, whose surface is about as thick as a bound neutron's tail is long.
2In one sentence
The finite square well is the potential V = 0 for |x| < a and V = V₀ for |x| > a; its energy eigenstates fall into a bound sector, 0 < E < V₀, with discrete energies and tails that decay outside, and an unbound sector, E > V₀, where every energy is allowed.
Example. An electron in a well 1.00 nm wide with 1.00 eV walls has just two bound energies, 0.191 eV and 0.703 eV. In the lower one ψ oscillates inside with k = 2.24 nm⁻¹ and decays outside with κ = 4.61 nm⁻¹; every energy above 1.00 eV is allowed too.
3Name the region before writing the equation
Put the origin at the centre of the well, so the walls sit at x = −a and x = +a, and name the three regions: I for x < −a, II for |x| < a, III for x > a. In each one V is a constant, so the time-independent Schrödinger equation −(ħ²/2m)ψ″ + Vψ = Eψ becomes ψ″ = (2m/ħ²)(V − E)ψ with a constant coefficient. That is the simplest second-order ODE there is: try ψ = e(λx), get λ² = 2m(V − E)/ħ², and the two roots give a general solution with two arbitrary constants. Three regions therefore carry six constants, and nothing inside any one region knows about the others. The walls are where V jumps; the equation is not solved across them but used to derive the conditions that join the pieces, which is the business of the matching-conditions topic. So the working rule for the whole week is: name the region, write its V, and only then write its solution.
4The sign of V − E decides what ψ does
Inside, V − E = −E. For E > 0 the roots are λ = ±ik and ψ = A sin kx + B cos kx, an oscillation with k = √(2mE)/ħ. Outside, the coefficient is (2m/ħ²)(V₀ − E), and its sign splits the spectrum. For 0 < E < V₀ it is positive, the roots are real, λ = ±κ, and ψ mixes e(κx) and e(−κx): the classically forbidden case, where a particle would need negative kinetic energy. For E > V₀ it is negative and ψ oscillates outside too, with q = √(2m(E − V₀))/ħ. An electron at 1.40 eV over a well with 1.00 eV walls has k = 6.06 nm⁻¹ inside and q = 3.24 nm⁻¹ outside, so its outside wavelength, 1.94 nm, is 1.87 times the inside one, just as a classical particle slows on climbing out. Below the floor nothing works. For E < 0, ψ″/ψ > 0 in every region; but a normalisable ψ, which can be taken real, reaches a largest |ψ| somewhere, and there ψ″ cannot share the sign of ψ, so no normalisable solution exists.
5One energy fixes both k and κ, and one number fixes the well
In the bound sector k and κ are not independent: adding their squares removes E, leaving k² + κ² = 2mV₀/ħ². Raise E and k grows while κ shrinks, so the inside wavelength shortens as the tail lengthens. For an electron in a 1.00 nm well with 1.00 eV walls, the ground state at 0.191 eV has κ = 4.61 nm⁻¹ and a tail length 1/κ = 0.217 nm; the second state, at 0.703 eV, has κ = 2.79 nm⁻¹ and 1/κ = 0.358 nm. At E = V₀, κ reaches zero, the tail never ends, and the bound sector stops. Now rescale with ξ = x/a and ε = E/V₀. The equation becomes d²ψ/dξ² = −z₀²εψ inside and d²ψ/dξ² = z₀²(1 − ε)ψ outside, with z₀ = a√(2mV₀)/ħ. Mass, width and depth survive only in that one dimensionless number, so two wells with equal z₀ have the same values of ε and the same shapes, however different their scales. This well has z₀ = 2.56; later this week z₀ sets the graphical solution and the count of bound states.
6Count the constants before solving anything
Six constants, two per region. The walls supply four conditions, ψ and ψ′ continuous at x = −a and at x = +a, and they apply in both sectors. The sectors differ in what normalisability does first. Below V₀, e(κx) grows without limit as x → +∞ and e(−κx) as x → −∞, so the first must be absent from region III and the second from region I. That leaves four constants against four linear, homogeneous conditions: a 4 × 4 system M(E)c = 0, whose only solution is c = 0 unless det M(E) = 0. That one equation in E holds only at isolated energies, the bound states, and leaves one overall factor free for normalisation to fix. Above V₀ nothing is discarded, because sin qx and cos qx stay bounded. Four conditions on six constants then leave a two-dimensional family of solutions at every energy: a continuum, each energy counted twice, for instance once for a wave arriving from the left and once from the right. None of them can be normalised; they belong to the barrier and scattering problems later in the course.
7Checks that hold before any energy is known
Dimensions: 2m(V₀ − E)/ħ² has units kg J/(J s)² = kg/(J s²) = m⁻², so κ is an inverse length, and z₀, a length times an inverse length, is a pure number. Deep walls: as V₀ → ∞ at fixed E, κ → ∞, the tails shrink to nothing, ψ(±a) → 0, and the infinite well of width 2a returns with Eₙ = n²π²ħ²/(8ma²). For 2a = 1.00 nm that gives 0.376 eV for n = 1, and the finite-well ground state, at 0.191 eV, sits below it as it must. The rim: as E → V₀ from below, κ → 0 and the state spreads without limit. Classical correspondence: a classical particle with 0 < E < V₀ bounces between x = −a and x = +a at any energy and never enters |x| > a; quantum mechanics keeps the confinement, makes the energies discrete, and lets a tail reach past the turning points. Above V₀ the classical particle crosses the well and speeds up inside, as k > q says. Normalisation: bound states can be normalised because their tails decay; unbound states cannot.
8Sharp walls are a fiction, but a useful one
No real potential jumps in zero distance; it climbs from 0 to V₀ over some distance d. The step is a fair model when ψ cannot tell the difference, and ψ resolves nothing much finer than its own wavelength or tail length, so the test is d ≪ 1/k and d ≪ 1/κ. The test depends on the scales. In the running example, an electron in a 1.00 nm well with 1.00 eV walls, the ground state has 1/κ = 0.217 nm, so even a wall one atomic layer thick, about 0.28 nm, fails it: κd ≈ 1.3, and the shape of the wall matters. Semiconductor quantum wells pass easily, because a small effective mass and a band-edge step of a few tenths of an eV stretch the tails past a nanometre, while the interfaces are grown abrupt to about one layer. A nucleus fails for the opposite reason: its surface takes about 2 fm to rise, comparable with a bound neutron's tail. The idealisation also leaves a mark of its own. At a sharp wall ψ″ jumps by (2m/ħ²)V₀ψ(±a) while ψ and ψ′ stay continuous, a jump in curvature that no smooth potential produces.
Key equations
What each relation says, and the units it is written in.
The finite square well
V(x) = 0 for |x| < a, and V(x) = V₀ for |x| > a
Centring the well on x = 0 makes V even, which halves the later work; putting the floor at V = 0 makes every allowed energy positive.
a is the half-width in m, so the well is 2a wide; V₀, in J or eV, is the wall height, and E is measured from the floor.
Inside the well, in both sectors
ψ″ = −k²ψ, with k = √(2mE)/ħ
Solutions A sin kx + B cos kx exist at every E > 0: inside, ψ oscillates whatever the energy, with local wavelength 2π/k.
k in m⁻¹, m in kg, E in J, ħ in J s. For an electron ħ²/2m = 0.0381 eV nm², so k = √(E/0.0381 eV) nm⁻¹.
Outside the well: the sign that sorts the sectors
ψ″ = (2m/ħ²)(V₀ − E)ψ for |x| > a
Positive constant: e(κx) and e(−κx), the bound sector. Negative: sin qx and cos qx, the unbound sector. At E = V₀ exactly, ψ is a straight line outside.
Write the constant as κ², κ = √(2m(V₀ − E))/ħ, if E < V₀; as −q², q = √(2m(E − V₀))/ħ, if E > V₀. Both in m⁻¹.
One energy sets both k and κ
k² + κ² = 2mV₀/ħ²
Raising E shortens the inside wavelength and lengthens the tail; at E = V₀, κ = 0 and the tail never ends. m, a and V₀ reach the solutions only through z₀.
Bound sector; every term in m⁻². Times a²: (ka)² + (κa)² = z₀², where z₀ = a√(2mV₀)/ħ is dimensionless.
Counting the constants
free constants = 6 − nd − 4
Below V₀: a homogeneous 4 × 4 system, solvable only where its determinant vanishes, so E is discrete. Above V₀: two constants free at every E, a continuum.
6: two per region. nd: discarded by normalisability, 2 below V₀ and 0 above. 4: ψ and ψ′ continuous at x = −a and x = +a.
When a sharp wall is a fair model
d × max(k, κ) ≪ 1
ψ cannot resolve structure much finer than its own wavelength or tail length, so a rise that quick looks like a step to it.
d, in m, is the distance over which the real potential climbs from 0 to V₀; k and κ in m⁻¹, so the product is a pure number.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Hold V₀ at 1.00 eV and raise E from 0.02 eV. The tails bend upward up to 0.18 eV and downward from 0.20 eV; between them, at 0.191 eV, lies the even ground state, whose tails decay instead. Above 1.00 eV nothing runs away: every energy there is allowed.
Live results
- INSIDE k = √(2mE)/ħ
- 3.62 nm⁻¹
- OUTSIDE κ, IF E < V₀
- 3.62 nm⁻¹
- OUTSIDE q, IF E > V₀
- 0.00 nm⁻¹
- WELL STRENGTH z₀ = a√(2mV₀)/ħ
- 2.56
Catch the common trap
The mistake this lesson is built to prevent.
Carrying over the infinite well's rule that ψ vanishes at the walls. Students set ψ(−a) = ψ(a) = 0, get k = nπ/(2a), and find energies in which V₀ never appears, which should already be alarming. At a finite wall nothing vanishes: V₀ is finite, so ψ″ is finite, ψ and ψ′ run continuously through x = ±a, and ψ(a) is simply where the outside tail begins. The walls still join the pieces, but by themselves they quantise nothing; what makes the energies discrete is the demand that ψ die away as |x| → ∞, which strikes out two constants before the joins are made. The other sector proves it. Above V₀ the walls are the same and the four matching conditions are the same, yet every energy is allowed, because nothing outside has to be discarded. If the walls alone quantised energies, they would quantise both sectors.
Test it: what changes?
- k grows, so the inside wavelength shortens
- κ shrinks, so the tail length 1/κ grows
- k² + κ² stays equal to 2mV₀/ħ²
- κ reaches zero at E = V₀
- Inside k is unchanged
- κ grows, so the tails shorten
- z₀ = a√(2mV₀)/ħ grows
- As V₀ → ∞, ψ(±a) → 0
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron meets a finite square well of half-width a = 0.400 nm with walls V₀ = 0.800 eV. For E = 0.300 eV and for E = 1.100 eV, say which sector the energy lies in, write the form of ψ in regions I, II and III, and find the length that characterises ψ in each region: the wavelength where it oscillates, the decay length 1/κ where it does not.
View solution →
- Sort first. Both energies are positive, so ψ oscillates inside in both cases. 0.300 eV < V₀ puts the first in the bound sector; 1.100 eV > V₀ puts the second in the unbound sector.
- E = 0.300 eV, region II: k = √(0.300/0.0381) nm⁻¹ = 2.806 nm⁻¹, so ψ = A sin kx + B cos kx with wavelength 2π/k = 2.24 nm.
- Regions I and III: V₀ − E = 0.500 eV, so κ = √(0.500/0.0381) nm⁻¹ = 3.623 nm⁻¹, with ψ = C e(κx) + D e(−κx) in I and F e(κx) + G e(−κx) in III; normalisability will later remove D and F. Decay length 1/κ = 0.276 nm. Check: k² + κ² = 7.87 + 13.12 = 21.0 nm⁻², which is V₀/(0.0381 eV nm²), as it must be.
- E = 1.100 eV: inside k = √(1.100/0.0381) nm⁻¹ = 5.373 nm⁻¹, wavelength 1.17 nm. Outside E − V₀ = 0.300 eV, so q = √(0.300/0.0381) nm⁻¹ = 2.806 nm⁻¹, with ψ = C sin qx + D cos qx in I and F sin qx + G cos qx in III, wavelength 2.24 nm. No constant is discarded.
- Classical check: the kinetic energy is 1.100 eV inside and 0.300 eV outside, a momentum ratio √(1.100/0.300) = 1.91. With λ = h/p the outside wavelength should be 1.91 times the inside one, and 2.24 nm/1.17 nm = 1.91.
Answer. At 0.300 eV: bound sector, oscillating inside with wavelength 2.24 nm and decaying outside over 1/κ = 0.276 nm. At 1.100 eV: unbound sector, oscillating everywhere, with wavelength 1.17 nm inside and 2.24 nm outside.
MediumExample 2
A GaAs layer 8.00 nm thick between Al₀.₃Ga₀.₇As barriers makes a finite well for a conduction electron, with V₀ = 0.240 eV and effective mass m* = 0.067 mₑ. Its ground state lies at E₁ = 0.0449 eV. Find k and κ for that state, test whether interfaces abrupt to one monolayer, d = 0.283 nm, justify treating the walls as sharp, and find the grading width at which that test would fail outright.
View solution →
- Replace mₑ by m*: ħ²/2m* = (0.0381 eV nm²)/0.067 = 0.569 eV nm². The lighter mass stretches every length at a given energy.
- Inside: k = √(0.0449/0.569) nm⁻¹ = 0.281 nm⁻¹, so 1/k = 3.56 nm and the wavelength is 22.4 nm. The 8.00 nm well holds well under half a wavelength, as a ground state should: ka = 0.281 × 4.00 = 1.12, below π/2.
- Outside: κ = √((0.240 − 0.0449)/0.569) nm⁻¹ = √0.343 nm⁻¹ = 0.586 nm⁻¹, a tail length 1/κ = 1.71 nm. Check: k² + κ² = 0.0789 + 0.343 = 0.422 nm⁻², which is V₀/(0.569 eV nm²).
- Test: kd = 0.281 × 0.283 = 0.080 and κd = 0.586 × 0.283 = 0.166. κ is the larger wavenumber, so κd decides, and at about one sixth the rise is short beside anything ψ can resolve. Sharp walls are a fair model.
- Failure: the test fails outright when κd ≈ 1, that is at d ≈ 1/κ = 1.71 nm, about six monolayers. Interdiffusion that smeared each interface over that distance would make the profile of the wall matter to this state.
Answer. k = 0.281 nm⁻¹ and κ = 0.586 nm⁻¹, a tail length of 1.71 nm. With d = 0.283 nm, κd = 0.17 and kd = 0.08, so sharp walls are fair; the model fails once grading approaches 1.7 nm, about six monolayers.
HardExample 3
A neutron in a one-dimensional model nucleus sees a finite square well of depth V₀ = 36.0 MeV and half-width a = 2.00 fm, with ħ²/2mₙ = 20.72 MeV fm². First rewrite the three-region equation in ξ = x/a and ε = E/V₀ and show that one parameter survives. Then find the half-width of an electron well with V₀ = 1.00 eV that is an exact scale model. Its ground state is found numerically at 0.183 eV: predict the neutron's ground-state energy and tail length.
View solution →
- With x = aξ, d²/dx² = (1/a²) d²/dξ². Inside, ψ″ = −(2mE/ħ²)ψ becomes d²ψ/dξ² = −(2mV₀a²/ħ²)εψ = −z₀²εψ; outside, d²ψ/dξ² = z₀²(1 − ε)ψ. The walls sit at ξ = ±1 in every well, so m, a and V₀ enter only through z₀ = a√(2mV₀)/ħ.
- Neutron: z₀ = a√(V₀/(ħ²/2mₙ)) = 2.00 fm × √(36.0 MeV/20.72 MeV fm²) = 2.00 × 1.318 = 2.636, a pure number as it must be.
- Electron model: equal z₀ requires a = z₀√((ħ²/2mₑ)/V₀) = 2.636 × √(0.0381 eV nm²/1.00 eV) = 2.636 × 0.1952 nm = 0.515 nm, a well 1.03 nm wide.
- Equal z₀ means equal ε. The electron gives ε₁ = 0.183 eV/1.00 eV = 0.183, so the neutron's ground state lies at E₁ = 0.183 × 36.0 MeV = 6.59 MeV above the floor, 29.4 MeV below the rim.
- Tails scale with a: κa = z₀√(1 − ε₁) = 2.636 × √0.817 = 2.383, so 1/κ = a/2.383 = 0.839 fm for the neutron and 0.216 nm for the electron, both 0.420a.
- Checks: ε₁ lies between 0 and the infinite-well value (π/2)²/z₀² = 0.355, as a finite-well ground state must. The two energies differ by a factor of 3.6 × 10⁷, yet drawn against ξ the two wavefunctions are identical. The scaling is exact for square wells; the real nucleus, with its diffuse surface, is only roughly one.
Answer. z₀ = 2.64, so the electron model needs a = 0.515 nm. The neutron's ground state lies 6.59 MeV above the floor with a tail length of 0.839 fm, 0.420 of the half-width, the same fraction as the electron's 0.216 nm.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
The sign of V₀ − E sorts bound from unbound.
Name the region, write its V, then its solution: oscillation inside with k = √(2mE)/ħ, decay outside with κ = √(2m(V₀ − E))/ħ when 0 < E < V₀. Normalisability removes two rising exponentials, so energies there are discrete; above V₀ nothing is removed and every energy is allowed.