The number of bound states, counted from z₀
One dimensionless number, z₀, counts every bound state.
Before you solve for a single energy, you can already say how many there are. Build z₀ from the depth, the width and the mass, divide by π/2, and the whole spectrum is counted: which states exist, which parity each has, and how much deeper or wider the well must be before the next one appears.
Build the model
One dimensionless number says how many states a well can hold.
1The big idea
The matching conditions fix each level of a finite well only through a transcendental equation, but the number of levels needs no root-finding. Measure E from the rim, write k and κ for the inside and outside wavenumbers, and the even and odd conditions collapse into one: ka = jπ/2 + arctan(κ/k), with j = 0, 1, 2, … and parity (−1)ʲ. The right-hand side is a budget for the phase ka that the wave turns through between the centre and the wall: jπ/2 is what its j nodes cost, and arctan(κ/k) is the extra turn needed before ψ′/ψ can match a decaying tail, a full π/2 at the floor of the well, where the wall is effectively impenetrable and ψ must all but vanish at it, and nothing at the rim, where the tail is flat. As E climbs from floor to rim, ka − arctan(κ/k) rises steadily from −π/2 to z₀ = (a/ħ)√(2mV₀), so it passes each jπ/2 below z₀ exactly once. That gives N = ⌊2z₀/π⌋ + 1 states, alternating even and odd, with a new one entering at E = 0 each time z₀ crosses a multiple of π/2. Because the sweep starts at −π/2 rather than at 0, the j = 0 crossing is always there: the shallowest one-dimensional well still binds one state, weakly, with a binding energy that falls as z₀⁴ and a tail far wider than the well. For a deep well the count approaches 2z₀/π, one state per 2πħ of classical phase space. The guarantee stops at one dimension. A spherically symmetric state in three dimensions must have u = rψ vanish at the centre, only the odd crossings survive, and a well with z₀ < π/2 binds nothing at all.
2In one sentence
The number of bound states of a finite square well of depth V₀ and half-width a is fixed by z₀ = (a/ħ)√(2mV₀) alone: N = ⌊2z₀/π⌋ + 1, one more than the integer part of 2z₀/π.
Example. An electron in a well 1.00 nm wide (a = 0.500 nm) and 0.500 eV deep has z₀ = 0.500 nm × √(0.500 eV ÷ 0.0381 eV nm²) = 1.81, so 2z₀/π = 1.15 and the well holds 2 bound states, one even and one odd.
3Both parity conditions are one phase condition
Measure energy from the rim, so a bound state has −V₀ < E < 0, with k = √(2m(E + V₀))/ħ inside and κ = √(−2mE)/ħ outside, both in m⁻¹ and tied by (ka)² + (κa)² = z₀². The even condition k tan(ka) = κ says tan(ka) = κ/k, so ka = nπ + arctan(κ/k). The odd condition −k cot(ka) = κ says tan(ka − π/2) = κ/k, so ka = nπ + π/2 + arctan(κ/k). Together they read ka = jπ/2 + arctan(κ/k) with j = 0, 1, 2, …, even j for even states and odd j for odd ones. Read it as a budget of phase from the centre to the wall. The jπ/2 is what the state's j nodes cost; the arctan, between 0 and π/2, is the extra turn needed before ψ′/ψ matches the decaying tail, and it falls short of the infinite well's π/2 because ψ need not reach zero at a finite wall. Check the limit: an infinitely high wall sends κ/k to infinity and the arctan to π/2, leaving ka = (j + 1)π/2, the infinite well of width 2a.
4Sweep the energy and count the crossings
Now let E climb from the floor of the well to the rim. The inside phase ka rises from 0 to z₀, while κ/k falls from infinity to 0, so arctan(κ/k) falls from π/2 to 0. Their difference ka − arctan(κ/k) therefore rises steadily from −π/2 to z₀. It passes each value jπ/2 exactly once if jπ/2 < z₀, and not at all otherwise. Counting needs no roots: every j = 0, 1, 2, … below 2z₀/π supplies one state, so N = ⌊2z₀/π⌋ + 1. The states arrive in order of j, so parity alternates even, odd, even, and state j has j nodes, all inside the well. For z₀ = 5.0, 2z₀/π = 3.18: j = 0 to 3 qualify and the well holds four states, the top one odd. Because the sweep is monotonic and always starts at −π/2, changing the well only slides the crossings along: none can merge with another or vanish in mid-spectrum, so states enter and leave only at the rim. Deepening the well binds each existing state more tightly and lets new ones in at the top.
5A new state enters at the rim with flat tails
At E = 0 the decay constant κ vanishes and the arctan with it, so state j appears exactly when z₀ = jπ/2, that is when V₀ = j²π²ħ²/(8ma²). Its inside wavefunction is cos(kx) or sin(kx) with k × 2a = jπ: the well is exactly j half-wavelengths wide for a particle of kinetic energy V₀, and ψ′ = 0 at each wall, so ψ continues outside as a constant. A constant cannot be normalised, so at threshold the state is not yet bound. Just past it κ is small and positive and the tail 1/κ is enormous; the state tightens as z₀ grows. Thresholds are evenly spaced in z₀, hence in width at fixed depth, while in depth they grow as j². For an electron with a = 0.50 nm they fall at V₀ = 0.376, 1.50, 3.38 and 6.02 eV; at a fixed depth of 1.00 eV they fall every 0.307 nm of half-width.
6Why one dimension always binds a state
The sweep starts at −π/2, not at 0, and that quarter-turn is the ground state's guarantee: however small z₀ is, the phase must still cross 0 before the rim, so j = 0 always exists. Physically, the even ground state leaves the centre with zero slope, by symmetry, and needs only enough curvature inside to meet a slowly decaying exponential at the wall. A shallow well can always supply that, because the tail may be as long as it needs. For z₀ ≪ 1, tan z ≈ z and z ≈ z₀ turn the even condition κa = z tan z into κa ≈ z₀², so EB ≈ (ħ²/2ma²)z₀⁴ = m(2aV₀)²/(2ħ²): halve the depth and the binding falls fourfold while the tail doubles. An electron in a 1.00 nm well only 5.0 meV deep has z₀ = 0.18, a binding energy close to 0.16 meV and a tail 1/κ of about 16 nm, with 94% of its probability outside the well. Bound, but barely inside.
7Three dimensions keep only the odd crossings
A spherical well of radius a and depth V₀ looks like the same problem, and for a spherically symmetric state it nearly is. Writing ψ = u(r)/r turns the radial equation into the one-dimensional Schrödinger equation on r > 0, with the same k inside and κ outside. The difference is the centre: ψ must stay finite at r = 0, so u(0) = 0, which admits sin(kr) and rules out cos(kr). Only the odd solutions of the one-dimensional well survive. Their count is the number of odd j below 2z₀/π, which is ⌊z₀/π + 1/2⌋, and the first needs z₀ > π/2, that is V₀a² > π²ħ²/(8m). Below that the sphere binds nothing. The reason is the same bending budget: u starts at zero with positive slope and must turn over, through at least a quarter wavelength, before it can meet a decaying tail. An electron in a sphere of radius 0.50 nm needs V₀ > 0.376 eV before it has a single bound state.
8Compare the count with classical phase space
For a deep well the count approaches 2z₀/π, and that number has a classical meaning. A particle at the rim energy crosses the width 2a with momentum ±√(2mV₀), so its orbit in the (x, p) plane is a rectangle of area 4a√(2mV₀), in J s. Divide by h = 2πħ and you get 2a√(2mV₀)/(πħ) = 2z₀/π: one quantum state for every h of classical phase space, the rule the old quantum theory was built on. The exact count exceeds this by between 0 and 1, and the excess comes from the arctan term again: its π/2 at the floor starts the sweep a quarter-turn below zero. For z₀ = 20 the estimate is 12.73 against N = 13, a 2% correction. For z₀ = 0.5 it is 0.32 against N = 1, and rounding the estimate would lose the only bound state. The + 1 is a quantum edge effect, largest exactly where classical intuition is weakest.
Key equations
What each relation says, and the units it is written in.
Depth parameter
z₀ = (a/ħ)√(2mV₀)
The only way depth, width and mass reach the count. For an electron, z₀ = a√(V₀/0.0381 eV nm²): doubling a doubles z₀, doubling V₀ raises it by √2.
a is the half-width in m, V₀ the depth in J, m the mass in kg, ħ = 1.055 × 10⁻³⁴ J s; z₀ is dimensionless.
Both parities in one condition
ka = jπ/2 + arctan(κ/k), j = 0, 1, 2, …
Even j is k tan(ka) = κ and odd j is −k cot(ka) = κ. The arctan is the extra phase the wall demands before ψ can join a decaying tail: π/2 at the floor, 0 at the rim, and π/2 always for infinite walls.
k = √(2m(E + V₀))/ħ and κ = √(−2mE)/ħ, both in m⁻¹, with −V₀ < E < 0; state j has parity (−1)ʲ and j nodes.
Threshold of state j
z₀ = jπ/2 ⇔ V₀ = j²π²ħ²/(8ma²)
Evenly spaced in z₀ and so in width, quadratic in depth: an electron well with a = 0.50 nm gains states at V₀ = 0.376, 1.50, 3.38 and 6.02 eV.
V₀ in J, a in m. At threshold E = 0 and κ = 0, so the tail is flat and cannot be normalised: state j counts only once z₀ > jπ/2.
Number of bound states
N = ⌊2z₀/π⌋ + 1
One state for each j with jπ/2 < z₀. Parities alternate from an even ground state, so the top state is even when N is odd and odd when N is even.
⌊ ⌋ takes the integer part; exact whenever 2z₀/π is not a whole number. N is a pure number and never less than 1.
Deep well: one state per h of phase space
N ≈ 2z₀/π = 4a√(2mV₀)/(2πħ)
The classical correspondence. N exceeds 2z₀/π by between 0 and 1, a 2% correction at z₀ = 20 and the whole answer for a shallow well.
4a√(2mV₀), in J s, is the area of the classical orbit at the rim energy: width 2a, momentum ±√(2mV₀).
Shallow well, z₀ ≪ 1
EB ≈ (ħ²/2ma²) z₀⁴ = m(2aV₀)²/(2ħ²)
The one state never leaves; it loosens as z₀⁴ while the depth falls only as z₀². The limit is the single bound state of a δ-function well of strength 2aV₀.
EB = −E is the binding energy in J; 2aV₀, in J m, is the area of the well, depth times width.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Take a to 0.05 nm and V₀ to 0.1 eV: z₀ drops to 0.08, yet the curve still rises from −1 across line 0 just short of the rim, so one state survives. Doubling a doubles the endpoint's height above line 0; doubling V₀ scales it by only √2. A sphere keeps only odd lines and binds nothing below line 1.
Live results
- DEPTH PARAMETER z₀
- 2.17
- BOUND STATES N
- 2
- PHASE-SPACE COUNT 2z₀/π
- 1.38
- DEPTH FOR THE NEXT STATE
- 4.18 eV
Catch the common trap
The mistake this lesson is built to prevent.
The tempting argument: squeezing a particle into a width 2a costs about ħ²π²/(2m(2a)²) of kinetic energy, the infinite-well ground energy, so a well shallower than that should hold nothing. In one dimension it is wrong, because a finite well does not squeeze the particle into 2a. Its walls are soft: as the well is made shallower, κ shrinks and the tails spread over 1/κ, so the region the particle actually occupies grows without limit and its kinetic cost falls faster than the depth, the binding energy going as z₀⁴ while V₀ goes as z₀². The criterion V₀ > π²ħ²/(8ma²) is exactly z₀ > π/2. That is the threshold for the second state in one dimension, and for the first state of a spherical well, where u = rψ is pinned to zero at the centre and the confinement cost is real.
Test it: what changes?
- z₀ grows as √V₀
- Existing states bind more tightly
- New states enter at the rim, E = 0
- Each newcomer's parity is opposite to the last
- Binding energy EB falls about fourfold
- Tail length 1/κ roughly doubles
- The single even state stays bound
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is trapped in a finite square well 0.80 nm wide and 3.0 eV deep. Taking ħ²/2m = 0.0381 eV nm² for the electron, find z₀, the number of bound states and their parities, and the depth at which one more state would appear.
View solution →
- The half-width is a = 0.40 nm, and z₀ = a√(V₀/(ħ²/2m)) = 0.40 nm × √(3.0 eV ÷ 0.0381 eV nm²) = 0.40 nm × 8.87 nm⁻¹ = 3.55. The units cancel, as a depth parameter's must.
- 2z₀/π = 2 × 3.55/π = 2.26. Its integer part is 2, so N = 2 + 1 = 3.
- The states are j = 0, 1 and 2, whose parities run even, odd, even. The next threshold, j = 3 at z₀ = 3π/2 = 4.71, lies above 3.55.
- State j = 3 enters when z₀ = 4.71. Since z₀ ∝ √V₀, the depth must rise by (4.71/3.55)² = 1.76, to V₀ = 3.0 eV × 1.76 = 5.3 eV. Directly: 0.0381 eV nm² × (4.712/0.40 nm)² = 5.29 eV.
Answer. z₀ = 3.55, so the well holds 3 bound states, even, odd and even; a fourth, odd, state appears once V₀ exceeds 5.3 eV.
MediumExample 2
A conduction electron in a thin semiconductor layer behaves as a particle of effective mass 0.067 mₑ in a finite square well 0.30 eV deep, whose width is the layer thickness. (a) How many bound states does a layer 10.0 nm thick hold? (b) For what range of thickness does it hold exactly two?
View solution →
- The kinetic constant scales inversely with mass: ħ²/2meff = 0.0381 eV nm² ÷ 0.067 = 0.569 eV nm². So z₀ = a√(0.30 eV ÷ 0.569 eV nm²) = a × 0.726 nm⁻¹.
- (a) A 10.0 nm layer has a = 5.00 nm, so z₀ = 3.63 and 2z₀/π = 2.31. N = 2 + 1 = 3: even, odd, even.
- (b) Exactly two states means the j = 1 threshold has been passed but the j = 2 threshold has not: π/2 < z₀ < π.
- Convert to half-width: a > (π/2) ÷ 0.726 nm⁻¹ = 2.16 nm and a < π ÷ 0.726 nm⁻¹ = 4.33 nm, so the thickness 2a lies between 4.33 nm and 8.65 nm.
- Check: the thresholds in thickness are evenly spaced, every 4.33 nm, and 10.0 nm ÷ 4.33 nm = 2.31 reproduces 2z₀/π from part (a), as it must.
Answer. (a) z₀ = 3.63, so 3 bound states. (b) Exactly two for thicknesses between 4.33 nm and 8.65 nm; every further 4.33 nm of thickness admits one more state.
HardExample 3
An electron sits in a shallow well 1.00 nm wide and only 20.0 meV deep. Show that it has a bound state, estimate the binding energy from the shallow-well limit and then find it accurately, find the tail length and the fraction of probability outside the well, and decide whether a spherical well of radius 0.500 nm and the same depth binds the electron.
View solution →
- a = 0.500 nm, so z₀ = 0.500 nm × √(0.0200 eV ÷ 0.0381 eV nm²) = 0.500 nm × 0.7245 nm⁻¹ = 0.362, and 2z₀/π = 0.231. The integer part is 0, so N = 1: one even state, however shallow.
- Shallow limit: κa ≈ z₀² = 0.131, so EB ≈ (ħ²/2ma²)z₀⁴ = (0.0381 eV nm² ÷ 0.250 nm²) × 0.01722 = 0.1524 eV × 0.01722 = 2.62 meV.
- Accurately: solve the even condition κa = z tan z with z = √(z₀² − (κa)²). Start from κa = 0.131 and iterate; the values run 0.1186, 0.1220, 0.1211, … and settle at κa = 0.1213 with z = 0.3414. Then EB = 0.1524 eV × 0.1213² = 2.24 meV. The estimate was 17% high because it treats the whole depth as kinetic energy inside, z = z₀; in fact z = 0.341 < z₀ = 0.362, and that shortfall outweighs the small gain from tan z exceeding z.
- Tail length: 1/κ = a/(κa) = 0.500 nm ÷ 0.1213 = 4.12 nm, over four times the width of the well itself.
- Fraction outside, from ψ = cos(kx) inside and a matched exponential outside: Pₒᵤₜ = [cos²z/(κa)] ÷ [1 + sin(2z)/(2z) + cos²z/(κa)] = 7.32 ÷ (1 + 0.924 + 7.32) = 0.792.
- Sphere: only odd solutions survive, so binding needs z₀ > π/2, that is V₀ > (π/2)² × 0.0381 eV nm² ÷ (0.500 nm)² = 0.376 eV, which is 18.8 times the actual depth. At 20.0 meV the sphere binds nothing.
Answer. One even state, bound by 2.24 meV (the z₀⁴ estimate gives 2.62 meV), with a 4.12 nm tail and 79% of its probability outside the well. The spherical well needs V₀ > 0.376 eV and binds nothing.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
N = ⌊2z₀/π⌋ + 1: count states before finding any.
Compute z₀ = (a/ħ)√(2mV₀) with the half-width a. State j, of parity (−1)ʲ, is bound once z₀ exceeds jπ/2, so a one-dimensional well always holds its even ground state. At z₀ = jπ/2 exactly the new state is not yet bound; a spherical well keeps only odd crossings.