Transcendental equations and their graphical solution
A circle and tangent branches locate every finite-well level.
This is where the finite well stops being algebra. Matching leaves one equation in E that no rearrangement can solve, so we rescale until every well looks alike, draw it once, read each level to two figures, and let a bracketed solver supply the rest: the routine for any bound state without a formula.
Build the model
No formula gives the energy, but a circle and a tangent curve pin it down.
1The big idea
Parity has already reduced the finite well to one condition per symmetry class: k tan(ka) = κ for even states and −k cot(ka) = κ for odd ones. Both k = √(2mE)/ħ and κ = √(2m(V₀ − E))/ħ depend on the unknown energy, which therefore sits inside a tangent and outside it at once, and no chain of algebra and inverse functions pulls it free: the equation is transcendental. The way through is to measure everything in units the well supplies. With z = ka and η = κa, the identity k² + κ² = 2mV₀/ħ², true at every energy, becomes a quarter-circle z² + η² = z₀² whose radius z₀ = (a/ħ)√(2mV₀) carries all of the well's data, while the conditions become fixed curves, η = z tan z and η = −z cot z, the same for every well ever built. Each branch climbs from zero to infinity across its own interval of width π/2 while the circle falls, so every branch the circle reaches is cut exactly once. The graph therefore proves that a level exists, fixes its parity, and hands a bracket to a bisection or Brent solver, which returns as many digits as anyone wants; E = V₀(z/z₀)² restores the units. The method needs V piecewise constant and symmetric. An asymmetric well loses the parity split and leaves one messier condition, and a smoothly varying V generally leaves no closed-form condition to plot, which is where integrating the equation outward, by shooting, takes over.
2In one sentence
A transcendental equation involves a non-algebraic function of its unknown; when the unknown sits both inside and outside it, as in z tan z = √(z₀² − z²), no closed form solves it, yet its roots are exact numbers, located on a graph and pinned down numerically.
Example. An electron in a well of half-width 0.500 nm and depth 0.610 eV has z₀ = 2.00. Its circle cuts z tan z at z = 1.030 and −z cot z at z = 1.896, so E = V₀(z/z₀)² puts the two levels at 0.162 eV and 0.548 eV.
3One equation in E, and no way to isolate it
Parity leaves one condition per symmetry class: k tan(ka) = κ for even states and −k cot(ka) = κ for odd ones. Written out in E, the even one reads √(2mE) tan(a√(2mE)/ħ) = √(2m(V₀ − E)). The unknown sits inside the tangent and outside it, and no sequence of algebraic steps and inverse functions separates the two: taking arctan gives ka = arctan(κ/k) + jπ, with E still on both sides. That makes it a transcendental equation of the kind with no closed form, the same obstruction as x = cos x, whose root 0.7391 is known to any precision and has no formula. The missing closed form says nothing about the physics. The condition is exact, its roots are sharp, and a calculator evaluates its own √ and arctan by iteration anyway. What is missing is a shortcut, so the job becomes finding the roots efficiently and knowing how many there are.
4Measure the well in its own units
Multiply both wavenumbers by the half-width: z = ka and η = κa, both dimensionless. The even condition becomes η = z tan z and the odd one η = −z cot z, curves that contain no property of the well at all. The well enters only through k² + κ² = 2mV₀/ħ², true at every energy, which becomes the quarter-circle z² + η² = z₀² with z₀ = (a/ħ)√(2mV₀). One number now describes the problem. An electron in a well of half-width 0.500 nm and depth 1.00 eV has z₀ = 2.56; so does a neutron in a well of half-width 2.00 fm and depth 34.0 MeV, and to the three figures their z₀ values share, the two have the ratios E/V₀ = 0.191 and 0.703, though their energies differ by a factor of about 3 × 10⁷. The reverse problem is easy: squaring and adding gives z₀ = z/|cos z| on an even branch and z₀ = z/|sin z| on an odd one, so choosing a level fixes the well in one line. Only the direction from well to levels needs the graph.
5Read the crossings off the z–η plane
Only the first quadrant matters, because κ > 0 is what makes the tails decay and ψ normalisable. There the even curve has branches on 0 ≤ z < π/2, π ≤ z < 3π/2, …, the odd curve fills π/2 ≤ z < π, 3π/2 ≤ z < 2π, …, and every branch climbs from η = 0 at its left end to infinity at its asymptote. The circle falls steadily from η = z₀ at z = 0 to η = 0 at z = z₀. A rising curve and a falling one cross at most once, and they must cross if the branch starts inside the circle, so each branch beginning below z₀ carries exactly one level, and the levels alternate even, odd, even up the well. For z₀ = 2.56 the circle cuts the first even branch near z ≈ 1.1 and the first odd branch near z ≈ 2.15, and ends before the second even branch begins at π. A careful sketch places a crossing to about ±0.02 in z, which, since E ∝ z², is about ±4% in energy at z = 1.1: good enough to count and to bracket, not to publish.
6From a bracket to as many digits as you want
The picture hands each level an interval: (n − 1)π/2 < zₙ < min(nπ/2, z₀). Do not root z tan z − √(z₀² − z²) as it stands, because it leaps from +∞ to −∞ at every asymptote. Multiply the even condition by cos z and the odd by sin z, neither of which vanishes inside its own interval, to get Fₑᵥₑₙ = z sin z − √(z₀² − z²) cos z and Fodd = z cos z + √(z₀² − z²) sin z. These are continuous, take opposite signs at the two ends of the bracket, and vanish only at the level. Bisection halves the interval at every step: shrinking a width of π/2 to 10⁻⁶ takes 21 steps, and a bracket read off a sketch to ±0.05 needs about 17. Brent's method, scipy's brentq, keeps that guarantee and converges faster near the root. Stop on the bracket width, not on |F|: an error δz in z becomes a relative error 2δz/z in E, so 10⁻⁴ in z at z ≈ 1 fixes E to 0.02%.
7Checks every root must pass
Substitute back: a root is right only if the branch and the circle give the same η, which is the same as z/|cos z| or z/|sin z| returning z₀. Dimensions come free, since z, η and z₀ are pure numbers and E = V₀(z/z₀)² carries the unit of V₀. Then push z₀ to its limits. As z₀ grows, each crossing climbs its branch towards the asymptote and zₙ approaches nπ/2, so E tends to n²π²ħ²/(2m(2a)²), the infinite well of width 2a; at z₀ = 30 the ground root is already 0.968 of π/2. As z₀ shrinks, z tan z ≈ z², the root tends to z₀(1 − z₀²/2) and η to z₀²: at z₀ = 0.100, z = 0.0995 and η = 0.0099, a level bound by only about z₀²V₀, 1% of the depth, with tails a hundred half-widths long. The classical limit is in the same numbers. At a given E/V₀ the tail length in half-widths is 1/η = 1/(z₀√(1 − E/V₀)), and z₀ grows as a√m, so in the same well and at the same E/V₀ a proton's tails are 43 times shorter than an electron's, and a grain of dust has none worth measuring.
Key equations
What each relation says, and the units it is written in.
Well-strength parameter
z₀ = (a/ħ)√(2mV₀)
The one number the roots depend on: wells with equal z₀ share the same ladder of E/V₀, whatever their size and whatever particle they hold.
a: half-width in m; V₀: depth in J; m in kg; ħ in J s. z₀ is dimensionless. Electron: z₀ = a√(V₀/(0.0381 eV nm²)).
The z₀ circle
(ka)² + (κa)² = z₀²
k² + κ² = 2mV₀/ħ² at every energy, so the unknown E becomes a point on a fixed quarter-circle; only η > 0 gives decaying, normalisable tails.
z = ka and η = κa are dimensionless; k = √(2mE)/ħ and κ = √(2m(V₀ − E))/ħ, both in m⁻¹.
Even-state condition
η = z tan z
Each branch climbs from η = 0 to infinity across its interval, so the falling circle can cut it at most once.
From k tan(ka) = κ with interior cos kx. Usable where tan z ≥ 0: 0 ≤ z < π/2, π ≤ z < 3π/2, …
Odd-state condition
η = −z cot z
Its branches fill the gaps between the even ones, so the crossings, and hence the levels, alternate even, odd, even up the well.
From −k cot(ka) = κ with interior sin kx. Usable where cot z ≤ 0: π/2 ≤ z < π, 3π/2 ≤ z < 2π, …
Bracket for the nth level
(n − 1)π/2 < zₙ < min(nπ/2, z₀)
The pole-free forms z sin z − η cos z (even) and z cos z + η sin z (odd) change sign across this interval exactly once, which is what bisection needs.
n = 1, 2, 3, … counts up from the ground state; odd n sit on even branches, even n on odd ones.
Energy from a root
E = V₀ (z/z₀)²
Units return only at this last step, which is why one universal plot serves every well with the same z₀.
E and V₀ in the same unit, z and z₀ dimensionless; the same as E = ħ²z²/(2ma²) with a in m, m in kg.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At z₀ = 4.00, walk z up from 1.00: the mismatch changes sign between 1.25 and 1.26 as the dots meet. It flips again between 1.57 and 1.58 with the dots far apart: that is the pole of tan z, not a root. Then lower z₀ below π and watch the third crossing vanish.
Live results
- BRANCH η AT TRIAL z
- 1.557
- CIRCLE η = √(z₀² − z²)
- 3.873
- MISMATCH, BRANCH − CIRCLE
- -2.316
- TRIAL E/V₀ = z²/z₀²
- 0.063
Catch the common trap
The mistake this lesson is built to prevent.
The commonest failure comes at the refinement step, not the drawing. A student codes f(z) = z tan z − √(z₀² − z²), finds a sign change and bisects. For z₀ = 4.00, f(1.30) = +0.90 and f(1.80) = −11.3, and bisection converges faithfully to z = 1.5708, which is π/2. Bisection promises a root only for a continuous function, and this sign change came from the pole of tan z, where f leaps from +∞ to −∞. The trap is convincing because the false answer, E = V₀(π/2)²/z₀² = 0.154V₀, is exactly the infinite-well level for that width, so it looks like physics. Two symptoms expose it: |f| grows instead of shrinking as the bracket narrows, and z tan z there is enormous rather than the circle's 3.68. The graph already puts the even root strictly inside (0, π/2), near 1.25; bracket within one branch, or root z sin z − √(z₀² − z²) cos z, and the solver returns 1.2524.
Test it: what changes?
- z₀ grows as √V₀
- Each root zₙ climbs towards nπ/2
- New level each time z₀ passes a multiple of π/2
- Eₙ nears the infinite-well level for width 2a
- Branch − circle mismatch flips sign
- No root there: the flip is tan z's pole
- The pole-free even form keeps its sign
- Bisecting this flip returns π/2, a false level
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is trapped in a symmetric square well 1.60 nm wide and 0.300 eV deep. Find z₀, decide which branches of the z–η plot its circle reaches, and state the interval of z, and of energy, that must hold each bound level. Take ħ²/2m = 0.0381 eV nm² for the electron.
View solution →
- The rescaled variables use the half-width, so a = 0.800 nm, not 1.60 nm. Then z₀ = (a/ħ)√(2mV₀) = a√(V₀/(ħ²/2m)) = 0.800 nm × √(0.300 eV ÷ 0.0381 eV nm²) = 0.800 × 2.806 = 2.245, a pure number as it must be.
- Branches start at z = 0 (even), π/2 = 1.571 (odd) and π = 3.142 (even). The circle ends at z = z₀ = 2.245, so it reaches the first two and misses the third: one even level and one odd level.
- The ground level lies on the first even branch, 0 < z₁ < π/2. The odd branch runs from π/2 towards its asymptote at π, but the circle stops at 2.245 first, so π/2 < z₂ < 2.245.
- Convert with E = V₀(z/z₀)². The boundary z = π/2 maps to 0.300 eV × (1.571/2.245)² = 0.147 eV, and z₀ itself maps to V₀. So 0 < E₁ < 0.147 eV and 0.147 eV < E₂ < 0.300 eV.
- Check the definition of a: using the full width gives z₀ = 4.49 and a phantom third level. Doubling a doubles z₀, just as quadrupling V₀ would, and a dimension check cannot see the difference.
Answer. z₀ = 2.245; the circle cuts one even and one odd branch, so 0 < E₁ < 0.147 eV (even) and 0.147 eV < E₂ < 0.300 eV (odd).
MediumExample 2
For the same electron well (a = 0.800 nm, V₀ = 0.300 eV, z₀ = 2.245), a sketch puts the odd crossing near z = 2.0. Refine it by bisection to four significant figures, then find E₂ and the tail length 1/κ, and compare that length with the half-width.
View solution →
- Root the pole-free odd function F(z) = z cos z + √(z₀² − z²) sin z, which vanishes exactly where −z cot z meets the circle. Bracket around the sketch reading: F(1.95) = +0.311 and F(2.10) = −0.375, opposite signs, and both ends lie inside (π/2, π), so no asymptote sits between them.
- Halve: F(2.025) = −0.018 puts the root in (1.95, 2.025); F(1.9875) = +0.150 narrows it to (1.9875, 2.025); F(2.00625) = +0.067 to (2.00625, 2.025); F(2.01563) = +0.025 to (2.01563, 2.025).
- The width started at 0.15 and halves each time, so reaching 10⁻⁴ takes 11 halvings in all; seven more converge on z₂ = 2.0211.
- Back-substitute: η = √(2.2449² − 2.0211²) = 0.977, and −z cot z at z = 2.0211 also gives 0.977, so branch and circle agree. Then E₂ = V₀(z/z₀)² = 0.300 eV × (2.0211/2.2449)² = 0.2432 eV.
- κ = η/a = 0.977 ÷ 0.800 nm = 1.221 nm⁻¹, so 1/κ = 0.819 nm, longer than the 0.800 nm half-width. A level at 81% of the depth leaks far past the walls, and it sits where the circle falls steeply, which is why a sketch reads its η poorly and the solver earns its keep.
Answer. z₂ = 2.021, so E₂ = 0.2432 eV, and its tails reach 1/κ = 0.819 nm, slightly longer than the 0.800 nm half-width.
HardExample 3
An electron well of total width 1.00 nm is to be built so that its first excited level sits exactly halfway up, at E₂ = V₀/2. Find the depth V₀ that does this. Then find every other bound level of that well to three significant figures, and the tail length of the highest one. Take ħ²/2m = 0.0381 eV nm².
View solution →
- At E = V₀/2, k = κ, so z = η. The odd condition η = −z cot z becomes cot z = −1, whose root on the odd branch π/2 < z < π is z₂ = 3π/4 = 2.356. This direction needs no solver.
- The circle through (3π/4, 3π/4) has z₀ = √2 × 3π/4 = 3.332, the same as z/|sin z|. With a = 0.500 nm, V₀ = z₀²(ħ²/2m)/a² = 11.10 × 0.0381 eV nm² ÷ 0.250 nm² = 1.692 eV, so E₂ = 0.846 eV.
- Now the forward problem. z₀ = 3.332 exceeds π = 3.142, so the circle reaches three branches: even on (0, π/2), odd on (π/2, π), which holds z₂, and even on (π, 3.332).
- Ground level: Fₑᵥₑₙ(z) = z sin z − √(z₀² − z²) cos z gives F(1.20) = −0.008 and F(1.25) = +0.212. Bisecting to 10⁻⁴ gives z₁ = 1.2018, so E₁ = 1.692 eV × (1.2018/3.3322)² = 0.220 eV.
- Third level: Fₑᵥₑₙ(π) = +√(z₀² − π²) = +1.111 and Fₑᵥₑₙ(z₀) = z₀ sin z₀ = −0.631. Bisection gives z₃ = 3.2937, so E₃ = 1.692 eV × (3.2937/3.3322)² = 1.653 eV, only 0.039 eV below the rim. Its η = √(z₀² − z₃²) = 0.505 makes 1/κ = a/η = 0.990 nm, twice the half-width.
- Checks: z/|cos z| returns 3.332 at both even roots, the parities run even, odd, even, and all three levels lie below V₀, with E₂ exactly at V₀/2 as designed.
Answer. V₀ = 1.69 eV; E₁ = 0.220 eV, E₂ = 0.846 eV and E₃ = 1.65 eV, with the top level's tails reaching 1/κ = 0.990 nm.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Read each level off the z₀ circle, then bisect.
With z = ka and η = κa, each level is where the circle z² + η² = z₀² meets η = z tan z (even) or η = −z cot z (odd). Bracket inside one branch, root the pole-free form, then E = V₀(z/z₀)². Valid for symmetric square wells.