Finite Square Well: Transcendental Quantisation

Derivation

Finite Square Well: Transcendental Quantisation

D-144 Home PU-202 Threads energy · waves · matter Depends on Separation of Variables and Stationary States, Continuity Conditions at Potential Steps

Statement

For a one-dimensional finite square well of depth V₀ and width 2a, a bound state (energy 0 < E < V₀) exists only when the interior wavenumber k = √(2mE)/ℏ and the exterior decay constant κ = √(2m(V₀−E))/ℏ satisfy the transcendental matching conditions k tan(ka) = κ (even parity) or k cot(ka) = −κ (odd parity). Equivalently, with dimensionless u = ka, v = κa, the states are the intersections of u² + v² = R² (with R² = 2mV₀a²/ℏ²) with the branch curves v = u tan u and v = −u cot u, giving a finite number of levels with exponentially decaying (evanescent) tails outside the well.

Why it matters

The finite well is the first quantum system where quantisation is not a tidy closed formula but the solution of a transcendental equation solved graphically or numerically. It teaches that discrete energy levels emerge from a matching condition, not from an imposed integer, and that the number of bound states is finite and controlled by a single dimensionless parameter R.

It is also the minimal model with barrier penetration: the wavefunction leaks into the classically forbidden region as an evanescent tail. This tail underlies quantum tunnelling, the exponential decay of surface states, the binding of the deuteron, and the mode structure of dielectric optical waveguides, which obey the identical eigenvalue equation.

Assumptions

The potential is exactly piecewise-constant: −V₀ for |x| < a and 0 outside. Drop this and the interior/exterior solutions are no longer pure sinusoids and exponentials, so the clean k, κ split fails and one needs the full differential equation or WKB.

The energy lies in the bound range 0 < E < V₀: above V₀ the exterior region is oscillatory (scattering continuum, no normalisable state); at or below the bottom there is no state at all.

The potential is symmetric about x = 0: this lets us split solutions into even and odd parity. Drop the symmetry and parity is no longer a good quantum number; you must match all four coefficients simultaneously rather than in two decoupled families.

The wavefunction and its first derivative are continuous at the walls: guaranteed because V has a finite (not infinite) jump, so ψ″ is finite and ψ′ cannot jump. At an infinite wall ψ′ is allowed to be discontinuous and only ψ = 0 is imposed.

Time-independent, single-particle, non-relativistic dynamics: the stationary Schrödinger equation applies. Relativistic corrections or many-body effects would modify the dispersion relation between E and k.

Derivation

−(ℏ²/2m) ψ″(x) + V(x) ψ(x) = E ψ(x)

Time-independent Schrödinger equation, the assumed prior result; we solve it region by region. A

|x| < a: ψ″ = −k² ψ, k ≡ √(2m(E+V₀))/ℏ

Inside the well V = −V₀, so E−V = E+V₀ > 0; the equation is that of a real oscillator. A

|x| > a: ψ″ = +κ² ψ, κ ≡ √(2m(−E))/ℏ

Measuring E from the well bottom convention shifts labels; taking the exterior potential as 0 and a bound state E < 0 gives a real positive κ². Here 0 < −E < V₀. A

ψ_in(x) = A cos(kx) + B sin(kx), ψ_out(x) = C e^−κ|x|

General solutions of the two ODEs. The growing exponential e^+κ|x| is discarded so ψ is normalisable (finite as |x|→∞). B

Even: ψ_in = A cos(kx); Odd: ψ_in = B sin(kx)

Because V(−x) = V(x), the Hamiltonian commutes with parity; eigenstates can be chosen even or odd, killing one of A, B. We treat the even case; the odd case is identical with cos→sin. B

Continuity of ψ at x = a: A cos(ka) = C e^−κa

Wavefunction-matching condition: ψ is continuous across a finite step. A

Continuity of ψ′ at x = a: −Ak sin(ka) = −Cκ e^−κa

Matching condition on the derivative; legal because the finite potential keeps ψ″ bounded, so ψ′ cannot jump. A

(7) ÷ (6): k tan(ka) = κ

Dividing the derivative match by the value match eliminates the unknown amplitudes A, C — the condition is on energy alone. This is the even-parity quantisation. B

Odd parity, same steps: −k cot(ka) = κ

Repeating steps 6–8 with ψ_in = B sin(kx) gives the sine analogue; sin′ = cos leads to cot. B

u ≡ ka, v ≡ κa ⇒ v = u tan u (even), v = −u cot u (odd)

Multiplying both matching conditions by a makes them dimensionless — the natural variables for the graphical solution. B

k² + κ² = 2mV₀/ℏ² ⇒ u² + v² = R², R = a√(2mV₀)/ℏ

Adding the definitions k² ∝ (E+V₀) and κ² ∝ (−E) cancels E, leaving a constant set only by the well. This constraint circle closes the system. B

Bound states = intersections of u²+v² = R² with v = u tan u and v = −u cot u

Simultaneous solution of the transcendental condition and the circle: a quarter-circle of radius R crosses each branch a finite number of times, fixing the allowed energies. A

Result

u tan u = v (even), −u cot u = v (odd), u² + v² = R²

Reading. With u = ka (interior half-oscillation count) and v = κa (tail decay over a half-width), the allowed levels are the finite set of points where the constraint quarter-circle of radius R = a√(2mV₀)/ℏ meets the tangent (even) and cotangent (odd) branches. A larger, deeper, or wider well means larger R and hence more intersections — more bound states. Because κ > 0 for every state, ψ always leaks outside as an evanescent tail e^−κ|x|, unlike the infinite well where ψ is strictly confined.

Units check. R = a√(2mV₀)/ℏ has dimensions [m]·√([kg][J])/[J·s] = [m]·√(kg·kg·m²s⁻²)/(kg·m²s⁻¹) = [m]·(kg·m·s⁻¹)/(kg·m²s⁻¹) = dimensionless, as required for the argument of tan. Likewise k, κ carry units of m⁻¹, so ka, κa are pure numbers.

Limiting cases

Infinite well (V₀→∞, R→∞): κ→∞ forces the tail to vanish, tan(ka)→∞ so ka → (2n−1)π/2 etc., recovering k_n = nπ/(2a) and E_n = n²π²ℏ²/(8ma²).
Shallow/narrow well (R < π/2): the quarter-circle meets only the first even branch — exactly one bound state. A symmetric 1D well always binds at least one state, however weak.
Threshold of the Nth state: a new level appears each time R crosses (N−1)π/2; at onset that state has E→0, κ→0 and an infinitely long, barely-bound tail.
Total count: number of bound states = ⌈2R/π⌉, alternating even, odd, even, ... in energy order (ground state always even).
Deep-but-finite: low-lying levels sit slightly below their infinite-well values because the tail lets the particle spread into the forbidden region, lowering kinetic energy.

Breaks when

E > V₀ (unbound continuum): κ becomes imaginary, the exterior solution is oscillatory rather than decaying, and no normalisable state exists — the transcendental quantisation is replaced by continuous scattering with transmission/reflection amplitudes.
Non-square (smooth) potential: if the walls are graded over a finite length, ψ_in is no longer a pure sinusoid and the simple tan/cot conditions do not hold; one must solve the full ODE or use WKB, where quantisation becomes ∮p dx ≈ (n+½)h.
Infinite walls: the derivative-matching step is illegal (ψ′ may jump), so step 7 collapses; only ψ = 0 at the walls survives and the transcendental equation degenerates to the closed-form infinite-well spectrum.
Asymmetric well (different depths on each side): parity is lost, the two exterior decay constants differ, and a threshold appears below which no bound state exists — unlike the symmetric case which always binds one.

Failure modes

Imposing ψ = 0 at the walls. Students carry over the infinite-well boundary condition; here ψ is nonzero at |x|=a and matches onto a tail. Setting ψ(a)=0 wrongly recovers the infinite-well levels.
Forgetting to drop the growing exponential. Keeping e^+κx outside gives a non-normalisable ψ and spurious "solutions."
Solving tan(ka) = κ/k without the circle constraint. The transcendental equation alone has infinitely many roots; only those also on u²+v²=R² are physical.
Sign/branch errors in cot. The odd condition is −ucot u=v with v>0; taking the positive branch places roots in the wrong interval.
Half-width vs full-width confusion. Using width L=2a in place of a (or vice versa) shifts R by a factor of 2 and miscounts the states.
Measuring E from the wrong reference. Mixing the "well bottom = 0" and "well top = 0" conventions corrupts the signs of k² and κ².

Discussion

The deep lesson is that quantisation here is a matching phenomenon. Nothing imposes integer nodes; instead, demanding that an oscillatory interior smoothly join a decaying exterior — both ψ and ψ′ continuous — over-constrains the problem, and only discrete energies satisfy it. The circle u²+v²=R² encodes the conservation of "total curvature budget": kinetic energy inside plus binding energy outside is fixed by the well, and the tan/cot branches enforce that the interior phase lands correctly at the wall.

The evanescent tail is the signature of the finite barrier. The particle has nonzero probability of being found where E < V, classically forbidden. This is not tunnelling through yet, but the same mathematics: bring a second well close and the two tails overlap, splitting each level into a symmetric/antisymmetric pair — the origin of covalent bonding and of energy bands in solids. The decay length 1/κ sets how strongly neighbouring wells couple.

The identical eigenvalue equation governs guided modes of a symmetric planar dielectric waveguide (or optical fibre step-index slab): there u and v become the transverse propagation and cladding-decay constants, R the normalised frequency ("V-number"), and the condition V < π/2 for single-mode operation is precisely the one-bound-state criterion. This is a concrete instance of how the Helmholtz and Schrödinger equations share structure: a confined oscillation matched to an external decay yields the same transcendental spectrum whether the field is a matter wave or light. Counting bound states is counting guided modes.

Common misconceptions. The finite well does not have "fewer but higher" levels than the infinite well — it has fewer levels, each slightly lower than the corresponding infinite-well level, because the wavefunction spreads into the tail and relaxes its curvature. And a symmetric 1D well binds at least one state no matter how shallow; the intuition that "a weak well may fail to trap anything" is a 3D result, not a 1D one.

Worked examples

Electron in a 0.30 nm-wide, 10 eV-deep well: how many bound states?

Well half-width a = 0.15 nm, depth V₀ = 10 eV. Compute R and count. A

R = a√(2mV₀)/ℏ

Symbolic form first (step 11). A

2mV₀ = 2(9.11×10⁻³¹ kg)(10×1.602×10⁻¹⁹ J) = 2.92×10⁻⁴⁸ kg·J

Convert eV to joules; substitute numbers. A

√(2mV₀) = 1.71×10⁻²⁴ kg·m·s⁻¹; R = (1.5×10⁻¹⁰)(1.71×10⁻²⁴)/(1.055×10⁻³⁴)

Divide by ℏ = 1.055×10⁻³⁴ J·s. A

R ≈ 2.43; count = ⌈2R/π⌉ = ⌈1.55⌉ = 2

Apply the state-count rule; 2R/π = 1.55. B

R ≈ 2.43 ⇒ 2 bound states (one even ground, one odd first-excited)

Reading. Since π/2 ≈ 1.57 < R < π ≈ 3.14, the quarter-circle crosses the first even branch and the first odd branch but not the second even branch — exactly two levels.

Units check. R came out pure-number, as required; every input was in SI before combining.

Ground-state energy of a well with R = 2 (electron, a = 0.20 nm).

Solve the even condition utan u = v = √(R²−u²) for the lowest root. B

u tan u = √(4 − u²)

Substitute the circle constraint v = √(R²−u²) into the even branch; one equation in u. B

Try u=1.0: 1.557 vs 1.732 (LHS<RHS); u=1.1: 2.16 vs 1.67 (LHS>RHS); root near u ≈ 1.03

Numerical root-bracketing between 1.0 and 1.1. B

u ≈ 1.03 ⇒ k = u/a = 1.03/(2.0×10⁻¹⁰ m) = 5.15×10⁹ m⁻¹

Convert dimensionless u back to a wavenumber. A

E+V₀ = ℏ²k²/2m = (1.055×10⁻³⁴)²(5.15×10⁹)²/(2·9.11×10⁻³¹) J

Interior dispersion (step 2 definition of k). A

E+V₀ = 1.62×10⁻¹⁹ J = 1.01 eV above the well bottom

Evaluate; this is the kinetic energy measured from −V₀. A

E + V₀ ≈ 1.0 eV (ground level, measured from well bottom)

Reading. The binding energy is V₀ − 1.0 eV. With R = 2 the well depth is V₀ = ℏ²R²/(2ma²) ≈ 3.8 eV, so the state sits ≈ 2.8 eV below the top and remains comfortably bound.

Units check. ℏ²k²/2m has units (J·s)²(m⁻¹)²/kg = J²s²m⁻²/kg = kg·m²s⁻² = J. ✓

Problems

Show that a symmetric finite well always has at least one bound state, and state the condition on R for the second (odd) state to appear.
Solution
The even branch v = utan u passes through the origin (u=0, v=0) with positive slope, while the quarter-circle u²+v²=R² starts at (0,R) and reaches (R,0). For any R>0 the two curves must cross once in 0<u<min(R,π/2), so at least one even state exists. The odd branch v = −ucot u is only positive for u>π/2, so the circle reaches it only when R > π/2. Hence the second state appears for R > π/2 ≈ 1.571.
A proton (m = 1.673×10⁻²⁷ kg) sits in a nuclear-scale well of half-width a = 2.0 fm and depth V₀ = 40 MeV. Find R and the number of bound states.
Solution
V₀ = 40×10⁶×1.602×10⁻¹⁹ = 6.41×10⁻¹² J. 2mV₀ = 2(1.673×10⁻²⁷)(6.41×10⁻¹²) = 2.14×10⁻³⁸; √ = 1.46×10⁻¹⁹ kg·m/s. R = (2.0×10⁻¹⁵)(1.46×10⁻¹⁹)/(1.055×10⁻³⁴) ≈ 2.78. Count = ⌈2R/π⌉ = ⌈1.77⌉ = 2 bound states (one even, one odd).
For the infinite-well limit, show that the even condition ktan(ka)=κ reproduces E_n = n²π²ℏ²/(8ma²) for odd n.
Solution
As V₀→∞, κ→∞, so ktan(ka)=κ→∞ requires tan(ka)→∞, i.e. ka → π/2, 3π/2, ... = nπ/2 with n odd. Then k = nπ/(2a) and E = ℏ²k²/(2m) = ℏ²n²π²/(8ma²). The odd-parity condition kcot(ka)=−κ similarly gives even n, together filling every integer — the full infinite-well spectrum for a well of full width 2a.
A well has exactly three bound states. Give the range of R, and state the parity sequence in order of increasing energy.
Solution
The Nth state appears when R crosses (N−1)π/2. Three states require the 3rd to exist but not the 4th: (3−1)π/2 < R ≤ (4−1)π/2, i.e. π < R ≤ 3π/2 (≈ 3.14 to 4.71). Parity alternates starting even: even, odd, even.
Estimate the evanescent decay length 1/κ for a state bound by 2.0 eV (i.e. V₀−E = 2.0 eV) for an electron, and comment on its size relative to atomic dimensions.
Solution
κ = √(2m(V₀−E))/ℏ. V₀−E = 2.0×1.602×10⁻¹⁹ = 3.20×10⁻¹⁹ J. 2m(V₀−E) = 2(9.11×10⁻³¹)(3.20×10⁻¹⁹) = 5.83×10⁻⁴⁹; √ = 7.64×10⁻²⁵. κ = 7.64×10⁻²⁵/1.055×10⁻³⁴ = 7.24×10⁹ m⁻¹. 1/κ ≈ 1.4×10⁻¹⁰ m = 0.14 nm — comparable to a Bohr radius (0.053 nm) and to a bond length, so the tail extends a fraction of an atomic diameter outside the well: significant, and the reason neighbouring wells couple.