A continuous spectrum, degenerate above zero
No states below zero, one at zero, two above.
This is where “the spectrum” stops meaning a list of levels. With nothing to confine the particle, you test each energy by asking which solutions stay bounded, then count them: none below zero, one at zero, two at every energy above. That count, and the reason behind it, runs every scattering problem still to come.
Build the model
Nothing below zero, one state at zero, and two at every energy above it.
1The big idea
The free particle's eigenvalue problem, −(ħ²/2m)ψ″ = Eψ on the whole line, has two independent solutions at every real E, so the equation alone decides nothing; admissibility decides the spectrum. The bound-state test, ψ → 0 at both ends, admits no free solution at any energy, so the test here is boundedness: a bounded solution can be wrapped into normalisable packets with energies as close to E as you like, and a growing one cannot. Apply it energy by energy. Below zero every combination of e(±κx) grows without limit on one side or the other: no states, as ⟨H⟩ = (ħ²/2m)∫|dψ/dx|² dx > 0 confirms without solving anything. At zero ψ = A + Bx, and only the constant survives: one state. Above zero e(ikx) and e(−ikx) both stay bounded, two states at each energy, the right- and left-moving motions that the reflection x → −x exchanges, matching the two real roots of p²/2m = E. Nothing in the test singles out an energy, so the spectrum is the whole half-line E ≥ 0. The levels of wells and oscillators came from demanding decay on both sides, two conditions only special energies can meet: quantisation belongs to confinement, not to the Schrödinger equation. It holds only while V = 0 everywhere: a constant V₀ moves the threshold to V₀, a step closes one side and leaves a nondegenerate continuum below its top, and a ring of length L breaks the continuum into levels whose spacing near any given energy shrinks as 1/L.
2In one sentence
The spectrum of a free particle is the set of energies at which −(ħ²/2m)ψ″ = Eψ has a solution that stays bounded as x → ±∞: every E ≥ 0, with two independent solutions at each E > 0 and only the constant at E = 0.
Example. An electron at E = 0.250 eV has two states, e(±ikx) with k = 2.56 nm⁻¹, moving at ±2.97 × 10⁵ m s⁻¹. At E = −0.250 eV the solutions e(±κx), with κ = 2.56 nm⁻¹, grow 13-fold per nanometre in one direction or the other, so there is no state.
3Change the test from normalisable to bounded
For bound states the admissibility test was ψ → 0 at both ends, so that ∫|ψ|² dx = 1 could be imposed. With V = 0 that test admits nothing: UQMI-05.02 showed that no solution of the free equation is normalisable at any energy, so applying it here would leave the spectrum empty. The working test is weaker: ψ must stay bounded as x → ±∞. The reason is operational. A bounded solution at energy E, multiplied by a broad smooth envelope, becomes a normalisable packet whose energies cluster around E, more tightly the broader the envelope, so E can be approached by genuine states as closely as you like. A solution that grows exponentially cannot be rescued that way: cut it off and normalise it, and 86% of its probability sits within 1/κ of the cut, so the state describes the cut, not the energy E. So E belongs to the spectrum when a bounded solution exists, and the number of independent bounded solutions is its degeneracy. In a confining potential the same test returns the old rule, because there the only bounded solutions are the ones that decay.
4Below zero, every solution explodes somewhere
For E < 0 put κ = √(−2mE)/ħ, and the equation becomes ψ″ = +κ²ψ, with the real solutions e(κx) and e(−κx). The general solution A e(κx) + B e(−κx) stays bounded as x → +∞ only if A = 0, and as x → −∞ only if B = 0, so the one bounded solution is ψ = 0. The growth is not subtle: for an electron at E = −0.10 eV, κ = 1.62 nm⁻¹, and e(κx) gains a factor of 1.1 × 10⁷ across 10 nm. A second argument needs no solving at all. For any normalisable ψ, integrate by parts: ⟨H⟩ = −(ħ²/2m)∫ψ*ψ″ dx = (ħ²/2m)∫|dψ/dx|² dx, the boundary term vanishing because ψ → 0 at both ends. The integrand is never negative, and it vanishes everywhere only if ψ is constant, which a normalisable function cannot be, so every state has ⟨H⟩ > 0. Classical mechanics says the same more bluntly: p²/2m = E has no real p when E < 0. With a potential the argument gives ⟨H⟩ > min V, which is why a constant V₀ simply lifts the threshold to V₀.
5At zero the second solution is a straight line
At E = 0 the equation reads ψ″ = 0: no curvature anywhere. Its solutions are ψ = A + Bx, with no exponential in sight. The constant is bounded and the line is not, so B = 0 and exactly one state survives, which is why E = 0 stands alone. Watch the travelling pair as k → 0: e(ikx) and e(−ikx) both tend to the same function, 1, so the two directions of motion merge into one state of zero momentum that is its own mirror image. The standing pair shows where the second solution went. cos kx tends to 1, while sin(kx)/k, which has slope 1 at the origin, tends to x; for k > 0 it is bounded by 1/k, and that bound diverges as k → 0. For an electron, 1/k is 6.2 nm at 1.0 meV and 195 nm at 1.0 μeV. The lone state carries no weight in practice, since a single energy point of a continuum holds zero probability in any packet. E = 0 is the bottom edge of the spectrum, and a normalisable packet can bring ⟨H⟩ as close to it as you like but never onto it: a Gaussian of width σ has ⟨H⟩ = ħ²/(8mσ²), which for an electron with σ = 10 nm is 95 μeV.
6Two states above zero, and why a bound state never has two
For E > 0 both e(ikx) and e(−ikx) have modulus 1 everywhere, so the boundedness test removes neither, and every positive energy carries two independent states. Reflection supplies the partner: V = 0 is unchanged by x → −x, so whenever ψ(x) solves the equation at E, so does ψ(−x), and reflection turns e(ikx) into the independent e(−ikx). Symmetry alone does not guarantee a pair, since cos kx is its own mirror image; the boundedness count is what leaves room for two. The basis is a choice: travelling waves labelled by momentum ±ħk, or standing waves cos kx and sin kx labelled by parity. The contrast with bound states is a theorem. For two solutions at one E, the Wronskian W = ψ₁ψ₂′ − ψ₂ψ₁′ has dW/dx = ψ₁ψ₂″ − ψ₂ψ₁″ = 0, because both obey ψ″ = (2m(V − E)/ħ²)ψ. If both vanish at infinity, the constant W is zero there and so everywhere, which makes ψ₂ a multiple of ψ₁: a one-dimensional bound state is never degenerate. For the plane waves W = −2ik, never zero, which is possible only because neither function vanishes at infinity. Classically the count is the number of real momenta with p²/2m = E: two above zero, one at zero, none below.
7Levels come from counting conditions, not from the equation
Every one-dimensional problem has the same bookkeeping. Far out on each side, where V levels off, the solution is a pair of exponentials, real where V > E and oscillating where V < E. A side where V exceeds E is closed: one of its exponentials grows and must be removed, which costs one condition. A side where V is below E is open: both waves are bounded and nothing is removed. Up to overall scale a solution at fixed E has one free ratio. Two closed sides impose two conditions on that one ratio, so they can be met together only at special energies: discrete, nondegenerate levels. One open side imposes one condition, which fixes the ratio at every energy: a continuum, each energy once. Two open sides impose none: a continuum, each energy twice. The free particle has both sides open at every E > 0, so nothing singles out any energy; the levels of the infinite well and of the harmonic oscillator came from closing both sides. That is the sense in which quantisation came from confinement: the differential equation is the same local statement everywhere, and the boundary makes the levels.
8Counting a continuum: states per unit energy
A continuum can still be counted by closing the line into a ring of length L, the periodic box of UQMI-05.03, and letting L grow. Then k = 2πn/L for every integer n, the energy (ħ²/2m)(2πn/L)² depends only on n², and the degeneracy is visible: n = 0 alone at E = 0, then n and −n together at every level above. The number of states up to energy E is N(E) = 2⌊L√(2mE)/(2πħ)⌋ + 1, close to L√(2mE)/(πħ) for large L, and its slope is the density of states g(E) = L/(πħv), with v = √(2E/m). Classical mechanics counts the same: the phase-space area with |p| ≤ √(2mE) on a length L is 2L√(2mE), and one state per area h reproduces N. For an electron on a 1.00 μm ring, 515 states lie below 0.100 eV, and near that energy g = 2.58 states per meV, arriving in pairs about 0.78 meV apart. Because g grows as L, that gap shrinks as 1/L and vanishes as L → ∞, while g/L stays fixed. g also rises as E(−1/2) toward the bottom, where a slow particle takes longest to cross, yet N stays finite.
Key equations
What each relation says, and the units it is written in.
Below zero: real exponentials
ψ(x) = A e(κx) + B e(−κx), with κ = √(−2mE)/ħ
Bounded as x → +∞ needs A = 0, and as x → −∞ needs B = 0. Nothing is left, so no negative energy belongs to the spectrum.
E < 0 in J, m in kg, ħ = 1.055 × 10⁻³⁴ J s; κ in m⁻¹, and 1/κ is the distance over which ψ grows by a factor e.
At zero: a solution with no curvature
ψ″ = 0, so ψ(x) = A + Bx
Boundedness forces B = 0 and leaves the constant alone: one state at E = 0, where e(ikx) and e(−ikx) have merged into the same function.
A carries the units of ψ and B those of ψ per metre. Bx grows without limit in both directions.
No normalisable free state has negative energy
⟨H⟩ = (ħ²/2m) ∫ |dψ/dx|² dx > 0
Integrate −ψ*ψ″ by parts; ψ → 0 kills the boundary term. Zero would need ψ constant, which is not normalisable, so the spectrum cannot reach below 0.
Integral over the whole line; ψ in m(−1/2), so |dψ/dx|² is in m⁻³ and ħ²/2m in J m², giving ⟨H⟩ in J.
Wronskian of two solutions at one energy
W = ψ₁ψ₂′ − ψ₂ψ₁′, with dW/dx = 0
If both solutions vanished at infinity, W would be 0 and ψ₂ a multiple of ψ₁: a 1D bound state is never degenerate. Plane waves never vanish, so two coexist.
ψ₁ and ψ₂ solve the same equation at the same E. For e(ikx) and e(−ikx), W = −2ik, in m⁻¹.
A constant potential moves the threshold
Spectrum E ≥ V₀, with k = √(2m(E − V₀))/ħ above it
Only the kinetic energy E − V₀ enters ψ″ = −k²ψ, so everything said about the free spectrum carries over with 0 replaced by V₀.
V₀ and E in J, k in m⁻¹. Below V₀ the same root is imaginary: real exponentials, and no state.
Density of states on a ring of length L
g(E) = dN/dE = L/(πħv), with v = √(2E/m)
It measures how dense the continuum is: g ∝ L, so near a given E the gap between levels shrinks as 1/L, and g rises as E(−1/2) at the bottom, where the particle is slowest.
L in m, v in m s⁻¹, g in J⁻¹, both directions counted. N(E) = L√(2mE)/(πħ) is phase-space area 2L√(2mE) divided by h.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Hold E at 0.10 eV and drag W out to 12 nm: both curves stay inside ±1 however far you look. Step E down to 0.00 and the dashed curve straightens into a line that leaves the frame; below zero both blow up, and the further below zero E goes, the nearer to x = 0 they leave the frame.
Live results
- BOUNDED SOLUTIONS
- 2
- WAVENUMBER k
- 1.62 nm⁻¹
- GROWTH CONSTANT κ
- 0.00 nm⁻¹
- ENVELOPE GAIN BY x = W
- 0.0 powers of 10
Catch the common trap
The mistake this lesson is built to prevent.
After a month of wells and oscillators it is natural to think the Schrödinger equation quantises energy by itself, so a free electron must also have levels, perhaps too closely spaced to resolve. The equation does no such thing: it has two solutions at every real E, and what picked out discrete values in the well and the oscillator was the demand that ψ vanish on both sides, two conditions only special energies can meet. Remove the confinement and nothing is left to satisfy at E > 0, so every positive energy is allowed exactly, not approximately. Levels return only if something closes the space, a box or ring of length L, and their spacing then comes from the size of the box, shrinking as 1/L near any given energy, not from free motion itself. A related slip reads e(−ikx) as a negative-energy solution. Both e(±ikx) have E = ħ²k²/2m > 0; the sign of k is the direction of motion, and that is the twofold degeneracy.
Test it: what changes?
- Bounded solutions: 2 above zero, 1 at zero, 0 below
- k shrinks to zero, then κ grows
- At E = 0, ψₒ straightens into a line
- Below zero, both curves grow faster as E falls
- Density of states g = L/(πħv) grows ∝ L
- Level gap near E shrinks as 1/L
- Levels above zero stay paired, n and −n
- g/L stays the same
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
In a long, uniform region an electron has constant potential energy V₀ = 0.300 eV. For E = 0.100 eV, 0.300 eV and 0.500 eV, write the general solution of the time-independent equation, say how many independent solutions stay bounded, and give k or κ. Check the count against classical mechanics, using ħ²/2m = 0.0381 eV nm² for the electron.
View solution →
- With V = V₀ the equation is ψ″ = −(2m(E − V₀)/ħ²)ψ: the free equation with E replaced by the kinetic energy E − V₀. The threshold therefore moves from 0 to V₀ = 0.300 eV, and the free-particle classification applies to E − V₀.
- E = 0.100 eV: E − V₀ = −0.200 eV, so κ = √(0.200 eV ÷ 0.0381 eV nm²) = 2.29 nm⁻¹ and ψ = A e(κx) + B e(−κx). Boundedness at +∞ needs A = 0 and at −∞ needs B = 0: no state. The growing exponential gains a factor e(2.29) = 9.9 every nanometre.
- E = 0.300 eV: ψ″ = 0, so ψ = A + Bx. The line is unbounded, so B = 0, and the constant is the one state.
- E = 0.500 eV: E − V₀ = +0.200 eV, so k = 2.29 nm⁻¹ and ψ = A e(ikx) + B e(−ikx), both terms bounded: two states, with wavelength 2π/k = 2.74 nm and speed √(2(E − V₀)/m) = 2.65 × 10⁵ m s⁻¹ in either direction.
- Checks: p²/2m = E − V₀ has no real p at 0.100 eV, one (p = 0) at 0.300 eV and two (±p) at 0.500 eV, the same 0, 1, 2. Dimensions: eV ÷ (eV nm²) = nm⁻², so the root is in nm⁻¹. Limit: V₀ → 0 returns the free threshold E = 0.
Answer. The spectrum is E ≥ 0.300 eV. At 0.100 eV there is no state (κ = 2.29 nm⁻¹); at 0.300 eV one, the constant; at 0.500 eV two, e(±ikx) with k = 2.29 nm⁻¹ and λ = 2.74 nm.
MediumExample 2
An electron moves freely round a thin ring of circumference L = 3.00 μm, modelled as a periodic box with ψ(x + L) = ψ(x). Count exactly the states with energy up to 20.0 meV and the distinct levels among them, and compare with the continuum estimate L√(2mE)/(πħ) and with the classical phase-space count. Then find the density of states near 20.0 meV and the gap from the highest level counted to the next.
View solution →
- Periodicity needs e(ikL) = 1, so k = 2πn/L with n = 0, ±1, ±2, …, and each state e(ikx)/√L is normalised on the ring: ∫|ψ|² dx over one circumference is L × (1/L) = 1. The energy is Eₙ = (ħ²/2m)(2πn/L)², which depends only on n².
- At 20.0 meV, kE = √(0.0200 eV ÷ 0.0381 eV nm²) = 0.7245 nm⁻¹, so |n| ≤ L kE/(2π) = 3000 nm × 0.7245 nm⁻¹ ÷ 2π = 345.9. That allows n = −345 to 345: N = 2 × 345 + 1 = 691 states, in 346 distinct levels, n = 0 alone and then 345 pairs ±n.
- Continuum estimate: L kE/π = 691.9. Classical count: the phase-space area with |p| ≤ ħkE = 7.6406 × 10⁻²⁶ kg m s⁻¹ along 3.00 μm is 2Lp = 4.5844 × 10⁻³¹ J s, and dividing by h = 6.6261 × 10⁻³⁴ J s gives 691.87, which rounds to the same 691.9, as it must, since 2LħkE/(2πħ) = L kE/π exactly. The extra figures are kept because the comparison is to a fraction of one state: the exact count of 691 falls below the estimate by less than one state.
- Density of states: v = √(2E/m) = 8.39 × 10⁴ m s⁻¹, so g = L/(πħv) = 3.00 × 10⁻⁶ m ÷ (π × 1.055 × 10⁻³⁴ J s × 8.39 × 10⁴ m s⁻¹) = 1.08 × 10²³ J⁻¹, or 17.3 states per meV. Units: m ÷ (J s × m s⁻¹) = J⁻¹.
- Gap: with ε₁ = (ħ²/2m)(2π/L)² = 1.671 × 10⁻⁷ eV, E₃₄₅ = 345² ε₁ = 19.89 meV and E₃₄₆ = 20.01 meV, just outside the count. The gap is (2 × 345 + 1)ε₁ = 0.1155 meV and holds two states, so 2 ÷ 0.1155 meV = 17.3 per meV, matching g. On a 3.00 mm ring the gap would be 1000 times smaller.
Answer. 691 states in 346 levels (n = 0 alone, then 345 pairs), against 691.9 from both the continuum formula and phase-space area over h. g = 1.08 × 10²³ J⁻¹ = 17.3 states per meV, and the top gap is 0.115 meV.
HardExample 3
An electron meets a potential step: V = 0 for x < 0 and V = V₀ = 0.400 eV for x > 0. Show that no state has E < 0. At E = 0.250 eV, show that exactly one bounded solution exists and find B/A, |C/A|² and the decay length of |ψ|² inside the step. At E = 0.600 eV, count the bounded solutions. State the spectrum with its degeneracy and compare with the classical motions. Use ħ²/2m = 0.0381 eV nm².
View solution →
- E < 0: on the left ψ = A e(κ₁x), since e(−κ₁x) grows as x → −∞, and on the right ψ = C e(−κ₂x), with κ₁, κ₂ > 0. Continuity of ψ at 0 gives A = C and continuity of ψ′ gives κ₁A = −κ₂C, so (κ₁ + κ₂)A = 0 and A = C = 0: no state, as ⟨H⟩ > min V = 0 already demanded.
- 0 < E < V₀: left ψ = A e(ikx) + B e(−ikx), right ψ = C e(−κx) + D e(κx), with D = 0 for boundedness. Three amplitudes and two matching conditions, A + B = C and ik(A − B) = −κC, leave one amplitude free: exactly one state at every such energy, a nondegenerate continuum, with B/A = (ik + κ)/(ik − κ).
- At E = 0.250 eV: k = √(0.250 ÷ 0.0381) nm⁻¹ = 2.56 nm⁻¹ and κ = √(0.150 ÷ 0.0381) nm⁻¹ = 1.98 nm⁻¹. Then B/A = (k² − κ² − 2ikκ)/(k² + κ²) = 0.250 − 0.968i, with |B/A| = 1: all that arrives is reflected, so the current on the left, (ħk/m)(|A|² − |B|²), is zero, as it must be beside a wave that only decays.
- C/A = 1 + B/A, so |C/A|² = 4k²/(k² + κ²) = 4E/V₀ = 2.50. Inside the step |ψ|² falls as e(−2κx), with decay length 1/(2κ) = 0.252 nm. Limits: as E → V₀, κ → 0 and the decay length diverges; as E → 0, |C/A|² → 0.
- E = 0.600 eV: k = √(0.600 ÷ 0.0381) nm⁻¹ = 3.97 nm⁻¹ on the left and k′ = √(0.200 ÷ 0.0381) nm⁻¹ = 2.29 nm⁻¹ on the right. All four exponentials are bounded, so four amplitudes minus two conditions leave two independent solutions, for instance one arriving from each side.
- Spectrum: every E > 0, nondegenerate below 0.400 eV and twofold above; even E = 0 now fails, since matching forces ψ = C(1 − κ₀x) on the left, with κ₀ = √(2mV₀)/ħ, which grows. Classically, below the step top the electron can only arrive from the left and turn back, one motion; above it, it can arrive from either side, two. Only one side is ever closed, so nothing is quantised.
Answer. No states for E ≤ 0; one at each 0 < E < 0.400 eV (at 0.250 eV, B/A = 0.250 − 0.968i, |C/A|² = 2.50, decay length 0.252 nm); two at each E > 0.400 eV, such as 0.600 eV.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Bounded solutions decide the spectrum: E ≥ 0, doubly degenerate above zero.
Test each energy for bounded solutions, then count them: none below zero, the constant at zero, e(±ikx) above. This needs V = 0 everywhere: a constant V₀ moves the threshold to V₀, a step leaves one state per energy below its top, and a ring of length L makes levels.
Next: Fourier decomposition: position space and momentum space →