University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.7
Bohr-Sommerfeld quantisation and the quantum bouncer
This is where the WKB machinery pays out: a whole spectrum from one integral and a count of the phase lost at each end of the motion. We trust it only after it reproduces two spectra known exactly, then aim it at a neutron bouncing on a mirror, a quantum system measured in micrometres and picoelectronvolts.
Build the model
Connect the measurement to the mechanism.
Bohr-Sommerfeld quantisation is what the connection formulas say about a trapped particle. Between the ends of the classical motion the WKB wave is a cosine of the accumulated phase ∫p dx/ħ, and each end fixes that cosine's offset on its own: a smooth turning point, through the Airy matching of UQMII-11.06, sets it at π/4, and a hard wall, where ψ must vanish, sets it at π/2. A bound state exists only where the two descriptions are the same function, so the phase between the ends must be (n − 1)π plus the two end phases: ∫p dx = (n − ½)πħ for two smooth ends, (n − ¼)πħ for one wall and nπħ for two.
The rule is correct to the first two orders of the expansion in ħ, yet it returns the oscillator and the infinite well exactly, and for a neutron above a mirror, V = mgz, it gives Eₙ = [3π(n − ¼)/2]²⁄³(mg²ħ²/2)¹⁄³: 0.76% below the exact Airy-zero energy at n = 1 and 0.06% below by n = 3. No fine-structure constant enters; the small parameter is ħ over the action of the orbit, of order 1/n, and the error goes as its square. The rule stops working where its derivation does: when an end lies within about one Airy length ℓ of a turning point, so no stretch of genuine WKB wave joins the two; when a wall is soft on the scale of the local wavelength; and when a barrier is thin enough to tunnel through, which the rule cannot see.
- Simple definition
- Bohr-Sommerfeld quantisation fixes bound-state energies by requiring the phase ∫p dx/ħ accumulated between the ends of the classical motion to equal (n − 1)π plus the phase lost at each end: π/4 at a smooth turning point, π/2 at a hard wall.
- Example
- A neutron above a mirror has one wall and one smooth turning point, so for n = 1 the integral of p dz from the mirror up to the turning point is ¾πħ. That gives E₁ = 1.396 peV at a height of 13.6 μm, 0.76% below the exact Airy value of 1.407 peV.
Each smooth end costs π/4, and together they make the ½. For V = ½mω²x² the integral is πE/ω, so E = (n − ½)ħω: the oscillator, exactly.
p(x) = √(2m(E − V(x))) in kg m s⁻¹; turning points a, b in m; ħ = 1.0546 × 10⁻³⁴ J s; each side is an action in J s
ψ = 0 at the wall turns the wave into a sine there, an offset of π/2 in place of π/4. This is the rule for the bouncer and for any particle pressed against a barrier.
hard wall at x = 0, where V rises far above E within a small fraction of the local wavelength; b in m, p in kg m s⁻¹
On a flat floor p = √(2mE) is constant and the sine is an exact solution, so pL = nπħ gives E = n²π²ħ²/2mL², the infinite well exactly.
walls at x = 0 and x = L, in m; ψ = 0 at each wall is exact, but the rule is exact only where p is constant between them
In units of ℓ and mgℓ every linear-potential bouncer is one problem; an electron in a field F uses eF in place of mg. WKB holds only several ℓ below a turning point.
m in kg, g in m s⁻² and ℓ in m; for a neutron with g = 9.81 m s⁻², ℓ = 5.868 μm and mgℓ = 0.6018 peV
From ∫₀(zₙ) √(2m(E − mgz)) dz = (2/3)√(2m)E³⁄²/mg = (n − ¼)πħ. Energies grow as (n − ¼)²⁄³, so the levels crowd together as n rises.
Eₙ in J; (mg²ħ²/2)¹⁄³ = mgℓ; turning heights zₙ = Eₙ/mg in m; the bracket is 2.320, 4.082, 5.517 for n = 1, 2, 3
Bohr-Sommerfeld is low by 0.76%, 0.15% and 0.062%. The Airy-zero expansion, quoted, puts the fractional gap near 5/(48t²) with t = 3π(n − ¼)/2, an order-ħ² effect.
aₙ are the zeros of the Airy function, quoted from a table: |a₁| = 2.3381, |a₂| = 4.0879, |a₃| = 5.5206
Two cosines that must be one function
Take a well with smooth turning points a and b and quote the connection formula of UQMII-11.06: joined to the decaying exponential beyond a, the allowed-region wave is ψ ≈ (2C/√p) cos(θₐ − π/4), where θₐ(x) = (1/ħ)∫ p dx′ from a to x. Starting from b instead gives ψ ≈ (2C′/√p) cos(θb − π/4), with θb measured from x to b. Their sum Θ = θₐ + θb is the whole phase from a to b and does not depend on x. The second cosine is then cos[(θₐ − π/4) − (Θ − π/2)], the same function as the first up to sign only if Θ − π/2 is a whole multiple of π. So ∫ₐᵇ p dx = (n − ½)πħ with n = 1, 2, 3, …, and C′ = (−1)(n−1) C. The cosine passes through n − 1 zeros between the turning points, so n counts states from the ground state up, and the ½ is the sum of the two π/4 offsets. What has been derived is a condition correct to the first two orders in ħ, resting on connection formulas that need V to be nearly linear across each turning point.
A hard wall costs π/2, not π/4
The π/4 is not a universal constant; it is what an Airy function contributes where V rises smoothly through E. At a wall where V jumps far above E there is no Airy region. The wave must vanish at the wall, x = 0, so in the allowed region ψ ∝ (1/√p) sin θ₀ = (1/√p) cos(θ₀ − π/2), with θ₀ the phase measured from the wall. Repeat the previous argument with π/2 at this end and π/4 at the smooth turning point b: the cosines agree only if Θ − 3π/4 = (n − 1)π, so ∫₀ᵇ p dx = (n − ¼)πħ. With two walls both ends give π/2 and the condition becomes nπħ. The general bookkeeping is (n − 1)π plus π/4 for each smooth end and π/2 for each wall. Hardness has a scale: V must rise, and the wave must die inside it, over a distance much shorter than the local wavelength. A glass mirror presents a neutron with a potential of order 100 neV, a tabulated value, about 7 × 10⁴ times the 1.4 peV of the lowest bouncing state, and the wave dies within 1/κ ≈ 14 nm of the surface, against ℓ = 5.9 μm.
The oscillator and the well come out exact
Test the rule where the answers are known. For V = ½mω²x² the turning points are ±A with A = √(2E/mω²), and ∫ √(2m(E − ½mω²x²)) dx from −A to A is half the area of the phase-space ellipse with semi-axes A and √(2mE): πE/ω. Setting it equal to (n − ½)πħ gives E = (n − ½)ħω, which is Quantum Mechanics I's (n′ + ½)ħω with n′ = n − 1, ground state included. Dimensions: E/ω is J s, an action, as the right side requires. That an approximation dropping every order beyond the second lands exactly here is checked against the ladder-operator spectrum; that every higher WKB correction vanishes for the oscillator is quoted, not derived. The infinite well is exact for a reason you can see. Its floor is flat, so p = √(2mE) is constant, the WKB sine is an exact solution with no ħ² correction to drop, and ψ = 0 at each wall is exact. Two walls give pL = nπħ and E = n²π²ħ²/2mL²: for an electron with L = 1.00 nm, 0.376 eV at n = 1, as QM I found.
The quantum bouncer by Bohr-Sommerfeld
Put a neutron above a horizontal mirror at z = 0, with V = mgz for z ≥ 0. The mirror is a hard wall and the top of the motion, zₜ = E/mg, is a smooth turning point, so the one-wall rule applies: ∫₀(zₜ) √(2m(E − mgz)) dz = (2/3)√(2m)E³⁄²/mg = (n − ¼)πħ. Solving, Eₙ = [3π(n − ¼)/2]²⁄³(mg²ħ²/2)¹⁄³. Dimensions: mg²ħ²/2 is kg × m² s⁻⁴ × J² s² = J³, so its cube root is an energy, and it equals mgℓ with the natural length ℓ = (ħ²/2m²g)¹⁄³. In these units every bouncer is one problem: Eₙ/mgℓ = zₙ/ℓ = 2.320, 4.082 and 5.517 for n = 1, 2 and 3. For a neutron, m = 1.675 × 10⁻²⁷ kg and g = 9.81 m s⁻² give ℓ = 5.868 μm and mgℓ = 0.6018 peV, so E₁ = 1.396 peV with its turning point 13.6 μm up. Only ħ, m and g appear: no fine-structure constant enters, and the expansion parameter is ħ divided by the action of the orbit. Doubling g multiplies every energy by 2²⁄³ = 1.59 and every turning height by 2⁻¹⁄³ = 0.79.
Against the Airy zeros: where the 0.8% is lost
This potential can also be solved exactly. For V = mgz the Schrödinger equation is Airy's equation in (z − zₜ)/ℓ, and the solution that decays upward is Ai((z − zₜ)/ℓ), as UQMII-11.05 showed. The mirror demands ψ(0) = 0, so Ai(−E/mgℓ) = 0 and Eₙ = |aₙ| mgℓ, with |aₙ| = 2.3381, 4.0879 and 5.5206 from a table of Airy zeros: the energies are derived, the three numbers quoted. Bohr-Sommerfeld sits below them by 0.76%, 0.15% and 0.062%. The π/4 is not to blame, because the Airy matching is exact for a linear potential. The error is made at the floor, where the WKB cosine is used to impose ψ(0) = 0. The condition ħ|dp/dz| ≪ p² of UQMII-11.04 becomes ½(ℓ/d)³⁄² ≪ 1 at depth d below the turning point, and at the floor d = zₙ: the ratio is 0.14 for n = 1, 0.061 for n = 2 and 0.039 for n = 3. The asymptotic expansion of the Airy zeros, quoted, puts the gap at about 5/(48t²) with t = 3π(n − ¼)/2, a relative correction of order ħ² that falls as 1/n².
The test: neutrons between a mirror and an absorber
The exact values for a neutron are E₁ = 1.407 peV with z₁ = 13.72 μm, and E₂ = 2.460 peV with z₂ = 23.99 μm. In 2002 a team at the Institut Laue-Langevin in Grenoble sent ultracold neutrons, moving horizontally at several metres per second, through a slit between a polished glass mirror and a rough absorber held above it, and counted the transmission as the slit height changed. Classically the count should grow smoothly as the gap opens. Instead it was consistent with zero below about 15 μm and switched on above, the step at which the lowest state first fits under the absorber, a result quoted here from the experiment. The step sits above z₁ because ψ does not stop at the turning point: the Airy tail decays over about ℓ, so a state is scraped away until the gap clears its turning point by a fraction of ℓ. The spacing E₂ − E₁ = 1.053 peV corresponds to 254.6 Hz, and later experiments drove transitions between such levels by vibrating the mirror, turning the bouncer into a probe of gravity at the micrometre scale.
Change one variable at a time
Make the relationship visible.
Raise E until the dot sits on the dashed level: at 1.40 peV, ∫p dz = 0.75πħ (the exact level is 1.407 peV). Carry on to 2.46 and 3.32 peV for n = 2 and 3. Then lower the absorber below 13.6 μm and the count of levels that fit falls to zero.
PHASE ∫p dz / πħ0.598
TURNING HEIGHT11.7 μm
NEAREST WKB LEVEL1.396 peV
LEVELS BELOW ABSORBER1
Live interpretationPHASE ∫p dz / πħ: 0.598. TURNING HEIGHT: 11.7 μm. NEAREST WKB LEVEL: 1.396 peV. LEVELS BELOW ABSORBER: 1
Catch the common trap
Explain before calculating.
The one-wall rule gives the neutron bouncer's ground state as E₁ = 1.396 peV. Which statement about that number is right?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA neutron sits in its lowest bouncing state above a horizontal mirror on the Moon, where g = 1.62 m s⁻². Use the one-wall Bohr-Sommerfeld rule to find its energy in peV and its classical turning height in μm, then check both against the Earth values by scaling. Take m = 1.675 × 10⁻²⁷ kg.
- The mirror is a hard wall and the top of the motion is a smooth turning point, so the integral of p dz from 0 to z₁ is (1 − ¼)πħ = ¾πħ. The bouncer integral then gives E₁ = t²⁄³ mgℓ with t = 3π(¾)/2 = 9π/8 = 3.534, and t²⁄³ = 2.3203.
- Energy scale: mg²ħ²/2 = (1.675 × 10⁻²⁷ kg)(1.62 m s⁻²)²(1.0546 × 10⁻³⁴ J s)²/2 = 2.445 × 10⁻⁹⁵ J³. Its cube root is mgℓ = 2.902 × 10⁻³² J = 0.18114 peV, so E₁ = 2.3203 × 0.18114 peV = 0.4203 peV.
- Turning height: z₁ = E₁/mg = 6.734 × 10⁻³² J ÷ (1.675 × 10⁻²⁷ kg × 1.62 m s⁻²) = 6.734 × 10⁻³² J ÷ 2.7135 × 10⁻²⁷ N = 2.482 × 10⁻⁵ m = 24.8 μm, which is 2.320ℓ with ℓ = 10.70 μm.
- Scaling check: E ∝ g²⁄³ and z ∝ g⁻¹⁄³. From Earth's 1.396 peV and 13.61 μm, the factor (1.62/9.81)²⁄³ = 0.3010 gives 0.4202 peV and (9.81/1.62)¹⁄³ = 1.823 gives 24.81 μm. Both agree with the direct results within the rounding of the Earth values.
- Bohr-Sommerfeld falls short of the exact Airy values by the same 0.76% as on Earth, because the fractional error depends only on n: the exact values are 0.4235 peV and 25.0 μm.
AnswerE₁ ≈ 0.420 peV at a turning height of 24.8 μm by Bohr-Sommerfeld, against the exact 0.424 peV and 25.0 μm. Weaker gravity lowers the level and lifts the neutron higher.
MediumA neutron moves between two horizontal mirrors, a floor at z = 0 and a ceiling at H = 10.0 μm. Decide which quantisation rule applies to its lowest state, find that state's energy by Bohr-Sommerfeld, show that the rule reduces to the infinite well plus mgH/2 when E ≫ mgH, and compare with a numerical solution of the Schrödinger equation, 2.552 peV. Use ℓ = 5.868 μm and mgℓ = 0.6018 peV.
- mgH = 0.6018 peV × 10.0/5.868 = 1.03 peV, while the infinite-well ground state π²ħ²/2mH² = 2.045 peV lies well above it. The lowest state has no turning point below the ceiling, so both ends are hard walls and the integral of p dz from 0 to H is nπħ.
- With ε = E/mgℓ and h = H/ℓ = 1.704, the integral is (2√(2m)/3mg)[E³⁄² − (E − mgH)³⁄²], and the condition becomes (2/3)[ε³⁄² − (ε − h)³⁄²] = nπ, valid when ε exceeds h.
- Limit E ≫ mgH: expanding (ε − h)³⁄² to second order in h leaves hε¹⁄² − h²/(4ε¹⁄²) = nπ, so ε ≈ (nπ/h)² + h/2. That is E ≈ n²π²ħ²/2mH² + mgH/2: the infinite well shifted by the average of mgz across the box, which is first-order perturbation theory.
- Start Newton's method from that estimate, ε₀ = 3.3984 + 0.8521 = 4.2505, or 2.558 peV. There (2/3)(8.7631 − 4.0632) = 3.1333, short of π by 0.0083, and the slope ε¹⁄² − (ε − h)¹⁄² = 0.4660 gives ε = 4.2505 + 0.0179 = 4.2684. A second step changes nothing at this precision.
- E₁ = 4.2684 × 0.6018 peV = 2.569 peV, which is 0.64% above the numerical 2.552 peV. First-order perturbation theory, 2.558 peV, happens to be closer at n = 1, while at n = 2 the rule's error drops to 0.01%.
AnswerTwo hard walls, so the integral of p dz across the gap is πħ: E₁ ≈ 2.57 peV by Bohr-Sommerfeld against 2.552 peV numerically, 0.6% high. For E ≫ mgH the rule reduces to n²π²ħ²/2mH² + mgH/2.
HardA real neutron mirror is a potential step of height V₀ = 100 neV at z = 0, not an infinite wall. Match the allowed-region wave to the decaying wave inside the mirror to find the phase the wall really contributes, rewrite the one-wall rule, and find the shift of the neutron's ground-state energy, E₁ = 1.407 peV. Compare it with the rule's own 0.76% error, and read it as a displaced floor. Use m = 1.675 × 10⁻²⁷ kg, mgℓ = 0.6018 peV and mg = 0.1026 peV μm⁻¹.
- Inside the mirror, z ≤ 0, ψ = D e(κz) with ħκ = √(2m(V₀ − E)). Just above it the wave is sin(θ(z) + δ), with θ measured from z = 0 and ħk = √(2mE) the momentum there. Matching ψ′/ψ at z = 0 gives k cot δ = κ, so tan δ = k/κ = √(E/(V₀ − E)); an infinite wall is δ = 0.
- Written as cos(θ − (π/2 − δ)), the wall now contributes π/2 − δ in place of π/2. Joining to the smooth turning point as before gives ∫₀(zₜ) p dz = (n − ¼)πħ − δħ, or (2/3)ε³⁄² = (n − ¼)π − δ with ε = E/mgℓ.
- Numbers for n = 1: with V₀ = 1.00 × 10⁵ peV, δ ≈ √(E/V₀) = √(1.407 × 10⁻⁵) = 3.75 × 10⁻³ rad. The phase changes with energy at the rate d[(2/3)ε³⁄²]/dε = ε¹⁄² = 1.529, so Δε = −δ/1.529 = −2.453 × 10⁻³ and ΔE = −2.453 × 10⁻³ × 0.6018 peV = −1.48 × 10⁻³ peV, or −0.105%.
- Displaced floor: 1/κ = ħ/√(2mV₀) = 14.4 nm. The phase k/κ is exactly what the wave would gain over a distance 1/κ, so the soft mirror acts like a hard one 14.4 nm lower. Lowering the floor under V = mgz lowers every level by mg/κ = 0.1026 peV μm⁻¹ × 0.0144 μm = 1.48 × 10⁻³ peV, the same shift.
- Because κ barely depends on E when V₀ ≫ E, the shift is the same 1.5 × 10⁻³ peV for every low level and for the exact Airy energies too. It is about a seventh of the 0.76% by which Bohr-Sommerfeld misses at n = 1, so a 100 neV mirror is a hard wall at the 0.1% level.
Answertan δ = √(E/(V₀ − E)), so the wall contributes π/2 − δ with δ = 3.75 × 10⁻³ rad. E₁ drops by 1.48 × 10⁻³ peV (0.105%), the same as a hard floor 14.4 nm lower: a seventh of the rule's own 0.76% error.