University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.2
The dimensionless Coulomb radial equation
This is the move that turns the hydrogen problem from a differential equation cluttered with ħ, μ, e and ε₀ into a question about one number. Measure length in the unit the binding energy sets, and the energy stops being an eigenvalue you hunt for: it becomes a coupling strength, ρ₀, whose allowed values the rest of the week will find.
Build the model
Connect the measurement to the mechanism.
The radial equation Quantum Mechanics I handed over, −(ħ²/2μ)u″ + [−Ze²/(4πε₀r) + ħ²l(l+1)/(2μr²)]u = Eu, carries ħ, μ, Ze²/4πε₀ and an unknown E, and hides how little of that matters. Four dimensional inputs built from three base dimensions admit one dimensionless combination: the ratio of 1/κ = ħ/√(−2μE), the length the binding energy sets, to the Coulomb length a/Z. Measure r in units of 1/κ and, as this lesson derives, the equation becomes d²u/dρ² = [1 − ρ₀/ρ + l(l+1)/ρ²]u, every constant, Z included, collapsed into one pure number ρ₀ = 2Z/(κa).
Because κ was built from E, the energy is no longer an eigenvalue on the right-hand side; it is ρ₀ in disguise, E = −4Z²ℛ/ρ₀², and the bound-state problem becomes the question of which ρ₀ admit a solution that vanishes at the origin and can be normalised. Two results follow before anything is solved: the allowed ρ₀ are the same numbers for every one-electron ion, so each spectrum is hydrogen's scaled by Z²μ, and the speed scale ħκ/μ = 2Zαc/ρ₀ shows the electron in the tightest states moving at about Zαc, so this non-relativistic, spinless, point-nucleus equation is right up to corrections of relative order (Zα)², quoted from later units, not derived. The construction stops at zero energy: above it κ is imaginary, the 1 becomes −1, and the continuum belongs to scattering.
- Simple definition
- The dimensionless Coulomb radial equation is the radial equation for u = rR with r measured in units of 1/κ = ħ/√(−2μE), which leaves l and one pure number, ρ₀ = 2Z/(κa), as its only parameters, the unknown energy being carried inside ρ₀.
- Example
- For hydrogen at a trial energy E = −3.40 eV, κ = √(2μ|E|)/ħ = 9.44 nm⁻¹ and ρ₀ = 2/(κa) = 4.00. A He⁺ ion with that same ρ₀ has E = −13.6 eV, four times deeper, because at fixed ρ₀ the energy scales as Z².
Real only when E is negative. At E = −13.598 eV in hydrogen 1/κ = 52.95 pm, the Bohr radius a for the reduced mass; a state bound four times more weakly has a ruler twice as long.
E in J, negative for a bound state; μ the reduced mass in kg; ħ in J s; κ in m⁻¹, so ρ is a pure number
The bracket is (Veff − E)/|E|. Only ρ₀ and l survive, so one family of solutions u(ρ; ρ₀, l) covers every energy, every one-electron ion and every reduced mass.
u = rR, zero at ρ = 0; l = 0, 1, 2, …; the bracket's terms are energy, Coulomb and centrifugal, all dimensionless
Twice the ratio of the energy's length 1/κ to the Coulomb length a/Z. The unknown E sits inside it, so finding the spectrum means finding which ρ₀ are allowed.
equivalently 2Z/(κa), where a = 4πε₀ħ²/(μe²) = 52.95 pm for hydrogen; e in C, ε₀ in F m⁻¹; ρ₀ is dimensionless
Derived here by inverting ρ₀ = 2Z/(κa). Later in this section the allowed values prove to be ρ₀ = 2n; quoted now, ρ₀ = 2 gives −13.598 eV.
ℛ = ħ²/(2μa²) = ½μc²α², 13.598 eV for hydrogen's reduced mass; E comes out in the unit ℛ is given in
No allowed region, no bound state: every l ≥ 1 level has E > −Z²ℛ/[l(l+1)], the floor of its effective well, which is −2.27 eV for d states of hydrogen.
zeros of 1 − ρ₀/ρ + l(l+1)/ρ², dimensionless; r± = ρ±/κ in m; real only when ρ₀ ≥ 2√(l(l+1))
At ρ₀ = 2 in hydrogen it is αc = 2.19 × 10⁶ m s⁻¹. With v/c of order Zα, relativistic corrections are smaller by (Zα)², a result quoted from fine structure.
α = e²/(4πε₀ħc) = 1/137.036, dimensionless; c in m s⁻¹; ħκ/μ is a speed in m s⁻¹
Two lengths, one ratio
Quantum Mechanics I supplies the radial equation for u = rR, quoted here: −(ħ²/2μ) d²u/dr² + [−Ze²/(4πε₀r) + ħ²l(l+1)/(2μr²)] u = E u, with u(0) = 0. Count what it contains before doing any algebra: ħ, the reduced mass μ, the coupling Ze²/(4πε₀), the energy E and the integer l. From ħ, μ and the coupling the only length is a/Z, with a = 4πε₀ħ²/(μe²) = 52.95 pm for hydrogen. From ħ, μ and E the only length is 1/κ = ħ/√(−2μE). Four dimensional inputs built from M, L and T leave exactly one dimensionless group, and the ratio of these two lengths is it, so apart from l the solution can depend on the constants only through that ratio, and on r only through r measured in one of the two lengths. For hydrogen at E = −3.40 eV, 1/κ = 105.9 pm, twice a; the parameter defined below as ρ₀ is twice this ratio, 4.00.
Rescaling r by 1/κ, term by term
Multiply the equation by −2μ/ħ² and write the negative energy as E = −ħ²κ²/(2μ): d²u/dr² = [κ² − 2μZe²/(4πε₀ħ²r) + l(l+1)/r²] u. Now put r = ρ/κ, so that d²/dr² = κ² d²/dρ², and divide through by κ². The energy term becomes exactly 1. The centrifugal term keeps its form, l(l+1)/ρ², because it was already a pure number over a length squared. The Coulomb term becomes ρ₀/ρ with ρ₀ = Zμe²/(2πε₀ħ²κ). Check its dimensions: μe²/ε₀ carries kg J m and ħ²κ carries J² s² m⁻¹, so the ratio is kg m² J⁻¹ s⁻², which is 1. The result, derived here in full, is d²u/dρ² = [1 − ρ₀/ρ + l(l+1)/ρ²] u, with the condition u = 0 at the origin carried over unchanged. It is the equation the rest of this section solves.
The energy becomes a coupling strength
Scaling by the Coulomb length instead, x = Zr/a, gives d²u/dx² = [|E|/(Z²ℛ) − 2/x + l(l+1)/x²] u with ℛ = ħ²/(2μa²), and E stays an eigenvalue in the usual place. Scaling by κ fixes the energy term at 1 and moves the unknown into the Coulomb coefficient; inverting ρ₀ = 2Z/(κa) gives E = −4Z²ℛ/ρ₀². The map is one-to-one for bound states, deep binding meaning small ρ₀ and a vanishing binding energy meaning ρ₀ → ∞. So 'which energies are allowed?' becomes 'how strong must the attraction be, measured against the binding, for a solution that vanishes at the origin to be normalisable?'. For every ρ₀ there is a solution that vanishes at the origin; only special values make it die away at infinity too. None of this applies above zero energy: there κ is imaginary, and writing E = ħ²k²/(2μ) and ρ = kr turns the 1 into −1, so u oscillates far out at every energy. Those continuum states belong to scattering.
One equation for every one-electron ion
Replacing e² by Ze² multiplies ρ₀ by Z at fixed κ and changes nothing else; the reduced mass has vanished into ρ₀ as well. So whatever list of allowed ρ₀ the coming topics find for each l, it is the same list for hydrogen, He⁺, Li²⁺, positronium and muonic hydrogen, and the scaling can be derived before any solution exists: at a given ρ₀, κ = 2Z/(ρ₀a) is proportional to Zμ, every length scales as 1/(Zμ) and every energy as Z²μ. Positronium tests the μ part. Its reduced mass is mₑ/2 against 0.999456 mₑ for hydrogen, so every positronium interval should be 0.500272 times the matching hydrogen one and the atom twice as large. Hydrogen's 1S–2S interval, measured at 2.466061 × 10¹⁵ Hz, then predicts 1.233702 × 10¹⁵ Hz; positronium's is measured at 1.233607 × 10¹⁵ Hz. They agree to 8 parts in 10⁵, a residue of order α².
Reading the bracket: where u can oscillate
The bracket equals (Veff − E)/|E|, with Veff the Coulomb plus centrifugal potential, so its sign tells u what to do. Where it is positive, d²u/dρ² has the sign of u and the curve bends away from the axis: classically forbidden. Where it is negative, u bends back toward the axis and oscillates: classically allowed. Its zeros solve ρ² − ρ₀ρ + l(l+1) = 0, giving the turning points ρ± = [ρ₀ ± √(ρ₀² − 4l(l+1))]/2. For l = 1 and ρ₀ = 4 they are 0.586 and 3.414, which with 1/κ = 2a are r = 1.17a and 6.83a. A bound state needs somewhere to oscillate, so ρ₀ must exceed 2√(l(l+1)), which reads E > −Z²ℛ/[l(l+1)], the bottom of the effective well. For l = 0 there is no inner turning point: the Coulomb term wins as ρ → 0 and the allowed region reaches the nucleus.
Checks: dimensions, limits, α and a measurement
Dimensions: ρ and ρ₀ are pure numbers, and E = −4Z²ℛ/ρ₀² is an energy because ℛ = ħ²/(2μa²) is. Limits: as the binding energy goes to zero, κ → 0 and ρ₀ → ∞, so a weakly bound state is enormous in r; as l grows the allowed region needs ρ₀ > 2√(l(l+1)), close to 2l + 1, so high-l states are weakly bound. Order in α: with a = ħ/(αμc), ħκ/μ = 2Zαc/ρ₀, near αc = 2.19 × 10⁶ m s⁻¹ for hydrogen's tightest states, so binding energies are of order α²μc² and the relativistic and spin corrections this equation omits are smaller by (Zα)², quoted from the fine-structure unit. Measurement: taking from later in this section that the ground state has ρ₀ = 2, E = −ℛ = −13.59829 eV, while the measured ionisation energy is 13.59843 eV. The gap, 1.5 × 10⁻⁴ eV or 1.1 × 10⁻⁵ of the whole, is of order α² = 5.3 × 10⁻⁵: fine structure and the Lamb shift, quoted from Dirac theory and QED.
Change one variable at a time
Make the relationship visible.
Set l = 2 and lower ρ₀ from 6: below 2√6 ≈ 4.9 the well no longer reaches the energy line and the band closes, so no d state of hydrogen lies below −2.27 eV. Raise ρ₀ instead and the band widens while the binding weakens.
HYDROGEN ENERGY −4ℛ/ρ₀²-1.511 eV
HYDROGEN RULER 1/κ = ρ₀a/2158.8 pm
BAND OPENS ABOVE ρ₀2.83
BAND WIDTH ρ₊ − ρ₋5.29
Live interpretationHYDROGEN ENERGY −4ℛ/ρ₀²: −1.511 eV. HYDROGEN RULER 1/κ = ρ₀a/2: 158.8 pm. BAND OPENS ABOVE ρ₀: 2.83. BAND WIDTH ρ₊ − ρ₋: 5.29
Catch the common trap
Explain before calculating.
The equation d²u/dρ² = [1 − ρ₀/ρ + l(l+1)/ρ²] u has been solved for hydrogen, giving a list of allowed ρ₀ for each l. Replace e² by 3e² to describe Li²⁺, ignoring the small change in reduced mass. What happens to the bound-state energies?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTreat hydrogen, with μ = 0.999456 mₑ and a = 52.95 pm, at a trial energy E = −5.00 eV, without asking yet whether that energy is allowed. Find κ and ρ₀, confirm ρ₀ by a second route, and locate, in ρ and in r, the point where the l = 0 bracket 1 − ρ₀/ρ changes sign.
- |E| = 5.00 eV = 8.011 × 10⁻¹⁹ J, so κ = √(2μ|E|)/ħ = √(2 × 9.1044 × 10⁻³¹ kg × 8.011 × 10⁻¹⁹ J)/(1.0546 × 10⁻³⁴ J s) = 1.1453 × 10¹⁰ m⁻¹. That is κ = 11.453 nm⁻¹, and the ruler is 1/κ = 87.3 pm.
- From the definition, ρ₀ = 2Z/(κa) = 2/(11.453 nm⁻¹ × 0.05295 nm) = 2/0.6064 = 3.298. The nanometres cancel, as they must for a pure number.
- Second route: E = −4ℛ/ρ₀² inverts to ρ₀ = 2√(ℛ/|E|) = 2√(13.598 eV/5.00 eV) = 2 × 1.6491 = 3.298, the same number.
- For l = 0 the bracket 1 − ρ₀/ρ vanishes at ρ = ρ₀ = 3.298, which is r = ρ₀/κ = 3.298 × 87.3 pm = 288 pm.
- Check with classical mechanics: an l = 0 electron turns where −e²/(4πε₀r) = E, at r = 1.440 eV nm/5.00 eV = 0.288 nm. The rescaling has kept the turning point exactly where it was.
Answerκ = 11.45 nm⁻¹ (1/κ = 87.3 pm) and ρ₀ = 3.30; the l = 0 bracket changes sign at ρ = 3.30, r = 288 pm, the classical turning point.
MediumHydrogen's Lyman-α line is measured at 121.567 nm. Using only the form of the dimensionless equation, with no solution of it, predict the corresponding line of He⁺ (Z = 2). Take the nuclear masses as 1836.15 mₑ for the proton and 7294.30 mₑ for the alpha particle, and compare with the measured 30.378 nm.
- With e² → Ze² the equation still reads d²u/dρ² = [1 − ρ₀/ρ + l(l+1)/ρ²] u. Neither Z nor μ appears in it; both sit inside ρ₀ = 2Z/(κa), with a ∝ 1/μ. So the allowed ρ₀ for each l, whatever they are, are the same numbers for He⁺ as for hydrogen, and corresponding lines join states with the same pair of ρ₀.
- At fixed ρ₀, κ = 2Z/(ρ₀a) ∝ Zμ and E = −ħ²κ²/(2μ) ∝ Z²μ. Every level, and so every line, scales by Z²μ(He⁺)/μ(H).
- μ(H) = mₑ/(1 + 1/1836.15) = 0.999456 mₑ and μ(He⁺) = mₑ/(1 + 1/7294.30) = 0.999863 mₑ, a ratio of 1.000407.
- The photon-energy ratio is 4 × 1.000407 = 4.001630, so λ = 121.567 nm/4.001630 = 30.379 nm. Without the reduced-mass factor the prediction would be 121.567 nm/4 = 30.392 nm.
- The measured 30.378 nm agrees with 30.379 nm to 5 parts in 10⁵, ten times better than Z² alone. The residue has the size of the order-(Zα)² fine structure, quoted from later in the course, which is four times larger relative to the line in He⁺ and so does not cancel in the ratio.
Answerλ(He⁺) = 30.379 nm predicted against 30.378 nm measured: Z² = 4 does almost everything, and the reduced mass supplies a further 4 parts in 10⁴.
HardFor l = 2, find the smallest ρ₀ for which the bracket 1 − ρ₀/ρ + 6/ρ² is negative anywhere, and turn it into a floor under every hydrogen d-state energy (ℛ = 13.598 eV, a = 52.95 pm). Then take ρ₀ = 6, which later in this section proves to be the lowest allowed value for l = 2, find its turning points in ρ and in r, and interpret the gap between its energy and the floor.
- The bracket W(ρ) = 1 − ρ₀/ρ + 6/ρ² has dW/dρ = ρ₀/ρ² − 12/ρ³, which vanishes at ρ = 12/ρ₀. There W = 1 − ρ₀²/12 + ρ₀²/24 = 1 − ρ₀²/24, which is negative only if ρ₀ > √24 = 4.899.
- With E = −4ℛ/ρ₀², the condition ρ₀ > √24 reads E > −4ℛ/24 = −ℛ/6 = −2.266 eV. That is the bottom of the l = 2 effective well, −ℛ/[l(l+1)]: no d state of hydrogen can lie deeper, whatever the solutions turn out to be.
- At ρ₀ = 6 the turning points solve ρ² − 6ρ + 6 = 0, so ρ = 3 ∓ √3 = 1.268 and 4.732.
- Here 1/κ = ρ₀a/2 = 3a = 159 pm, so r₋ = 3.80a = 201 pm and r₊ = 14.20a = 752 pm.
- E = −4ℛ/36 = −ℛ/9 = −1.511 eV, which is 0.755 eV above the floor. A state confined between 201 pm and 752 pm cannot rest at the bottom of the well; the gap is the zero-point cost of that confinement, paid as radial kinetic energy and as the state's spread up the walls. As a check, the 3d radial density peaks at r = 9a = 477 pm (quoted from later in this section), inside the allowed band.
AnswerThe band opens only for ρ₀ > 2√6 = 4.90, so every hydrogen d level lies above −2.266 eV. At ρ₀ = 6 the turning points are ρ = 1.27 and 4.73, r = 201 pm and 752 pm, and E = −1.511 eV sits 0.755 eV above the floor.