University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.4

The power series for v(ρ) and its recursion

Most derivations hurry through this step and pay for it later. Here you run the recursion by hand, watch it hand back a perfectly good solution at every energy, and pin down exactly how an endless series blows up, including the power of ρ that the quick textbook argument gets wrong.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

With the behaviour at both ends factored off, all that is left of the hydrogen radial problem is v(ρ), which obeys ρv″ + 2(l + 1 − ρ)v′ + [ρ₀ − 2(l + 1)]v = 0. The model is the plainest one available: write v as a power series and let the equation set the coefficients. Each term of the equation either keeps the power of ρ or lowers it by one, so matching powers gives a one-step recursion, and because its denominator (j + 1)(j + 2l + 2) never vanishes, the whole series follows from c₀ at any ρ₀ whatever. That is the first result, and it is a negative one: the differential equation, even with regularity at the origin built in, quantises nothing.

The spectrum has to come from the other end. For large j the recursion tends to 2/j, the ratio between coefficients of e^(2ρ), and beyond j = ρ₀/2 − l − 1 every coefficient has one sign, so nothing can cancel. An endless series therefore grows like e^(2ρ) and u like e^ρ — the very branch discarded at infinity, back through the series. The crude ratio argument stops there.

It sees the exponential rate but is blind to the power of ρ in front, and keeping its 1/j correction, confirmed by the quoted large-argument form of Kummer's function, gives u ∼ ρ^(−ρ₀/2)e^ρ rather than the ρ^(l+1)e^ρ usually written. Everything is exact inside the non-relativistic, spinless, point-nucleus model; its limits belong to later units.

Simple definition: The Coulomb recursion is the rule cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)] that fixes every coefficient of the power series v(ρ) = Σ cⱼρʲ from the first one, c₀, at any value of the energy parameter ρ₀.
Example: For l = 2 and ρ₀ = 9.0 the denominators are 6, 14 and 24, so from c₀ = 1 the recursion gives c₁ = −3/6 = −0.5, c₂ = 0.0357 and c₃ = 0.00149. Since ρ₀/2 − l − 1 = 1.5 is not a whole number, no numerator ever vanishes and the series runs on forever.

The recursion (derived here)cⱼ₊₁ = [2(j + l + 1) − ρ₀] cⱼ / [(j + 1)(j + 2l + 2)]

One seed, c₀, fixes the whole chain, and since the denominator is never zero the chain runs at every value of ρ₀.

j = 0, 1, 2, …; ρ = κr and ρ₀ are dimensionless, so every ratio cⱼ₊₁/cⱼ is a pure number

The coefficients in closed formcⱼ = c₀ 2ʲ (a)ⱼ / [(b)ⱼ j!], with a = l + 1 − ρ₀/2 and b = 2l + 2

Iterating the recursion reproduces term by term the series of Kummer's M(a, b, 2ρ), so v = c₀M(a, b, 2ρ) is derived rather than assumed.

(a)ⱼ = a(a + 1)…(a + j − 1) and (a)₀ = 1; c₀ carries the unit and the normalisation

The recursion measured against e^(2ρ)(j + 1) cⱼ₊₁ / (2cⱼ) = 1 − (l + 1 + ρ₀/2)/(j + 2l + 2)

The leading 1 is the growth rate e^(2ρ). The 1/j shortfall sets the power of ρ in front of it, which the crude 2/j argument cannot see.

The left side is exactly 1 for every term of e^(2ρ) = Σ(2ρ)ʲ/j!; all quantities are dimensionless

Kummer's function at large argument (quoted)M(a, b, z) ≈ [Γ(b)/Γ(a)] e^z z^{(a − b)} as z → ∞

It confirms the power found by matching the 1/j term, and its amplitude 1/Γ(a) vanishes only at a = 0, −1, −2, …, where the series stops.

Quoted from tables of confluent hypergeometric functions (Abramowitz and Stegun 13.1.4); here z = 2ρ

An endless series makes u growu ≈ c₀ [Γ(2l + 2)/Γ(l + 1 − ρ₀/2)] 2^{(−l−1−ρ₀/2)} ρ^(−ρ₀/2) e^ρ

Not ρ^(l+1)e^ρ. At ρ₀ = 5, l = 0, summing the series numerically gives u/(ρ^(−2.5)e^ρ) = 0.0380 at ρ = 300, closing on the predicted 0.0374.

From u = ρ^(l+1) e^(−ρ) v and the quoted Kummer form; the power of ρ does not depend on l

Limiting case: the Coulomb term switched offρ₀ = 0, l = 0: u = c₀ sinh ρ

It grows as (c₀/2)e^ρ: power ρ⁰ = ρ^(−ρ₀/2) and amplitude 1/2, exactly as the large-ρ form predicts, and nothing like ρe^ρ.

ρ₀ is proportional to Z, so ρ₀ = 0 removes the attraction; sinh ρ solves u″ = u with u(0) = 0

Match powers of ρ, one index shift at a time

Start from the equation for v derived in the previous topic, ρv″ + 2(l + 1 − ρ)v′ + [ρ₀ − 2(l + 1)]v = 0, and put v = Σ cⱼρʲ with j = 0, 1, 2, …. It splits into four sums: ρv″ = Σ j(j − 1)cⱼρ^(j−1), 2(l + 1)v′ = Σ 2(l + 1)j cⱼρ^(j−1), −2ρv′ = Σ(−2j)cⱼρʲ and [ρ₀ − 2(l + 1)]v = Σ[ρ₀ − 2(l + 1)]cⱼρʲ. The first two lower the power by one, so relabel j → j + 1 in them, and together they become Σ(j + 1)(j + 2l + 2)cⱼ₊₁ρʲ. Every sum now runs over ρʲ, and a power series vanishes for all ρ only if each coefficient does: (j + 1)(j + 2l + 2)cⱼ₊₁ + [ρ₀ − 2(j + l + 1)]cⱼ = 0. The lowest power, ρ⁽⁻¹⁾, comes only from j = 0 in the first two sums, with coefficient 0 × (2l + 1)c₀ = 0, so it imposes nothing and c₀ is free. That is the regular solution. The other root of the indicial equation, s = −(2l + 1), would give u ∼ ρ^(−l), which was rejected at the origin.

One seed, one chain, and a chain that never breaks

Solved for the next coefficient, the recursion reads cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)]. Compare the oscillator of Quantum Mechanics I, whose Hermite recursion stepped by two and needed two seeds, a₀ and a₁, for its even and odd chains. Here each term of the equation shifts the power by zero or one, so there is one chain and one seed; the second constant a second-order equation owes you was spent at the origin, discarding ρ^(−l). The denominator is 2l + 2 at j = 0 and only grows, so the chain cannot break of its own accord, and every cⱼ is fixed by c₀. Iterating gives cⱼ = c₀ 2ʲ (a)ⱼ/[(b)ⱼ j!], with a = l + 1 − ρ₀/2, b = 2l + 2 and (a)ⱼ = a(a + 1)…(a + j − 1). That is term by term the series of Kummer's M(a, b, 2ρ), quoted in the previous topic, so the recursion has derived v = c₀M(a, b, 2ρ) rather than assumed it. The dimensions check at once: ρ, ρ₀ and every ratio cⱼ₊₁/cⱼ are pure numbers, and all the cⱼ share whatever unit normalisation gives c₀.

Every energy gets a solution

Nothing in the recursion restricts ρ₀. The definitions of κ and ρ₀ give ρ₀ = 2√(R_μ/|E|), with R_μ = μe⁴/(32π²ε₀²ħ²) = 13.598 eV for hydrogen's reduced mass; so take l = 0 and ρ₀ = 5, which is the arbitrary energy E = −4R_μ/25 = −2.176 eV. From c₀ = 1: c₁ = (2 − 5)/2 = −1.5, c₂ = (4 − 5)/6 × c₁ = 0.25, c₃ = (6 − 5)/12 × c₂ = 0.0208 and c₄ = 3/20 × c₃ = 0.003125, and so on without end. Successive terms of the series shrink as cⱼ₊₁ρ/cⱼ → 2ρ/j, so it converges at every ρ: v is a smooth function, regular at the origin, that satisfies the equation exactly. Any other ρ₀ does the same. So the differential equation, with the condition at ρ = 0 already built in, quantises nothing, just as the oscillator's equation did not; the levels must come from the one condition not yet imposed, normalisability at infinity. The signs hint at what goes wrong. The numerator 2(j + l + 1) − ρ₀ is negative only for j below ρ₀/2 − l − 1 = 1.5, so c₁ is negative and from c₂ onwards every coefficient is positive.

The tail of the series is the tail of e^(2ρ)

At large j the recursion becomes cⱼ₊₁/cⱼ ≈ 2j/j² = 2/j, the same as the ratio 2/(j + 1) between successive coefficients of e^(2ρ) = Σ(2ρ)ʲ/j!. The tail is what counts: at a given ρ the largest terms of Σ(2ρ)ʲ/j! sit near j ≈ 2ρ, so large ρ is governed by large j, and any finite set of early terms adds only a polynomial. The argument can be made a bound. For any small δ the ratio eventually exceeds (2 − δ)/(j + 1), and past j = ρ₀/2 − l − 1 all the coefficients share one sign, so nothing cancels and |v| grows at least as fast as e^{(2ρ − δρ)}. Then u = ρ^(l+1)e^(−ρ)v grows at least as fast as e^{(ρ − δρ)}: the growing solution e^(+ρ), thrown away when the asymptotic factor was chosen, has come back through the series, and ∫|u|²dρ diverges. The approach is steady but slow. At ρ₀ = 5 and l = 0 the ratio at j = 20 is 37/462 = 0.0801 against 2/21 = 0.0952, and at j = 100 it is 197/10302 = 0.01912 against 0.01980.

The power of ρ that the quick argument misses

The crude argument keeps only the leading 2/j and concludes v ∼ e^(2ρ), so u ∼ ρ^(l+1)e^ρ. But any series whose coefficients are proportional to 2ʲjˢ/j! has that same leading ratio whatever s is, and sums at large ρ to a constant times (2ρ)ˢe^(2ρ), because its terms crowd near j ≈ 2ρ. The leading ratio is blind to s; the next term is not. Written exactly, cⱼ₊₁/cⱼ = [2/(j + 1)][1 − (l + 1 + ρ₀/2)/(j + 2l + 2)], while jˢ coefficients give [2/(j + 1)](1 + 1/j)ˢ ≈ [2/(j + 1)](1 + s/j). Matching the 1/j terms gives s = −(l + 1 + ρ₀/2), so v ∼ ρ^{(−l−1−ρ₀/2)}e^(2ρ) and u ∼ ρ^(−ρ₀/2)e^ρ: l cancels. That matching is a heuristic, derived here. The quoted large-z form of Kummer's function, M(a, b, z) ≈ [Γ(b)/Γ(a)]e^z z^(a−b), confirms it, since a − b = −(l + 1 + ρ₀/2), and supplies the amplitude, which carries 1/Γ(a) and so is zero only at a = 0, −1, −2, …, where the series stops. Summed on a computer at ρ₀ = 5, l = 0, u/(ρe^ρ) has fallen to 8 × 10⁻¹¹ by ρ = 300, while u/(ρ^(−2.5)e^ρ) settles at 0.038.

What is derived, what is quoted, and what is left out

Derived here: the recursion, its closed form and the identification v = c₀M(a, b, 2ρ); the fact that it runs at every ρ₀; the exponential rate of an endless u, with the one-sign bound that rules out cancellation; and the exact limit u = c₀ sinh ρ at ρ₀ = 0, l = 0. Estimated by a heuristic and confirmed by a quoted formula: the power ρ^(−ρ₀/2). Quoted from tables of confluent hypergeometric functions: Kummer's large-argument form with its 1/Γ(a) amplitude. Inside its model, non-relativistic, spinless and with a point nucleus, the argument is exact, and no power of the fine-structure constant α is dropped: α sits inside ρ₀, because a_μ = ħ/(μcα). The check against measurement comes from the first escape. For l = 0 the series stops at once when ρ₀ = 2, because then c₁ = 0, and ρ₀ = 2√(R_μ/|E|) places that at E = −R_μ = −13.5982873 eV. The measured ionisation energy of hydrogen is 13.5984346 eV; the gap of 1.47 × 10⁻⁴ eV, about one part in 10⁵, is of order α² times the binding, the size of the relativistic and QED corrections the model leaves out.

Change one variable at a time

Make the relationship visible.

Interactive model

Energy parameter ρ₀5.0

Orbital quantum number l0

Coefficient index j20

With l = 0 and ρ₀ = 5.0, slide j out to 40: q climbs but is still 0.08 short of 1, and that 1/j shortfall sets the power of ρ. Now set ρ₀ = 6.0: the zero lands on j = 2, so c₃ = 0, the chain stops and u no longer grows. Nudge ρ₀ to 6.1 and it runs on forever.

RATIO cⱼ₊₁/cⱼ AT j0.0801

e^(2ρ) RATIO 2/(j + 1)0.0952

ZERO j* = ρ₀/2 − l − 11.50

IF ENDLESS, POWER OF ρ IN u-2.50

Live interpretationRATIO cⱼ₊₁/cⱼ AT j: 0.0801. e^(2ρ) RATIO 2/(j + 1): 0.0952. ZERO j* = ρ₀/2 − l − 1: 1.50. IF ENDLESS, POWER OF ρ IN u: −2.50

Catch the common trap

Explain before calculating.

Misconception check

The usual slip is to say that the series diverges unless it terminates, and that the divergence is what quantises the energy. It never diverges. Successive terms stand in the ratio cⱼ₊₁ρ/cⱼ → 2ρ/j, which goes to zero at every fixed ρ, so the series converges on the whole half-line and v is a smooth, finite solution at every ρ₀; the recursion, whose denominator never vanishes, guarantees it. What fails is a property of the function far away: the sum grows like e^(2ρ), so u grows like e^ρ and ∫|u|²dρ is infinite. Quantisation is a boundary condition imposed at infinity, as normalisability was for the oscillator, not a failure of convergence. Enclose the atom in a finite sphere and the same unending series would be quantised instead, by demanding u = 0 at the wall.

For l = 2 the series for v is run at ρ₀ = 7.0, a value at which it never ends. Which statement correctly describes u(ρ) = ρ^(l+1)e^(−ρ)v(ρ) at large ρ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHydrogen with l = 1 is given the trial energy E = −1.00 eV, which is not one of its levels. Using ρ₀ = 2√(R_μ/|E|) with R_μ = 13.598 eV, find ρ₀, run the recursion from c₀ = 1 as far as c₄, and say from which coefficient onwards they all share one sign.

ρ₀ = 2√(13.598/1.00) = 2 × 3.6875 = 7.375.
With l = 1 the recursion reads cⱼ₊₁ = [2(j + 2) − ρ₀]cⱼ/[(j + 1)(j + 4)], so the denominators are 4, 10, 18 and 28.
c₁ = (4 − 7.375)/4 = −0.8438; c₂ = (6 − 7.375)/10 × c₁ = −0.1375 × (−0.8438) = 0.1160; c₃ = (8 − 7.375)/18 × c₂ = 0.03472 × 0.1160 = 4.028 × 10⁻³; c₄ = (10 − 7.375)/28 × c₃ = 0.09375 × 4.028 × 10⁻³ = 3.78 × 10⁻⁴.
The numerator 2(j + 2) − ρ₀ is negative only for j below ρ₀/2 − 2 = 1.69, so the ratios at j = 0 and j = 1 are negative and every later ratio is positive: c₁ is negative, and from c₂ onwards all the coefficients are positive.
Nothing stops the chain, because 1.69 is not a whole number. At j = 10 the ratio is (24 − 7.375)/(11 × 14) = 0.108, only 59% of e^(2ρ)'s 2/11 = 0.182, since the shortfall (l + 1 + ρ₀/2)/(j + 2l + 2) = 5.69/14 is still large.

Answerρ₀ = 7.375; c₁ = −0.844, c₂ = 0.116, c₃ = 4.03 × 10⁻³, c₄ = 3.78 × 10⁻⁴. All coefficients from c₂ on are positive, and the series never ends.

MediumSwitch the Coulomb attraction off by setting ρ₀ = 0, and take l = 0. Sum the recursion in closed form, identify u(ρ), and use the result to test two candidate large-ρ laws at ρ = 10: ρ^(−ρ₀/2)e^ρ and ρ^(l+1)e^ρ.

With ρ₀ = 0 and l = 0 the recursion is cⱼ₊₁ = 2(j + 1)cⱼ/[(j + 1)(j + 2)] = 2cⱼ/(j + 2). Iterating from c₀: c₁ = c₀, c₂ = 2c₀/3, and in general cⱼ = 2ʲc₀/(j + 1)!.
Sum: v = c₀Σ(2ρ)ʲ/(j + 1)! = [c₀/(2ρ)]Σ(2ρ)^(j+1)/(j + 1)! = c₀(e^(2ρ) − 1)/(2ρ).
Then u = ρe^(−ρ)v = c₀(e^ρ − e^(−ρ))/2 = c₀ sinh ρ. Check: with ρ₀ = 0 and l = 0 the radial equation is u″ = u, which sinh ρ satisfies, and sinh 0 = 0 as the regular solution requires.
At large ρ, u → (c₀/2)e^ρ. The power of ρ is ρ⁰ = ρ^(−ρ₀/2), and the amplitude matches the quoted large-ρ form, Γ(2)/Γ(1) × 2⁽⁻¹⁾ = 1/2.
At ρ = 10: sinh 10 = 1.101 × 10⁴, while ρ^(l+1)e^ρ = 10e¹⁰ = 2.203 × 10⁵. The naive law is too large by 20.0 = 2ρ, a factor that grows without limit.

Answeru = c₀ sinh ρ, growing as (c₀/2)e^ρ with power ρ⁰, as ρ^(−ρ₀/2)e^ρ predicts. At ρ = 10 the form ρ^(l+1)e^ρ is already 20 times too large.

HardFor hydrogen with l = 0 the series stops at once when ρ₀ = 2. Try ρ₀ = 2.002 instead, writing ε = (ρ₀ − 2)/2 = 0.001. Find the growing part of u to first order in ε, estimate the radius where it overtakes the part that decays, and say how far out that point moves when ε falls to 10⁻⁴. Take a_μ = 0.05295 nm and R_μ = 13.598 eV.

c₁ = (2 − ρ₀)/2 × c₀ = −εc₀. For j ≥ 1 the numerator is 2j − 2ε ≈ 2j, so cⱼ₊₁ ≈ 2j cⱼ/[(j + 1)(j + 2)], which gives cⱼ ≈ −εc₀ 2ʲ/[j(j + 1)j!]; for example c₂ = −εc₀/3 = −3.33 × 10⁻⁴ c₀.
So v ≈ c₀[1 − εΣ(2ρ)ʲ/(j(j + 1)j!)]. The terms crowd near j ≈ 2ρ, where 1/[j(j + 1)] ≈ 1/(2ρ)², so the sum is about e^(2ρ)/(4ρ²) and u ≈ c₀[ρe^(−ρ) − (ε/4)e^ρ/ρ]. The growing piece carries ρ⁽⁻¹⁾ = ρ^(−ρ₀/2) to this order.
Cross-check with the quoted Kummer form: the amplitude is Γ(2)/Γ(−ε) × 2^(−2−ε), and Γ(−0.001) = −1000.6, so it is −ε/4 to first order, as found.
The two parts match when ρe^(−ρ) = (ε/4)e^ρ/ρ, that is 2ρ − 2 ln ρ = ln(4/ε) = ln 4000 = 8.294. Iterating ρ = 4.147 + ln ρ gives ρ ≈ 5.93. Summing the series numerically puts the zero of u, a spurious node, at ρ = 5.80; the estimate is 2% high because it keeps only the leading term.
In metres: κ = 2/(ρ₀a_μ), so r = ρ/κ = 5.80 × 1.001 × a_μ = 5.80 × 1.001 × 0.05295 nm = 0.307 nm. Beyond it u plunges negative as −(ε/4)c₀e^ρ/ρ, though the trial energy −4R_μ/ρ₀² = −13.571 eV lies only 27 meV above −R_μ.
For ε = 10⁻⁴, ln 40000 = 10.597 and the estimate gives ρ ≈ 7.28; the summed series gives 7.19. Ten times closer buys only 1.4 more in ρ, because the crossing grows like ½ ln(1/ε).

AnswerGrowing part ≈ −(ε/4)c₀e^ρ/ρ. It overtakes the decaying part at ρ ≈ 5.9 by estimate (a spurious node at ρ = 5.80, r ≈ 0.31 nm, from the summed series). At ε = 10⁻⁴ the node moves only to ρ ≈ 7.2.