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Low-lying radial distributions of the Coulomb model: normalisations, moments and the failure of cross-l orthogonality of reduced functions

Physical scope

An exact computation of the reduced radial functions, their normalisations and low moments for the states with n = 1 and n = 2 of the declared one-centre Coulomb model, together with the explicit check that reduced radial functions of different angular momentum are not orthogonal in L^2(dr). The example is a conditional model computation in terms of the declared parameters; it reports no measured radius or density, identifies no size with a laboratory length, and asserts no orbit or trajectory.

Example

Adopt the declared one-centre Coulomb model The declared Coulomb and two-body hydrogen model: one-centre operator, relative problem, reduced mass and scope exclusions clauses (a)-(c) with parameters $ℏ > 0$ , $m > 0$ , coupling $κ > 0$ and $a : = a_{κ} = ℏ^{2} / (m κ)$ , and the radial reading of The one-centre Coulomb model: position and radial distributions of the bound states, with the ground-state density and its scales. In the notation of that item, $u_{n l}$ denotes the reduced regular solution with $ψ_{n l m} (x) = u_{n l} (r) Y_{l m} (Ω) / r$ and $p_{n l} = ∣ u_{n l} ∣^{2} / ∥ u_{n l} ∥_{2}^{2}$ the radial density of the normalised bound state. For the three reduced solutions below choose the normalisation $∥ u_{n l} ∥_{2} = 1$ and the phase for which the coefficient $c_{n l}$ multiplying the displayed Laguerre polynomial-exponential expression is positive real. Then:

(a) $u_{10} (r) = 2 a^{- 3 / 2} r e^{- r / a}$ and $u_{20} (r) = 2^{- 3 / 2} a^{- 3 / 2} (2 r - \frac{r^{2}}{a}) e^{- r / (2 a)}$ and $u_{21} (r) = 2^{- 3 / 2} 3^{- 1 / 2} a^{- 5 / 2} r^{2} e^{- r / (2 a)}$ ;

(b) $∥ u_{10} ∥_{2} = ∥ u_{20} ∥_{2} = ∥ u_{21} ∥_{2} = 1$ , so for these states the reduced solution is already normalised and $p_{n l} = ∣ u_{n l} ∣^{2}$ ;

(c) the moments $\int_{0}^{\infty} r p_{10} (r) d r = \frac{3 a}{2}, \int_{0}^{\infty} r^{2} p_{10} (r) d r = 3 a^{2}, \int_{0}^{\infty} r p_{20} (r) d r = 6 a, \int_{0}^{\infty} r p_{21} (r) d r = 5 a,$ and $\int_{0}^{\infty} \frac{p_{20} (r)}{r} d r = \int_{0}^{\infty} \frac{p_{21} (r)}{r} d r = \frac{1}{4 a};$

(d) the same- $l$ pair is orthogonal, $\int_{0}^{\infty} u_{10} (r) u_{20} (r) d r = 0$ , while the different- $l$ pair is not: $\int_{0}^{\infty} u_{20} (r) u_{21} (r) d r = - \frac{\sqrt{3}}{2} \neq 0$ , so the reduced radial functions of different angular momentum are not orthogonal in $L^{2} ((0, \infty), d r)$ , the orthogonality of the full bound states being carried by the angular factor;

(e) the closed forms of (a) agree with the standard radial functions $R_{n l} = u_{n l} / r$ of the cited treatments for these states, and the value $\int r p_{n l}$ for the circular state $n = 2$ , $l = 1$ agrees with the general circular-orbit formula $⟨ r ⟩_{n, n - 1} = n (n + \frac{1}{2}) a$ at $n = 2$ ;

(f) Scope. The computation is conditional on the declared model and the cited mathematics; $a$ is a model scale, the moments are moments of the model's radial reading and not measured radii, and no measurement, orbit or trajectory is asserted.

Verification

Given: The declared one-centre Coulomb model with $ℏ, m, κ > 0$ , $a = ℏ^{2} / (m κ)$ , $γ = 2 / a$ ; the reduced solutions $u_{n l}$ of the model with $ψ_{n l m} = u_{n l} (r) Y_{l m} (Ω) / r$ ; the Axiom of Countable Choice.

[L1] Exponential moments. For every $λ > 0$ and integer $k \geq 0$ , $\int_{0}^{\infty} r^{k} e^{- λ r} d r = k! / λ^{k + 1}$ (Moments of the exponential law on the half-line).

[L2] Reduced solutions. $u_{n l} (r) = c_{n l} r^{l + 1} e^{- r / (n a)} L_{n - l - 1}^{(2 l + 1)} (2 r / (n a))$ with $c_{n l} \neq 0$ and $L_{j}^{(a)}$ the Laguerre polynomial $L_{j}^{(a)} (x) = \sum_{i = 0}^{j} (- 1)^{i} (\binom{j + a}{j - i}) x^{i} / i!$ ; for $j = 0$ one has $L_{0}^{(a)} = 1$ and for $j = 1$ one has $L_{1}^{(a)} (x) = 1 + a - x$ (Kummer recurrence and termination clause (d), The one-centre Coulomb model: position and radial distributions of the bound states, with the ground-state density and its scales clause (c)).

[L3] Radial reading. For a radial set the position law of the normalised bound state is $\int_{a}^{b} p_{n l} (r) d r$ with $p_{n l} = ∣ u_{n l} ∣^{2} / ∥ u_{n l} ∥_{2}^{2}$ , and $\int_{0}^{\infty} p_{n l} d r = 1$ (The one-centre Coulomb model: position and radial distributions of the bound states, with the ground-state density and its scales clause (a), Position and momentum outcome laws of the massive spin-s model clause (b)).

Proof technique: direct; insert the Laguerre data into the closed forms, evaluate the resulting elementary integrals with the exponential-moments lemma, and compare with the cited radial functions.

Given: The data of the statement and [L1]-[L3].

1.1givenL2algebra

[Closed forms] For $n = 1$ , $l = 0$ : $j = 0$ , $k = 1 / (n a) = 1 / a$ , so [L2] gives $u_{10} (r) = c r e^{- r / a}$ . For $n = 2$ , $l = 0$ : $j = 1$ , $L_{1}^{(1)} (x) = 2 - x$ and $k = 1 / (2 a)$ , so $u_{20} (r) = c r e^{- r / (2 a)} (2 - r / a)$ . For $n = 2$ , $l = 1$ : $j = 0$ , $L_{0}^{(3)} = 1$ and $k = 1 / (2 a)$ , so $u_{21} (r) = c r^{2} e^{- r / (2 a)}$ . The constants $c$ are fixed by the normalisation condition $∥ u_{n l} ∥_{2} = 1$ , evaluated in the next step, giving the displayed coefficients of (a).

2.1step 1.1L1algebra

[Normalisations] By [L1] with $λ = 2 / a$ : $\int_{0}^{\infty} r^{2} e^{- 2 r / a} d r = 2 / (2 / a)^{3} = a^{3} / 4$ and $\int_{0}^{\infty} r^{4} e^{- 2 r / a} d r = 24 / (2 / a)^{5} = 3 a^{5} / 4$ ; hence $\int_{0}^{\infty} (2 a^{- 3 / 2} r e^{- r / a})^{2} d r = 4 a^{- 3} \cdot a^{3} / 4 = 1$ , so $c = 2 a^{- 3 / 2}$ for $u_{10}$ and $∥ u_{10} ∥_{2} = 1$ . By [L1] with $λ = 1 / a$ : $\int_{0}^{\infty} r^{2} e^{- r / a} d r = 2 a^{3}$ , $\int_{0}^{\infty} r^{3} e^{- r / a} d r = 6 a^{4}$ , $\int_{0}^{\infty} r^{4} e^{- r / a} d r = 24 a^{5}$ ; expanding $(2 - r / a)^{2} = 4 - 4 r / a + r^{2} / a^{2}$ gives $\int_{0}^{\infty} r^{2} (2 - r / a)^{2} e^{- r / a} d r = 8 a^{3} - 24 a^{3} + 24 a^{3} = 8 a^{3}$ , so $c = 2^{- 3 / 2} a^{- 3 / 2}$ yields $∥ u_{20} ∥_{2} = 1$ ; and $c^{2} \int r^{4} e^{- r / a} d r = c^{2} \cdot 24 a^{5} = 1$ gives $c = 24^{- 1 / 2} a^{- 5 / 2} = 2^{- 3 / 2} 3^{- 1 / 2} a^{- 5 / 2}$ for $u_{21}$ . This is (b).

2.2step 1.1L1algebra

[Moments of the ground state] With $p_{10} = 4 a^{- 3} r^{2} e^{- 2 r / a}$ , [L1] gives $\int_{0}^{\infty} r p_{10} (r) d r = 4 a^{- 3} \cdot 3! / (2 / a)^{4} = 4 a^{- 3} \cdot 6 a^{4} / 16 = 3 a / 2$ and $\int_{0}^{\infty} r^{2} p_{10} (r) d r = 4 a^{- 3} \cdot 4! / (2 / a)^{5} = 4 a^{- 3} \cdot 24 a^{5} / 32 = 3 a^{2}$ . Thus the variance is $3 a^{2} - (3 a / 2)^{2} = 3 a^{2} / 4$ .

2.3step 1.1L1algebra

[Moments of the $n = 2$ states] With $p_{20} = 2^{- 3} a^{- 3} r^{2} (2 - r / a)^{2} e^{- r / a}$ and $p_{21} = 24^{- 1} a^{- 5} r^{4} e^{- r / a}$ , [L1] gives $\int r p_{20} = 2^{- 3} a^{- 3} \int r^{3} (2 - r / a)^{2} e^{- r / a} d r = 2^{- 3} a^{- 3} \cdot a^{4} \int x^{3} (4 - 4 x + x^{2}) e^{- x} d x = 2^{- 3} a^{- 3} \cdot a^{4} (4 \cdot 3! - 4 \cdot 4! + 5!) = 2^{- 3} a^{- 3} \cdot 48 a^{4} = 6 a$ ; and $\int r p_{21} = 24^{- 1} a^{- 5} \cdot 5! a^{6} = 5 a$ . Moreover $\int r^{- 1} p_{20} = 2^{- 3} a^{- 3} \int r (2 - r / a)^{2} e^{- r / a} d r = 2^{- 3} a^{- 3} (4 \cdot 1! a^{2} - 4 \cdot 2! a^{2} + 3! a^{2}) = 2^{- 3} a^{- 3} \cdot 2 a^{2} = a^{- 1} / 4$ , and $\int r^{- 1} p_{21} = 24^{- 1} a^{- 5} \cdot 3! a^{4} = a^{- 1} / 4$ . This is (c).

2.4step 1.1L1algebra

[Overlaps] With $u_{10} = 2 a^{- 3 / 2} r e^{- r / a}$ and $u_{20} = 2^{- 3 / 2} a^{- 3 / 2} (2 r - r^{2} / a) e^{- r / (2 a)}$ one has $\int u_{10} u_{20} d r = 2^{- 1 / 2} a^{- 3} \int (2 r^{2} - r^{3} / a) e^{- 3 r / (2 a)} d r$ ; by [L1] with $λ = 3 / (2 a)$ , $\int r^{2} e^{- 3 r / (2 a)} d r = 2 (2 a / 3)^{3} = 16 a^{3} / 27$ and $\int r^{3} e^{- 3 r / (2 a)} d r = 6 (2 a / 3)^{4} = 32 a^{4} / 27$ , so the bracket is $2 \cdot 16 a^{3} / 27 - (32 a^{4} / 27) / a = 0$ : the same- $l$ pair is orthogonal. With $u_{21} = 2^{- 3 / 2} 3^{- 1 / 2} a^{- 5 / 2} r^{2} e^{- r / (2 a)}$ , the cross overlap is $\int u_{20} u_{21} d r = 2^{- 3} 3^{- 1 / 2} a^{- 4} \int (2 r^{3} - r^{4} / a) e^{- r / a} d r = 2^{- 3} 3^{- 1 / 2} a^{- 4} (2 \cdot 6 a^{4} - 24 a^{4}) = 2^{- 3} 3^{- 1 / 2} (- 12) = - 3 / (2 \sqrt{3}) = - \sqrt{3} / 2$ , nonzero. This is (d).

3.1step 1.1step 2.3L1L2L3∎

[Cross-checks and scope] Since $R_{n l} = u_{n l} / r$ , step 1.1 gives $R_{10} = 2 a^{- 3 / 2} e^{- r / a}$ , $R_{20} = 2^{- 3 / 2} a^{- 3 / 2} (2 - r / a) e^{- r / (2 a)}$ and $R_{21} = 2^{- 3 / 2} 3^{- 1 / 2} a^{- 3 / 2} (r / a) e^{- r / (2 a)}$ , which agree with the cited tabulated radial functions, and the value $5 a$ of step 2.3 agrees with the circular-orbit formula $⟨ r ⟩_{n, n - 1} = n (n + \frac{1}{2}) a$ at $n = 2$ ; this is (e). Only the declared model, the cited mathematics and the exponential-moments lemma were used, and $a$ remains a model scale, so the scope statement (f) holds.

Remarks

Why cross- $l$ overlaps need not vanish. The full bound states $ψ_{n l m}$ and $ψ_{n^{'} l^{'} m^{'}}$ are orthogonal in $L^{2} (R^{3})$ whenever the angular factors $Y_{l m}, Y_{l^{'} m^{'}}$ are orthogonal, even if the reduced radial functions overlap; the nonzero value $- \sqrt{3} / 2$ computed in (d) is exactly such a case. No orthogonality of reduced functions across angular momenta is claimed anywhere on this pair.
Moments are readings, not measurements. The mean radius and the inverse radius are moments of the model's radial reading; the inverse-radius value $1 / (4 a)$ for the two $n = 2$ states is a consistency check with the model's virial-type relation and is not compared with any measured quantity here.

Low-lying radial distributions of the Coulomb model: normalisations, moments and the failure of cross-l orthogonality of reduced functions

Example

Verification

Remarks

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