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Ground radial probability and units

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Let $γ > 0$ , let $C = - Δ - γ / ∣ \cdot ∣$ on $D (C) = H^{2} (R^{3}; C)$ be as above, and put $a_{0} : = 2 / γ > 0$ . Define $ψ_{100} (x) : = (\frac{γ^{3}}{8 π})^{1 / 2} e^{- r / a_{0}} (x = r Ω, r = ∣ x ∣) .$ Then:

(a) $ψ_{100} \in D (C)$ , $∥ ψ_{100} ∥_{L^{2}} = 1$ , and $C ψ_{100} = - \frac{γ^{2}}{4} ψ_{100}$ ;

(b) for every Borel set $B \subseteq R^{3}$ , writing $d x$ for Lebesgue measure, $\int_{B} ∣ ψ_{100} (x) ∣^{2} d x = \int_{0}^{\infty} \frac{γ^{3}}{8 π} r^{2} e^{- 2 r / a_{0}} (\int_{S^{2}} 1_{B} (r Ω) d σ (Ω)) d r,$ so the radial probability density $p (r) : = \frac{γ^{3}}{2} r^{2} e^{- 2 r / a_{0}} = 4 r^{2} a_{0}^{- 3} e^{- 2 r / a_{0}}$ is a probability density on $(0, \infty)$ : $p \geq 0$ and $\int_{0}^{\infty} p (r) d r = 1$ ;

(c) the mean of $p$ is $\int_{0}^{\infty} r p (r) d r = 3 a_{0} / 2$ , and the unique maximum (mode) of $p$ is at $r = a_{0}$ ;

(d) the spatial density $∣ ψ_{100} (x) ∣^{2} = \frac{γ^{3}}{8 π} e^{- 2 r / a_{0}}$ is strictly decreasing in $r$ and maximal at $x = 0$ . For every $ε > 0$ there is a remainder $R_{ε}$ such that

\int_{∣ x ∣ < ε} ∣ ψ_{100} (x) ∣^{2} d x = \frac{1}{6} (\frac{2 ε}{a_{0}})^{3} + R_{ε}, ∣ R_{ε} ∣ \leq \frac{1}{8} (\frac{2 ε}{a_{0}})^{4} .

Thus this probability is positive and tends to $0$ as $ε ↓ 0$ , and the spatial probability measure has no point atoms, including at the density maximum $x = 0$ ;

(e) parametrisation and dimensions. If $ℏ, m, κ > 0$ and $γ : = 2 m κ / ℏ^{2}$ , then $a_{0} = ℏ^{2} / (m κ)$ and the operator

C_{ℏ, m, κ} : = - \frac{ℏ^{2}}{2 m} Δ - \frac{κ}{∣ \cdot ∣}, D (C_{ℏ, m, κ}) = H^{2} (R^{3}; C),

has eigenvalues $- \frac{m κ^{2}}{2 ℏ^{2} n^{2}} = \frac{ℏ^{2}}{2 m} (- \frac{γ^{2}}{4 n^{2}})$ ( $n \geq 1$ ); under the dimensional conventions $[ℏ] = J s$ , $[m] = k g$ , $[κ] = J m$ , the constants $a_{0}^{2}$ and $m κ^{2} / (2 ℏ^{2})$ have the dimensions of $m^{2}$ and $J$ respectively.

Clauses (a)–(d) are mathematical statements about the normalised ground vector and its distribution; clause (e) is a dimensional-consistency statement about the parametrisation and is not a physical claim, measurement or identification of any quantity with nature.

Facts & Assumptions

Given: $γ > 0$ , $a_{0} = 2 / γ$ , the operator $C$ and its eigenfunctions as in the previous items of this page.

[F1]

The regular solution at $k = γ / 2$ , $l = 0$ is, up to a nonzero scalar, $u (r) = r e^{- γ r / 2} = r e^{- r / a_{0}}$ , and $A_{0} u = - (γ^{2} / 4) u$ with transported function $ψ = e^{- r / a_{0}} Y_{00}$ ; the eigenvalue site is exactly the ground level of $A_{0}$ ; moreover the negative eigenvalues of $C$ are exactly $- γ^{2} / (4 n^{2})$ , $n \geq 1$ , with $n^{2}$ -dimensional eigenspaces spanned by the transported regular solutions (Kummer recurrence and termination (d), (e), Laguerre bound energy exhaustion (b), (c), Transported radial endpoint realization (e), Negative-subspace completeness and degeneracy (a)). Nonnegative radial eigenvalues are excluded by Coulomb continuum spectral density (d); the exact block-domain identity then excludes nonnegative eigenvalues of $C$ as well.

[F2]

$Y_{00} = 1 / \sqrt{4 π}$ and $σ (S^{2}) = 4 π$ ; the polar decomposition of Lebesgue measure holds (Sphere ONB, Sphere integrals in angle coordinates, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F3]

For $a > 0$ and $n \in {0, 1, 2, 3}$ one has $\int_{0}^{\infty} r^{n} e^{- a r} d r = n! / a^{n + 1}$ ; this follows by induction from the derivative of $r^{n} e^{- a r}$ and the fundamental theorem, the boundary terms at $0$ and $+ \infty$ vanishing (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The exponential function is smooth and $(\exp)^{'} = \exp$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F4]

The $L^{2}$ norm of a Borel function is computed by the polar decomposition, and $∣ Y_{00} ∣^{2} = 1 / (4 π)$ (The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions, Hilbert space).

Proof

1.1F1F2F3F4algebra

[The normalized ground vector] By [F1] the ground eigenfunction of $A_{0}$ is $u (r) = c r e^{- r / a_{0}}$ and the corresponding $C$ -eigenvector is $c e^{- r / a_{0}} Y_{00}$ , up to the scalar $c \neq 0$ . By [F4] and [F2], $∥ e^{- r / a_{0}} Y_{00} ∥_{L^{2}}^{2} = \frac{1}{4 π} \int_{0}^{\infty} r^{2} e^{- 2 r / a_{0}} 4 π d r = \int_{0}^{\infty} r^{2} e^{- 2 r / a_{0}} d r = 2 / (2 / a_{0})^{3} = a_{0}^{3} / 4$ by [F3]; hence the normalized ground vector is $ψ_{100} = (4 / a_{0}^{3})^{1 / 2} e^{- r / a_{0}} Y_{00} = (γ^{3} / (8 π))^{1 / 2} e^{- r / a_{0}}$ , since $Y_{00} = 1 / \sqrt{4 π}$ and $a_{0} = 2 / γ$ . By [F1] it lies in $D (C)$ and $C ψ_{100} = - (γ^{2} / 4) ψ_{100}$ . This is (a).

2.1step 1.1F2F3F4algebra

[The radial law] For a Borel set $B$ , [F2] and [F4] give $\int_{B} ∣ ψ_{100} ∣^{2} d x = \frac{γ^{3}}{8 π} \int_{B} e^{- 2 r / a_{0}} d x = \frac{γ^{3}}{8 π} \int_{0}^{\infty} r^{2} e^{- 2 r / a_{0}} \int_{S^{2}} 1_{B} (r Ω) d σ (Ω) d r$ , which is the displayed identity of (b); the factor $4 π$ enters only when the angular indicator is identically one, giving the radial marginal; the density $p (r) = \frac{γ^{3}}{2} r^{2} e^{- 2 r / a_{0}}$ is nonnegative and integrates to $1$ by the value of the integral computed in step 1.1.

3.1step 2.1F3algebra

[Mean and mode] By [F3] together with [F2], $\int_{0}^{\infty} r p (r) d r = \frac{γ^{3}}{2} \int_{0}^{\infty} r^{3} e^{- 2 r / a_{0}} d r = \frac{γ^{3}}{2} \cdot \frac{6}{(2 / a_{0})^{4}} = \frac{γ^{3}}{2} \cdot \frac{6 a_{0}^{4}}{16} = \frac{3 γ^{3} a_{0}^{4}}{16} = \frac{3 a_{0}}{2}$ because $γ a_{0} = 2$ . For the mode, differentiate $q (r) : = r^{2} e^{- 2 r / a_{0}}$ using [F3]'s ingredients: $q^{'} (r) = 2 r e^{- 2 r / a_{0}} - (2 / a_{0}) r^{2} e^{- 2 r / a_{0}} = (2 / a_{0}) r (a_{0} - r) e^{- 2 r / a_{0}}$ , which is positive for $0 < r < a_{0}$ , zero at $r = a_{0}$ and negative for $r > a_{0}$ ; hence $q$ , and with it $p$ , has its unique maximum at $r = a_{0}$ . This is (c).

3.2step 2.1F3algebra

[Density versus point probability] Set $t = 2 ε / a_{0}$ . Step 2.1 gives $P (∣ x ∣ < ε) = \frac{1}{2} \int_{0}^{t} s^{2} e^{- s} d s = 1 - e^{- t} (1 + t + t^{2} / 2)$ by differentiation and the fundamental theorem. Since $0 \leq 1 - e^{- s} = \int_{0}^{s} e^{- v} d v \leq s$ , the difference from $t^{3} / 6$ has absolute value at most $t^{4} / 8$ . Taking $R_{ε} : = P (∣ x ∣ < ε) - t^{3} / 6$ gives the exact remainder bound in (d), with the dimensionless scale $t = 2 ε / a_{0}$ ; the probability is strictly positive for every $ε > 0$ and tends to zero. The density $(π a_{0}^{3})^{- 1} e^{- 2 r / a_{0}}$ is strictly decreasing in $r$ , while every singleton has zero probability because its Lebesgue measure is zero. This proves (d), with no point atom at the density maximum.

4.1step 1.1step 3.2F1algebra

[Parametrisation and dimensions] Setting $γ = 2 m κ / ℏ^{2}$ gives $a_{0} = 2 / γ = ℏ^{2} / (m κ) > 0$ , and the operator identity $C_{ℏ, m, κ} = \frac{ℏ^{2}}{2 m} C$ on $H^{2}$ holds because $\frac{ℏ^{2}}{2 m} \cdot \frac{2 m κ}{ℏ^{2}} = κ$ ; hence the eigenvalues of $C_{ℏ, m, κ}$ are $\frac{ℏ^{2}}{2 m} \cdot (- \frac{γ^{2}}{4 n^{2}}) = - \frac{m κ^{2}}{2 ℏ^{2} n^{2}}$ by [F1], including exclusion of every nonnegative eigenvalue. For the dimensional statement, substituting the dimensional conventions and the dependent unit identity $J = k g m^{2} s^{- 2}$ , one computes $[ℏ^{2} / (m κ)] = J^{2} s^{2} k g^{- 1} J^{- 1} m^{- 1} = J s^{2} k g^{- 1} m^{- 1} = m$ , so $a_{0}^{2}$ has dimension $m^{2}$ , and $[m κ^{2} / ℏ^{2}] = k g J^{2} m^{2} J^{- 2} s^{- 2} = k g m^{2} s^{- 2} = J$ . This is (e).

5.1step 1.1step 2.1step 3.1step 3.2step 4.1∎

[Conclusion] Clauses (a), (b), (c), (d), (e) are steps 1.1, 2.1, 3.1, 3.2 and 4.1.

Ground radial probability and units

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources