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Qm Math Oscillator — Examples

1 · Prerequisites

2 · Summary

The companion page to qm-math-oscillator records three explicit mathematical checks behind its spectral, coherent and differential results. All three items are mathematics: they carry domain: mathematics, depend only on the items of the parent page and on mathematical suppliers, and make no physical or empirical claim.

Ground energy units reads the spectrum ${ℏ ω (n + \frac{1}{2})}$ of the realised oscillator in units of $ℏ ω$ : the ground energy is $1 / 2$ , every gap is $1$ , the ground eigenvector is the Gaussian $ψ_{0} (x) = π^{- 1 / 4} ℓ^{- 1 / 2} e^{- x^{2} / (2 ℓ^{2})}$ solving $- \frac{ℏ^{2}}{2 m} ψ_{0}^{''} + \frac{m ω^{2}}{2} x^{2} ψ_{0} = \frac{ℏ ω}{2} ψ_{0}$ pointwise, and the isotropic $d$ -dimensional ground energy is $d / 2$ in the same units, so $\frac{3}{2} ℏ ω$ for $d = 3$ . None of these numbers is identified with a measured energy.

Poisson number distribution computes the coefficients of the coherent vector in the number basis: $p_{n} = ∣ ⟨ g_{n}, ψ_{z} ⟩ ∣^{2} = e^{- ∣ z ∣^{2}} ∣ z ∣^{2 n} / n!$ has total mass $1$ , mean $∣ z ∣^{2}$ and variance $∣ z ∣^{2}$ , so the coefficient distribution is the Poisson distribution with parameter $λ = ∣ z ∣^{2}$ , with every $p_{n}$ positive when $z \neq 0$ . The statement is about the coefficients of a vector in an orthonormal basis, not about measured counts.

Formal nonnormalizable solution exhibits the classical formal second solution $u_{2} (y) = e^{- y^{2} / 2} \int_{0}^{y} e^{t^{2}} d t$ of the ground-state equation $u^{''} = (y^{2} - 1) u$ : it has constant Wronskian $g_{0} u_{2}^{'} - g_{0}^{'} u_{2} = π^{- 1 / 4}$ against the Gaussian $g_{0}$ , satisfies $∣ u_{2} (y) ∣^{2} \geq e^{y^{2} - 4 y + 2}$ for $y \geq 1$ and is therefore not in $L^{2} (R, d y)$ , and every $C^{2} \cap L^{2}$ solution of the equation is a constant multiple of $g_{0}$ . This is the analytic reason the $L^{2}$ eigenfunction at the bottom of the spectrum is the normalizable Gaussian and not the formal growing solution.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Ground energy units

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix $m, ω, ℏ > 0$ and let $ℓ = \sqrt{ℏ / (m ω)}$ , $ψ_{n}$ , $H_{o s c}$ be as in Oscillator scaling and exact realization. Then:

(a) the spectrum of $H_{o s c}$ is exactly ${ℏ ω (n + \frac{1}{2}) : n \in N}$ ; measured in units of $ℏ ω$ the eigenvalues are $n + \frac{1}{2}$ , the ground energy is

$\frac{E_{0}}{ℏ ω} = \frac{1}{2},$

and every consecutive gap is $1$ unit: $\frac{E_{n + 1} - E_{n}}{ℏ ω} = 1$ ;

(b) the ground eigenvector is the Gaussian

$ψ_{0} (x) = π^{- 1 / 4} ℓ^{- 1 / 2} e^{- x^{2} / (2 ℓ^{2})},$

and it satisfies the eigenvalue equation pointwise:

$- \frac{ℏ^{2}}{2 m} ψ_{0}^{''} (x) + \frac{m ω^{2}}{2} x^{2} ψ_{0} (x) = \frac{ℏ ω}{2} ψ_{0} (x) (x \in R);$

(c) for the isotropic $d$ -dimensional oscillator of Multidimensional oscillator completeness(d) with common frequency $ω$ , the ground energy is $\frac{d}{2} ℏ ω$ in the same units, in particular $\frac{3}{2} ℏ ω$ for $d = 3$ , and all gaps within one coordinate direction are $ℏ ω$ ;

(d) no other value of the ground energy is compatible with the displayed spectrum: $E_{0}$ is the infimum of $σ (H_{o s c})$ .

No physical identification of $ℏ ω$ with a measured energy is made; the statements are about the operator $H_{o s c}$ built from the positive parameters.

Facts & Assumptions

Given: positive reals $m, ω, ℏ$ ; $ℓ = \sqrt{ℏ / (m ω)}$ ; the oscillator operator $H_{o s c} = \frac{ℏ ω}{2} D A D^{- 1}$ with basis $(ψ_{n})$ .

[F1]

$H_{o s c}$ is self-adjoint with complete orthonormal eigenbasis $(ψ_{n})$ , $H_{o s c} ψ_{n} = ℏ ω (n + \frac{1}{2}) ψ_{n}$ , and $σ (H_{o s c}) = {ℏ ω (n + \frac{1}{2}) : n \in N}$ (Oscillator scaling and exact realization(b)).

[F2]

$ψ_{0} = D g_{0}$ with $g_{0} (y) = π^{- 1 / 4} e^{- y^{2} / 2}$ , $ℓ = \sqrt{ℏ / (m ω)}$ , and for every $f = D (p g_{0})$ one has $H_{o s c} f = - \frac{ℏ^{2}}{2 m} f^{''} + \frac{m ω^{2}}{2} x^{2} f$ pointwise (Oscillator scaling and exact realization(a),(c), Hermite recurrence(a)).

[F3]

$(e^{- x^{2} / (2 ℓ^{2})})^{'} = - \frac{x}{ℓ^{2}} e^{- x^{2} / (2 ℓ^{2})}$ and $(e^{- x^{2} / (2 ℓ^{2})})^{''} = (- \frac{1}{ℓ^{2}} + \frac{x^{2}}{ℓ^{4}}) e^{- x^{2} / (2 ℓ^{2})}$ by the product rule, the chain rule 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 $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The exponential function is smooth and $(\exp)^{'} = \exp$ , The complex exponential by its power series).

[F4]

$\frac{ℏ^{2}}{2 m ℓ^{2}} = \frac{ℏ ω}{2}$ and $\frac{ℏ^{2}}{2 m ℓ^{4}} = \frac{m ω^{2}}{2}$ , because $ℓ^{2} = ℏ / (m ω)$ (Oscillator scaling and exact realization).

Verification

1.1F1algebra

[(a) levels] By [F1], each $ψ_{n}$ is a unit eigenvector with eigenvalue $ℏ ω (n + \frac{1}{2})$ , and $σ (H_{o s c})$ is exactly the set of these numbers. Hence $E_{0} = ℏ ω / 2$ , so $E_{0} / (ℏ ω) = \frac{1}{2}$ , and $E_{n + 1} - E_{n} = ℏ ω$ for every $n$ , so $(E_{n + 1} - E_{n}) / (ℏ ω) = 1$ .

1.2F1algebra

[(c) $d$ -dimensional ground energy] In the isotropic $d$ -dimensional operator $H$ of Multidimensional oscillator completeness(d), the eigenvalues are $ℏ ω (M + \frac{d}{2})$ , $M \in N$ , so the smallest is $ℏ ω \cdot \frac{d}{2}$ : the ground energy is $\frac{d}{2} ℏ ω$ , equal to $\frac{3}{2} ℏ ω$ for $d = 3$ , and the increment of $M$ by one raises the energy by $ℏ ω$ .

2.1step 1.1F2F3F4algebra

[(b) ground eigenfunction] By [F2], $ψ_{0} (x) = ℓ^{- 1 / 2} π^{- 1 / 4} e^{- x^{2} / (2 ℓ^{2})}$ . Using [F3], $- \frac{ℏ^{2}}{2 m} ψ_{0}^{''} + \frac{m ω^{2}}{2} x^{2} ψ_{0} = [\frac{ℏ^{2}}{2 m ℓ^{2}} + (- \frac{ℏ^{2}}{2 m ℓ^{4}} + \frac{m ω^{2}}{2}) x^{2}] ψ_{0}$ , and by [F4] the coefficient of $x^{2}$ vanishes while the constant term is $\frac{ℏ ω}{2}$ ; hence $H_{o s c} ψ_{0} = \frac{ℏ ω}{2} ψ_{0}$ pointwise, in agreement with $H_{o s c} ψ_{0} = ℏ ω (0 + \frac{1}{2}) ψ_{0}$ from [F1].

3.1step 1.1F1∎

[(d) minimality] By [F1] the spectrum is the closed discrete set ${ℏ ω (n + \frac{1}{2}) : n \in N}$ , whose infimum is its least element $ℏ ω / 2 = E_{0}$ ; no smaller value belongs to the spectrum.

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Poisson number distribution

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(g_{n})_{n \in N}$ be the Hermite family of $L^{2} (R, d y)$ , $N = a^{†} a$ the number operator of Closed ladder domains, and for $z \in C$ let $ψ_{z} = e^{- ∣ z ∣^{2} / 2} \sum_{n} \frac{z^{n}}{\sqrt{n!}} g_{n}$ be the coherent vector of Coherent-state convergence. Define

$p_{n} : = ∣ ⟨ g_{n}, ψ_{z} ⟩ ∣^{2} = e^{- ∣ z ∣^{2}} \frac{∣ z ∣^{2 n}}{n!} (n \in N) .$

Then:

(a) $(p_{n})_{n \in N}$ is a probability distribution on $N$ : $p_{n} \geq 0$ and $\sum_{n} p_{n} = 1$ ;

(b) $⟨ N ⟩ : = ⟨ N ψ_{z}, ψ_{z} ⟩ = \sum_{n} n p_{n} = ∣ z ∣^{2}$ ;

(d) the distribution is the Poisson distribution with parameter $λ = ∣ z ∣^{2}$ , and it is nontrivial for every $z \neq 0$ : $p_{n} > 0$ for all $n$ .

Facts & Assumptions

Given: the coherent vector $ψ_{z}$ with coefficients $c_{n} = e^{- ∣ z ∣^{2} / 2} z^{n} / \sqrt{n!}$ in the orthonormal family $(g_{n})$ ; the numbers $p_{n} = ∣ c_{n} ∣^{2}$ .

[F1]

$⟨ g_{n}, g_{m} ⟩ = δ_{n m}$ and $c_{n} = ⟨ ψ_{z}, g_{n} ⟩$ with $ψ_{z} = e^{- ∣ z ∣^{2} / 2} \sum_{n} \frac{z^{n}}{\sqrt{n!}} g_{n}$ ; $⟨ N ψ_{z}, ψ_{z} ⟩ = ∣ z ∣^{2}$ ; $ψ_{z} \in D (N^{k})$ for every $k$ (Coherent-state convergence(a),(b),(e), Hermite orthogonality(a)).

[F2]

For $t = ∣ z ∣^{2} \geq 0$ , $\sum_{n \geq 0} \frac{t^{n}}{n!} = e^{t}$ and the series converges absolutely (The complex exponential by its power series, The complex exponential series converges absolutely for every complex argument).

[F3]

For an orthonormal family, $⟨ N ψ_{z}, ψ_{z} ⟩ = \sum_{n} n ∣ c_{n} ∣^{2}$ and, for $ψ_{z} \in D (N^{2})$ , $⟨ (N - ∣ z ∣^{2} I)^{2} ψ_{z}, ψ_{z} ⟩ = \sum_{n} (n - ∣ z ∣^{2})^{2} ∣ c_{n} ∣^{2}$ ; these are the coefficient formulas of the diagonal operator on the orthonormal family (Coherent-state convergence and The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions).

Verification

1.1F1F2algebra

[(a) probability distribution] $p_{n} = ∣ c_{n} ∣^{2} = e^{- ∣ z ∣^{2}} ∣ z ∣^{2 n} / n! \geq 0$ , and $\sum_{n} p_{n} = e^{- ∣ z ∣^{2}} \sum_{n} \frac{(∣ z ∣^{2})^{n}}{n!} = e^{- ∣ z ∣^{2}} e^{∣ z ∣^{2}} = 1$ by [F2].

1.2F1F3

[(b) mean] By [F3] and [F1], $\sum_{n} n p_{n} = \sum_{n} n ∣ c_{n} ∣^{2} = ⟨ N ψ_{z}, ψ_{z} ⟩ = ∣ z ∣^{2}$ .

1.3F2F3algebra

[(c) variance] Expanding $(N - ∣ z ∣^{2} I)^{2} = N^{2} - 2 ∣ z ∣^{2} N + ∣ z ∣^{4} I$ and using [F3], $\sum_{n} (n - ∣ z ∣^{2})^{2} p_{n} = e^{- ∣ z ∣^{2}} \sum_{n} (n^{2} - 2 ∣ z ∣^{2} n + ∣ z ∣^{4}) \frac{∣ z ∣^{2 n}}{n!}$ . With $t = ∣ z ∣^{2}$ , the three series are $\sum_{n} n^{2} \frac{t^{n}}{n!} = (t^{2} + t) e^{t}$ , $\sum_{n} n \frac{t^{n}}{n!} = t e^{t}$ and $\sum_{n} \frac{t^{n}}{n!} = e^{t}$ , obtained from the absolutely convergent exponential series by reindexing ( $\sum_{n \geq 1} \frac{n t^{n}}{n!} = t \sum_{m \geq 0} \frac{t^{m}}{m!}$ and $\sum_{n \geq 2} \frac{n (n - 1) t^{n}}{n!} = t^{2} \sum_{m \geq 0} \frac{t^{m}}{m!}$ ); multiplying by $e^{- t}$ gives $(t^{2} + t) - 2 t^{2} + t^{2} = t = ∣ z ∣^{2}$ .

2.1step 1.1step 1.2step 1.3∎

[(d) Poisson form] By (a)-(c) the distribution has total mass $1$ , mean $∣ z ∣^{2} = λ$ and variance $∣ z ∣^{2} = λ$ , and its weights are $e^{- λ} λ^{n} / n!$ , which is the Poisson distribution with parameter $λ$ ; if $z \neq 0$ then $λ > 0$ and $p_{n} > 0$ for every $n$ , so no outcome is excluded.

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Formal nonnormalizable solution

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $g_{0} (y) = π^{- 1 / 4} e^{- y^{2} / 2}$ (Hermite recurrence(a)) and define

$F (y) : = \int_{0}^{y} e^{t^{2}} d t, u_{2} (y) : = e^{- y^{2} / 2} F (y) (y \in R) .$

Then:

(a) $F$ is $C^{1}$ with $F^{'} = e^{y^{2}}$ , and $u_{2}$ is $C^{2}$ and solves the ground-state eigenvalue equation

$u_{2}^{''} (y) = (y^{2} - 1) u_{2} (y) (y \in R),$

which is the equation $(- \partial_{y}^{2} + y^{2}) u = u$ satisfied by $g_{0}$ (Hermite orthogonality(b));

(b) the Wronskian is constant and nonzero,

$W : = g_{0} u_{2}^{'} - g_{0}^{'} u_{2} = π^{- 1 / 4},$

so $g_{0}$ and $u_{2}$ are linearly independent solutions;

(c) $u_{2} \notin L^{2} (R, d y)$ : for $y \geq 1$ one has $∣ u_{2} (y) ∣^{2} \geq e^{y^{2} - 4 y + 2}$ , which is not integrable, so $\int_{R} ∣ u_{2} ∣^{2} d y = + \infty$ ;

(d) every $C^{2}$ function $u : R \to C$ (with complex smoothness and derivatives understood componentwise as in Complex Lp classes and Euclidean test-function conventions) that belongs to $L^{2} (R, d y)$ and solves $u^{''} = (y^{2} - 1) u$ is a constant multiple of $g_{0}$ : the quantity $W (u) : = g_{0} u^{'} - g_{0}^{'} u$ is constant for such $u$ ; if this constant is nonzero then $u (y) = g_{0} (y) (C + c π^{1 / 2} F (y))$ for some constants with $c \neq 0$ , which satisfies $∣ u (y) ∣ \geq K e^{y^{2} / 2 - 2 y + 1}$ for some $K > 0$ and all sufficiently large $y$ , and is not in $L^{2}$ ; hence the constant is zero and $u / g_{0}$ is constant.

Thus $u_{2}$ is the classical formal second solution of the ground-state equation, and its failure to be square-integrable is what forces the $L^{2}$ eigenfunction to be the Gaussian.

Facts & Assumptions

Given: $g_{0} (y) = π^{- 1 / 4} e^{- y^{2} / 2}$ , $F (y) = \int_{0}^{y} e^{t^{2}} d t$ , $u_{2} = e^{- y^{2} / 2} F$ , and, in part (d), a componentwise $C^{2}$ function $u : R \to C$ with $u \in L^{2} (R, d y)$ .

[F1]

$g_{0}^{'} = - y g_{0}$ and $g_{0}^{''} = (y^{2} - 1) g_{0}$ , so $(- \partial_{y}^{2} + y^{2}) g_{0} = g_{0}$ ; $g_{0} > 0$ everywhere (Hermite recurrence(a),(d), Hermite orthogonality(b), 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 $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The exponential function is smooth and $(\exp)^{'} = \exp$ , The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

[F2]

$F (y) = \int_{0}^{y} e^{t^{2}} d t$ uses oriented limits (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ). For each $y_{0} \in R$ , choose $a < \min (0, y_{0})$ and $b > \max (0, y_{0})$ and put $G (x) = \int_{a}^{x} e^{t^{2}} d t$ on $[a, b]$ . Oriented additivity gives $F (x) = G (x) - G (0)$ on this interval, and the first FTC applied to the continuous integrand gives $F^{'} (y_{0}) = G^{'} (y_{0}) = e^{y_{0}^{2}}$ (For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive); hence $F \in C^{1} (R)$ . For $t \in R$ , $e^{t^{2}} \geq e^{0} = 1$ by strict increase and the defining series at zero (The exponential function is strictly increasing, The complex exponential by its power series), so $F (y) \geq y \geq 0$ for $y \geq 0$ .

[F5]

For every real $x$ , $e^{x} > 0$ and $e^{- x} = 1 / e^{x}$ (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ); the real exponential is strictly increasing (The exponential function is strictly increasing). The complex exponential extends the real one and satisfies $e^{z + w} = e^{z} e^{w}$ for all $z, w \in C$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[F3]

If $G$ is differentiable with integrable derivative on $[a, b]$ , then $\int_{a}^{b} G^{'} = G (b) - G (a)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ). If a nonnegative measurable $h$ satisfies $h \geq 1$ on $[a, \infty)$ for some real $a$ , then for every integer $N \geq 1$ , $1_{[a, a + N]} \leq h$ ; monotonicity of the nonnegative integral and the identity $\int 1_{E} d λ = λ (E)$ for indicators (from the simple-integral definition and its agreement with the nonnegative integral) give $\int h d λ \geq λ ([a, a + N]) = N$ , so $h$ is not integrable (The nonnegative Lebesgue integral, The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions, Monotonicity and nonnegative homogeneity of the nonnegative integral, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F4]

For a complex-valued function, $C^{2}$ means its real and imaginary parts are $C^{2}$ , with derivatives taken componentwise (Complex Lp classes and Euclidean test-function conventions).

Verification

1.1givenF1F2algebra

[(a) the formal solution] By [F2], $F \in C^{1}$ with $F^{'} = e^{y^{2}}$ , so $u_{2}$ is $C^{2}$ with, using the product rule and [F1], $u_{2}^{'} = e^{- y^{2} / 2} (F^{'} - y F) = e^{- y^{2} / 2} (e^{y^{2}} - y F)$ and $u_{2}^{''} = e^{- y^{2} / 2} (2 y e^{y^{2}} - F - y F^{'} - y (e^{y^{2}} - y F)) = (y^{2} - 1) u_{2}$ because the $e^{y^{2}}$ terms cancel. Since $(- \partial_{y}^{2} + y^{2}) g_{0} = g_{0}$ by [F1], the equation $u^{''} = (y^{2} - 1) u$ is exactly the ground-state eigenvalue equation.

1.2givenF2F3F5algebra

[(c) nonintegrability] For $y \geq 1$ , $F (y) = \int_{0}^{y} e^{t^{2}} d t \geq \int_{y - 1}^{y} e^{t^{2}} d t \geq e^{(y - 1)^{2}}$ because $t \geq y - 1 \geq 0$ on the interval and the exponential is increasing by [F5]. Hence $∣ u_{2} (y) ∣^{2} = e^{- y^{2}} F (y)^{2} \geq e^{- y^{2}} e^{2 (y - 1)^{2}} = e^{y^{2} - 4 y + 2}$ for $y \geq 1$ ; the lower bound exceeds $1$ for $y \geq 4$ , so [F3] implies $\int_{R} ∣ u_{2} ∣^{2} d y = + \infty$ . Hence $u_{2} \notin L^{2} (R, d y)$ .

2.1step 1.1F1F5algebra

[(b) the Wronskian] Using $g_{0}^{'} = - y g_{0}$ from [F1] and the first formula of step 1.1, $W = g_{0} u_{2}^{'} - g_{0}^{'} u_{2} = π^{- 1 / 4} e^{- y^{2} / 2} \cdot e^{- y^{2} / 2} (e^{y^{2}} - y F) + y π^{- 1 / 4} e^{- y^{2} / 2} \cdot e^{- y^{2} / 2} F = π^{- 1 / 4} (1 - y F e^{- y^{2}} + y F e^{- y^{2}}) = π^{- 1 / 4}$ , a nonzero constant; hence $g_{0}, u_{2}$ are linearly independent (if $a g_{0} + b u_{2} = 0$ , applying $W (g_{0}, v) : = g_{0} v^{'} - g_{0}^{'} v$ gives $b π^{- 1 / 4} = 0$ , hence $b = 0$ , and then $a = 0$ because $g_{0} > 0$ ).

3.1step 1.1step 1.2F1F3F4F5algebra∎

[(d) all complex $L^{2}$ solutions are multiples of $g_{0}$ ] Let $u : R \to C$ be componentwise $C^{2}$ by [F4] and solve $u^{''} = (y^{2} - 1) u$ . Put $W (u) : = g_{0} u^{'} - g_{0}^{'} u$ . Applying the real product rule to the real and imaginary parts gives $W (u)^{'} = g_{0} u^{''} - g_{0}^{''} u = g_{0} (y^{2} - 1) u - (y^{2} - 1) g_{0} u = 0$ using [F1]. Applying [F3] to the real and imaginary parts on any finite interval gives $W (u) (y) = W (u) (0) = c$ , so $W (u)$ is constant. Since $g_{0} > 0$ on all of $R$ , $(u / g_{0})^{'} = W (u) / g_{0}^{2} = c / g_{0}^{2}$ , and $1 / g_{0}^{2} = π^{1 / 2} e^{y^{2}}$ is continuous; applying [F3] separately to the real and imaginary parts and integrating from $0$ to $y$ gives $u (y) = g_{0} (y) (C + c π^{1 / 2} F (y))$ for $C = u (0) / g_{0} (0)$ . If $c \neq 0$ , choose $Y \geq 1$ so large that $∣ c ∣ \sqrt{π} F (y) \geq 2 ∣ C ∣$ for all $y \geq Y$ ; such $Y$ exists by step 1.2. The reverse triangle inequality gives $∣ C + c \sqrt{π} F (y) ∣ \geq \frac{1}{2} ∣ c ∣ \sqrt{π} F (y)$ , hence for $y \geq Y$ , $∣ u (y) ∣ \geq \frac{1}{2} ∣ c ∣ π^{1 / 2} g_{0} (y) e^{(y - 1)^{2}} = \frac{1}{2} ∣ c ∣ π^{1 / 4} e^{y^{2} / 2 - 2 y + 1}$ by the lower bound on $F$ , which tends to $+ \infty$ ; hence $∣ u (y) ∣^{2} \geq 1$ on some half-line $[a, \infty)$ , and [F3] gives $u \notin L^{2}$ . Therefore if $u \in L^{2}$ then $c = 0$ and $u = C g_{0}$ .

Qm Math Oscillator — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Ground energy units

Example

Facts & Assumptions

Verification

Poisson number distribution

Example

Facts & Assumptions

Verification

Formal nonnormalizable solution

Example

Facts & Assumptions

Verification

Sources