LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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Exact free Gaussian propagation

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Fix $m > 0$ , $ℏ > 0$ , $σ > 0$ and real $x_{0}, p_{0}$ , and let

$ψ_{0} (x) : = (2 π σ^{2})^{- 1 / 4} \exp (- \frac{(x - x_{0})^{2}}{4 σ^{2}}) \exp (i \frac{p_{0} (x - x_{0})}{ℏ}), x \in R .$

Let $H_{0}$ be the free operator of clause (b) of Real multiplication operators and free plus bounded-potential Schrödinger operators with $c : = 2 π^{2} ℏ^{2} / m$ and $b : = 0$ , i.e.

$H_{0} : = F_{2}^{*} M_{2 π^{2} ℏ^{2} ξ^{2} / m} F_{2}, D (H_{0}) = {f \in L^{2} (R) : ξ^{2} F_{2} f \in L^{2} (R)},$

which is self-adjoint with $D (H_{0}) = W^{2, 2} (R; C)$ and $H_{0} f = - \frac{ℏ^{2}}{2 m} Δ f$ for $f \in S (R)$ . Let $U$ be the momentum transform of Momentum Fourier Jacobian: the unitary momentum representation, the surjective isometry $L^{2} (R, d x) \to L^{2} (R, d p)$ with $U = D_{ℏ} \circ F_{2}$ on $S (R)$ . For $t \in R$ put $τ : = \frac{ℏ t}{2 m σ^{2}}$ and

$ψ_{t} (x) : = (2 π σ^{2})^{- 1 / 4} (1 + i τ)^{- 1 / 2} \exp (- \frac{(x - x_{0} - p_{0} t / m)^{2}}{4 σ^{2} (1 + i τ)} + i \frac{p_{0} (x - x_{0})}{ℏ} - i \frac{p_{0}^{2} t}{2 m ℏ}),$

where $z^{- 1 / 2} : = \exp (- \frac{1}{2} Log z)$ is the principal branch of Complex logarithms, the principal logarithm, and principal and multivalued complex powers, legitimate at $z = 1 + i τ$ because $Re (1 + i τ) = 1 > 0$ . Let $Q f (x) = x f (x)$ and $P f (x) = - i ℏ f^{'} (x)$ act on $S (R)$ . Then:

(a) $ψ_{0} \in S (R)$ , $∥ ψ_{0} ∥_{2} = 1$ , and for every $t$ one has $ψ_{t} \in S (R) \subseteq D (H_{0})$ ; the map $t \mapsto ψ_{t}$ is continuously differentiable $R \to L^{2} (R)$ with $ψ_{t} \in D (H_{0})$ and $\frac{d}{d t} ψ_{t} = - \frac{i}{ℏ} H_{0} ψ_{t}$ , and consequently $ψ_{t} = e^{- i t H_{0} / ℏ} ψ_{0}$ , where $e^{- i t H_{0} / ℏ} : = e^{i t T}$ with $T : = - H_{0} / ℏ$ denotes the unitary group generated by the self-adjoint operator $T$ (A self-adjoint operator generates a strongly continuous unitary group).

(b) Position density. $∣ ψ_{t} (x) ∣^{2} = (2 π σ^{2} (1 + τ^{2}))^{- 1 / 2} \exp (- (x - x_{0} - p_{0} t / m)^{2} / (2 σ^{2} (1 + τ^{2})))$ ; it is a probability density of mean $x_{0} + p_{0} t / m$ and variance $σ^{2} (1 + τ^{2})$ .

(c) Momentum density. $∣ U ψ_{t} (p) ∣^{2} = \sqrt{2 / π} (σ / ℏ) \exp (- 2 σ^{2} (p - p_{0})^{2} / ℏ^{2}) = ∣ U ψ_{0} (p) ∣^{2}$ for almost every $p$ ; it is a probability density of mean $p_{0}$ and variance $ℏ^{2} / (4 σ^{2})$ , independent of $t$ .

(d) Standard deviations. With $Δ Q : = ∥ (Q - ⟨ Q ψ_{t}, ψ_{t} ⟩) ψ_{t} ∥_{2}$ and $Δ P : = ∥ (P - ⟨ P ψ_{t}, ψ_{t} ⟩) ψ_{t} ∥_{2}$ one has $Δ Q = σ \sqrt{1 + τ^{2}}$ , $Δ P = ℏ / (2 σ)$ and

$Δ Q Δ P = \frac{ℏ}{2} \sqrt{1 + τ^{2}} \geq \frac{ℏ}{2},$

with equality exactly at $t = 0$ ; the lower bound $ℏ / 2$ is the one supplied by clause (d) of Weak-form Robertson uncertainty on a common domain with $c = ℏ$ .

(e) Scope. These are exact conditional calculations for the free model defined by $H_{0}$ ; no measurement, instrument or laboratory preparation is described, and nothing here asserts that a physical wave packet is Gaussian.

Facts & Assumptions

Given: $m, ℏ, σ > 0$ , $x_{0}, p_{0} \in R$ , the vector $ψ_{0}$ above, the operator $H_{0}$ and the transform $U$ .

[F1]

$S (R)$ consists of the $C^{\infty}$ functions all of whose polynomial-weighted derivatives are bounded; every Schwartz function lies in $L^{1} \cap L^{2}$ (Schwartz space and its seminorms).

[F2]

$U : L^{2} (R, d x) \to L^{2} (R, d p)$ is a surjective complex-linear isometry with $U = D_{ℏ} F_{2}$ on $S (R)$ , where $D_{ℏ} g (p) = (2 π ℏ)^{- 1 / 2} g (p / (2 π ℏ))$ ; being a surjective isometry, $U$ preserves inner products and $U^{*} = U^{- 1}$ (Momentum Fourier Jacobian: the unitary momentum representation).

[F3]

If $Re z > 0$ then $\int_{R} e^{- z x^{2}} d x = \sqrt{π} z^{- 1 / 2}$ for the branch $z^{- 1 / 2} = \exp (- \frac{1}{2} Log z)$ , and for $G_{1} (z) : = \int_{R} e^{- z x^{2}} d x$ one has $G_{1}^{'} (z) = - \int_{R} x^{2} e^{- z x^{2}} d x$ (Complex Gaussian continuation(a),(b)).

[F4]

$H_{0}$ is self-adjoint, $D (H_{0}) = W^{2, 2} (R; C) = {f \in L^{2} : ξ^{2} F_{2} f \in L^{2}}$ with equivalent norms on this set, and $H_{0} f = - \frac{ℏ^{2}}{2 m} Δ f$ for $f \in S (R)$ (Real multiplication operators and free plus bounded-potential Schrödinger operators(b)). Here $F_{2}$ is the Plancherel transform extending the $L^{1}$ transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x ξ} d x$ ; it is a surjective isometry preserving inner products (Plancherel theorem, Fourier transform on complex L1 classes).

[F5]

For a self-adjoint operator $T$ the operators $V (t) : = e^{i t T}$ form a strongly continuous one-parameter unitary group with $V (t) D (T) = D (T)$ , $T V (t) = V (t) T$ on $D (T)$ , and $\lim_{s \to 0} \frac{1}{s} (V (s) x - x) = i T x$ exactly for $x \in D (T)$ (A self-adjoint operator generates a strongly continuous unitary group).

[F6]

Dominated convergence: if $f_{n} \to f$ pointwise and $∣ f_{n} ∣ \leq g$ for a fixed integrable $g$ , then $\int f_{n} \to \int f$ (Dominated convergence). A continuous real function on an interval has a primitive and $\int_{a}^{b} G^{'} = G (b) - G (a)$ for it (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ); the integral over $R$ is the limit of the compact-interval integrals when it exists (Improper integrals over unbounded intervals); $(f g)^{'} = f^{'} g + f g^{'}$ (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 $g (c) \neq 0$ ); for differentiable $u, v$ with integrable derivatives on $[a, b]$ , $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ ). Substitutions: an invertible linear $T : R \to R$ scales Lebesgue measure by $∣ \det T ∣$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not), and $λ_{1} (E + h) = λ_{1} (E)$ for every Lebesgue measurable $E$ and $h \in R$ (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation); hence for every nonnegative measurable $H$ and every real $x_{0}$ , $\int_{R} H (x - x_{0}) d x = \int_{R} H (y) d y$ , by approximation with simple functions and monotone convergence (Monotone convergence for the integral).

[F7]

For every unit $f \in S (R)$ and every $c > 0$ , with $Q f (x) = x f (x)$ and $P_{c} f (x) = - i c f^{'} (x)$ , one has $Δ Q Δ P_{c} \geq c / 2$ , the differences being centered norms (Weak-form Robertson uncertainty on a common domain(d)).

Proof

1.1F1F3F6algebra

[The initial packet is a unit Schwartz vector] The derivative $\partial_{x}^{k} ψ_{0}$ equals $P_{k} (x) ψ_{0} (x)$ for a polynomial $P_{k}$ : this follows by induction with the product rule [F6] applied to the three factors $(2 π σ^{2})^{- 1 / 4}$ , $e^{- (x - x_{0})^{2} / (4 σ^{2})}$ and the bounded factor $e^{i p_{0} (x - x_{0}) / ℏ}$ , each differentiation of the last factor multiplying it by the constant $i p_{0} / ℏ$ . Since $x \mapsto x^{j} P_{k} (x) e^{- (x - x_{0})^{2} / (4 σ^{2})}$ is bounded for all $j, k$ (the exponential decays faster than any polynomial), $ψ_{0} \in S (R)$ [F1]. By [F6] $\int_{R} ∣ ψ_{0} (x) ∣^{2} d x = (2 π σ^{2})^{- 1 / 2} \int_{R} e^{- y^{2} / (2 σ^{2})} d y = (2 π σ^{2})^{- 1 / 2} \sqrt{π} (2 σ^{2})^{1 / 2} = 1$ , using [F3] with $z = 1 / (2 σ^{2})$ and the substitution $x = x_{0} + y$ .

1.2F3F6algebra

[The complex shift identity] For $Re a > 0$ and $b \in R$ one has $\int_{R} e^{- a y^{2}} e^{i b y} d y = \sqrt{π / a} e^{- b^{2} / (4 a)}$ with $\sqrt{π / a} : = \sqrt{π} a^{- 1 / 2}$ : setting $F (b) : = \int_{R} e^{- a y^{2} + i b y} d y$ (absolutely convergent since $∣ e^{- a y^{2} + i b y} ∣ = e^{- Re (a) y^{2}}$ is integrable) and $G_{R} (b) : = \int_{- R}^{R} e^{- a y^{2} + i b y} d y$ , the fundamental theorem and [F6] give $\frac{d}{d b} G_{R} (b) = i \int_{- R}^{R} y e^{- a y^{2} + i b y} d y$ , and since $∣ y ∣ e^{- Re (a) y^{2}}$ is integrable, $G_{R} \to F$ pointwise with the difference quotients converging, so $F$ is differentiable with $F^{'} (b) = i \int_{R} y e^{- a y^{2} + i b y} d y$ ; integrating $\partial_{y} (e^{- a y^{2} + i b y}) = (- 2 a y + i b) e^{- a y^{2} + i b y}$ over $[- R, R]$ gives $\int_{- R}^{R} (- 2 a y + i b) e^{- a y^{2} + i b y} d y = e^{- a R^{2} + i b R} - e^{- a R^{2} - i b R}$ , of modulus at most $2 e^{- Re (a) R^{2}} \to 0$ , so [F6] as $R \to \infty$ gives $- 2 a F_{y} (b) + i b F (b) = 0$ with $F_{y} (b) : = \int_{R} y e^{- a y^{2} + i b y} d y$ , hence $F_{y} = \frac{i b}{2 a} F$ and $F^{'} = - \frac{b}{2 a} F$ ; the product rule [F6] gives $\frac{d}{d b} [F (b) e^{b^{2} / (4 a)}] = 0$ , so the product is constant (fundamental theorem applied to real and imaginary parts [F6]) and its value at $b = 0$ is $F (0) = \int_{R} e^{- a y^{2}} d y = \sqrt{π} a^{- 1 / 2}$ by [F3].

1.3F2F4algebra

[The free operator in momentum coordinates] Let $B$ be the multiplication operator $(B φ) (p) = \frac{p^{2}}{2 m} φ (p)$ on $D (B) : = {φ \in L^{2} (d p) : p^{2} φ \in L^{2}}$ ; then $U^{*} B U = H_{0}$ with $D (H_{0}) = U^{*} D (B)$ , and $H_{0} U^{*} = U^{*} B$ , because $D_{ℏ}^{*} = D_{ℏ}^{- 1}$ and $(D_{ℏ}^{- 1} M_{g} D_{ℏ} h) (ξ) = g (2 π ℏ ξ) h (ξ)$ for every Borel $g$ and every $h$ in the domain of $D_{ℏ}^{- 1} M_{g} D_{ℏ}$ , so with $g (p) = p^{2} / (2 m)$ one has $U^{*} B U = F_{2}^{*} D_{ℏ}^{*} M_{p^{2} / (2 m)} D_{ℏ} F_{2} = F_{2}^{*} M_{2 π^{2} ℏ^{2} ξ^{2} / m} F_{2} = H_{0}$ as operators [F2, F4]; for the domains, $f \in D (H_{0})$ means $ξ^{2} F_{2} f \in L^{2}$ , which by $U f (p) = (2 π ℏ)^{- 1 / 2} F_{2} f (p / (2 π ℏ))$ is equivalent to $p^{2} U f \in L^{2}$ (the substitution $ξ = p / (2 π ℏ)$ and the constants $(2 π ℏ)^{\pm 2}$ preserve $L^{2}$ -membership), i.e. to $U f \in D (B)$ , so $D (H_{0}) = U^{*} D (B)$ , the identity $U^{*} B U = H_{0}$ holds on it, and multiplying on the right by $U^{*}$ gives $H_{0} U^{*} = U^{*} B U U^{*} = U^{*} B$ .

2.1step 1.1step 1.2F2F4F6algebra

[Fourier transform of the initial packet] Since $ψ_{0} \in L^{1} \cap L^{2}$ [step 1.1], $F_{2} ψ_{0}$ is computed by the $L^{1}$ integral [F4]; substituting $x = x_{0} + y$ [F6] and using $ψ_{0} (x_{0} + y) = (2 π σ^{2})^{- 1 / 4} e^{- y^{2} / (4 σ^{2})} e^{i p_{0} y / ℏ}$ gives $F_{2} ψ_{0} (ξ) = (2 π σ^{2})^{- 1 / 4} e^{- 2 π i x_{0} ξ} \int_{R} e^{- y^{2} / (4 σ^{2})} e^{i (p_{0} / ℏ - 2 π ξ) y} d y = (2 π σ^{2})^{- 1 / 4} e^{- 2 π i x_{0} ξ} 2 σ \sqrt{π} e^{- σ^{2} (p_{0} / ℏ - 2 π ξ)^{2}}$ by step 1.2 with $a = 1 / (4 σ^{2})$ and $b = p_{0} / ℏ - 2 π ξ$ ; since $ψ_{0} \in S (R)$ , $U ψ_{0} = D_{ℏ} F_{2} ψ_{0}$ [F2], so $φ_{0} : = U ψ_{0}$ satisfies $φ_{0} (p) = (2 π ℏ)^{- 1 / 2} F_{2} ψ_{0} (p / (2 π ℏ)) = K e^{- i x_{0} p / ℏ} e^{- σ^{2} (p - p_{0})^{2} / ℏ^{2}}$ with $K : = (2 π ℏ)^{- 1 / 2} (2 π σ^{2})^{- 1 / 4} 2 σ \sqrt{π}$ and $K^{2} = \sqrt{2 / π} σ / ℏ$ .

2.2step 1.1F1F4algebra

[The packet is Schwartz and its modulus is the spreading Gaussian] Putting $z_{t} : = 1 / (1 + i τ)$ (so $Re z_{t} = 1 / (1 + τ^{2}) > 0$ ) and $c_{t} : = x_{0} + p_{0} t / m$ , one has $∣ ψ_{t} (x) ∣ = (2 π σ^{2})^{- 1 / 4} ∣ 1 + i τ ∣^{- 1 / 2} e^{- Re (z_{t}) (x - c_{t})^{2} / (4 σ^{2})} = (2 π σ^{2})^{- 1 / 4} (1 + τ^{2})^{- 1 / 4} e^{- (x - c_{t})^{2} / (4 σ^{2} (1 + τ^{2}))}$ ; the remaining $x$ -dependent phase is $\exp (- i Im (z_{t}) (x - c_{t})^{2} / (4 σ^{2}) + i p_{0} (x - x_{0}) / ℏ)$ , which has modulus one. By the product rule, each $x$ -derivative of $ψ_{t}$ is a polynomial in $x$ times $ψ_{t}$ , so the displayed decaying modulus makes every polynomial-weighted derivative bounded and $ψ_{t} \in S (R)$ ; in particular $ψ_{t} \in L^{1} \cap L^{2}$ and $ψ_{t} \in D (H_{0})$ by [F4].

3.1step 1.2step 2.1step 2.2F2F4F6algebra

[Momentum coordinates of the packet] With $A_{t} : = (2 π σ^{2})^{- 1 / 4} (1 + i τ)^{- 1 / 2}$ , $a_{t} : = 1 / (4 σ^{2} (1 + i τ))$ and $b_{ξ} : = p_{0} / ℏ - 2 π ξ$ , substituting $x = c_{t} + y$ [F6] in the $L^{1}$ integral [F4] of the packet $ψ_{t} = A_{t} e^{- i p_{0}^{2} t / (2 m ℏ)} e^{- a_{t} (x - c_{t})^{2}} e^{i (p_{0} / ℏ) (x - x_{0})}$ [step 2.2] gives $F_{2} ψ_{t} (ξ) = A_{t} \sqrt{π / a_{t}} e^{i p_{0}^{2} t / (2 m ℏ)} e^{- 2 π i ξ c_{t}} e^{- b_{ξ}^{2} / (4 a_{t})}$ by step 1.2 with $a = a_{t}$ (legal since $Re a_{t} > 0$ ); since $ψ_{t} \in S (R)$ , $U ψ_{t} = D_{ℏ} F_{2} ψ_{t}$ [F2], and with $p = 2 π ℏ ξ$ , $\sqrt{π / a_{t}} = 2 σ \sqrt{π} (1 + i τ)^{1 / 2}$ , $A_{t} (1 + i τ)^{1 / 2} = (2 π σ^{2})^{- 1 / 4}$ and $1 / (4 a_{t}) = σ^{2} (1 + i τ)$ one gets $U ψ_{t} (p) = K e^{i p_{0}^{2} t / (2 m ℏ)} e^{- i p c_{t} / ℏ} e^{- σ^{2} (p - p_{0})^{2} / ℏ^{2}} e^{- i t (p - p_{0})^{2} / (2 m ℏ)}$ with $K$ as in step 2.1.

3.2step 2.2F3F6algebra

[Position density, mean and variance] By step 2.2, $∣ ψ_{t} (x) ∣^{2} = (2 π σ^{2} (1 + τ^{2}))^{- 1 / 2} e^{- (x - c_{t})^{2} / (2 d_{t}^{2})}$ with $c_{t} = x_{0} + p_{0} t / m$ and $d_{t} : = σ \sqrt{1 + τ^{2}}$ ; substituting $x = c_{t} + d_{t} u$ and using $\int_{R} e^{- u^{2} / 2} d u = \sqrt{2 π}$ (that is [F3] with $z = 1 / 2$ ) gives $\int_{R} ∣ ψ_{t} ∣^{2} = 1$ , while the centering $\int_{R} (x - c_{t}) ∣ ψ_{t} (x) ∣^{2} d x = d_{t} \int_{R} u \frac{e^{- u^{2} / 2}}{\sqrt{2 π}} d u = 0$ and $\int_{R} (x - c_{t})^{2} ∣ ψ_{t} (x) ∣^{2} d x = d_{t}^{2} \int_{R} u^{2} \frac{e^{- u^{2} / 2}}{\sqrt{2 π}} d u = d_{t}^{2}$ , since $\int_{R} u^{2} e^{- u^{2} / 2} d u = - G_{1}^{'} (1 / 2) = \sqrt{2 π}$ by [F3]; hence the density has mean $c_{t} = x_{0} + p_{0} t / m$ and variance $σ^{2} (1 + τ^{2})$ , which is clause (b).

4.1step 2.1step 3.1F2algebra

[The packet is the free evolution of the momentum Gaussian] With $θ_{t} (p) : = t p^{2} / (2 m ℏ)$ and $φ_{t} : = e^{- i θ_{t}} φ_{0}$ , steps 2.1 and 3.1 give $U ψ_{t} (p) = e^{- i t p^{2} / (2 m ℏ)} φ_{0} (p) = φ_{t} (p)$ for every $p$ , because the two exponents differ by $i (p_{0}^{2} t / (2 m ℏ) - p (x_{0} + p_{0} t / m) / ℏ - t (p - p_{0})^{2} / (2 m ℏ) + p x_{0} / ℏ + t p^{2} / (2 m ℏ)) = i t (p_{0}^{2} - 2 p p_{0} - (p - p_{0})^{2} + p^{2}) / (2 m ℏ) = 0$ ; hence $U ψ_{t} = φ_{t}$ and, $U$ being a surjective isometry [F2], $ψ_{t} = U^{*} φ_{t}$ .

4.2step 2.1step 3.1step 3.2F3F6algebra

[Momentum density, mean and variance] By steps 2.1 and 3.1, $∣ U ψ_{t} (p) ∣^{2} = ∣ φ_{0} (p) ∣^{2} = K^{2} e^{- 2 σ^{2} (p - p_{0})^{2} / ℏ^{2}} = \sqrt{2 / π} \frac{σ}{ℏ} e^{- (p - p_{0})^{2} / (2 w^{2})}$ with $w : = ℏ / (2 σ)$ and $K^{2} = \sqrt{2 / π} σ / ℏ$ ; substituting $p = p_{0} + w u$ as in step 3.2 gives total mass $1$ , mean $p_{0}$ and variance $w^{2} = ℏ^{2} / (4 σ^{2})$ , and the density does not depend on $t$ ; this is clause (c).

5.1step 1.3step 2.1step 4.1F2F6algebra

[The momentum path solves the momentum equation] Put $q (p) : = p^{2} / (2 m ℏ)$ and $φ_{t} (p) : = e^{- i t q (p)} φ_{0} (p)$ . The Gaussian formula of step 2.1 gives $q φ_{0} \in L^{2} (d p)$ and $p^{2} φ_{0} \in L^{2} (d p)$ , so $φ_{t} \in D (B)$ for every $t$ . For fixed $t$ and $h \neq 0$ , $\frac{φ_{t + h} (p) - φ_{t} (p)}{h} = φ_{t} (p) \frac{e^{- i h q (p)} - 1}{h}$ tends pointwise to $- i q (p) φ_{t} (p) = - (i / ℏ) \frac{p^{2}}{2 m} φ_{t} (p)$ . Since $∣ e^{- i s} - 1 ∣ \leq ∣ s ∣$ , the difference between the quotient and its limit has squared modulus at most $4 ∣ q (p) φ_{0} (p) ∣^{2}$ , an integrable function; dominated convergence [F6] therefore gives convergence in $L^{2} (d p)$ . Also $φ_{t + h} \to φ_{t}$ and $B φ_{t + h} \to B φ_{t}$ in $L^{2} (d p)$ by pointwise convergence and the respective integrable square majorants $4 ∣ φ_{0} ∣^{2}$ and $4 ∣ B φ_{0} ∣^{2}$ . Thus $t \mapsto φ_{t}$ is continuously differentiable with $φ_{t}^{'} = - (i / ℏ) B φ_{t}$ . Consequently $ψ_{t} = U^{*} φ_{t}$ [step 4.1] is $C^{1}$ in $L^{2}$ , lies in $D (H_{0}) = U^{*} D (B)$ , and $\frac{d}{d t} ψ_{t} = U^{*} φ_{t}^{'} = - \frac{i}{ℏ} U^{*} B φ_{t} = - \frac{i}{ℏ} H_{0} U^{*} φ_{t} = - \frac{i}{ℏ} H_{0} ψ_{t}$ , using that $U^{*}$ is an isometry [F2] and $H_{0} U^{*} = U^{*} B$ from step 1.3; at $t = 0$ the formula of the Statement reduces to $ψ_{0}$ .

6.1step 1.1step 5.1F4F5F6algebra

[Uniqueness of the Cauchy problem] Let $V (t) : = e^{i t T}$ with $T : = - H_{0} / ℏ$ [F5]; since $ψ_{0} \in D (H_{0}) = D (T)$ [step 1.1, F4], one has $V (t) ψ_{0} \in D (T)$ for all $t$ and $t \mapsto V (t) ψ_{0}$ is $C^{1}$ with $\frac{d}{d t} V (t) ψ_{0} = i T V (t) ψ_{0} = - \frac{i}{ℏ} H_{0} V (t) ψ_{0}$ , because [F5] gives $\frac{1}{h} (V (t + h) - V (t)) ψ_{0} = V (t) \frac{1}{h} (V (h) ψ_{0} - ψ_{0}) \to V (t) (i T ψ_{0})$ as $ψ_{0} \in D (T)$ and $V (t)$ is bounded, and $T V (t) = V (t) T$ on $D (T)$ by [F5]; by step 5.1, $w_{t} : = ψ_{t} - V (t) ψ_{0}$ is $C^{1}$ with $w_{t} \in D (H_{0})$ and $w_{t}^{'} = - \frac{i}{ℏ} H_{0} w_{t}$ , and $w_{0} = 0$ since $V (0) = I$ and the Statement's formula at $t = 0$ is the initial packet; for the real function $t \mapsto ∥ w_{t} ∥^{2}$ one has $\frac{∥ w_{t + h} ∥^{2} - ∥ w_{t} ∥^{2}}{h} = 2 Re ⟨ w_{t}, \frac{w_{t + h} - w_{t}}{h} ⟩ + \frac{∥ w_{t + h} - w_{t} ∥^{2}}{h} \to 2 Re ⟨ w_{t}, w_{t}^{'} ⟩$ , the last term tending to $0$ ; now, since the inner product is conjugate-linear in its second argument, $⟨ w_{t}, w_{t}^{'} ⟩ = ⟨ w_{t}, - \frac{i}{ℏ} H_{0} w_{t} ⟩ = \frac{i}{ℏ} ⟨ w_{t}, H_{0} w_{t} ⟩$ . The number $⟨ w_{t}, H_{0} w_{t} ⟩ = ⟨ H_{0} w_{t}, w_{t} ⟩$ is real because $H_{0}$ is self-adjoint, hence symmetric, on $D (H_{0})$ [F4], so $Re ⟨ w_{t}, w_{t}^{'} ⟩ = 0$ and $∥ w_{t} ∥^{2}$ has vanishing derivative, hence is constant by the fundamental theorem [F6] and equals $∥ w_{0} ∥^{2} = 0$ ; thus $ψ_{t} = V (t) ψ_{0} = e^{- i t H_{0} / ℏ} ψ_{0}$ for every $t$ , which is clause (a).

7.1step 3.2step 4.2step 6.1F2F6algebra

[Operator standard deviations] For $ψ \in S (R)$ one has $U (P ψ) = M_{p} U ψ$ with $(M_{p} g) (p) = p g (p)$ , because $F_{2} (P ψ) (ξ) = \int_{R} (- i ℏ ψ^{'} (x)) e^{- 2 π i x ξ} d x = 2 π ℏ ξ F_{2} ψ (ξ)$ by integration by parts, the boundary terms $[ψ (x) e^{- 2 π i x ξ}]$ vanishing at $\pm \infty$ as $ψ \in S (R)$ [F6], and applying $D_{ℏ}$ with $p = 2 π ℏ ξ$ gives the claim [F2]; the unit vector $ψ_{t}$ of step 6.1 is the packet of steps 3.2 and 4.2, and for it this gives $b_{t} : = ⟨ P ψ_{t}, ψ_{t} ⟩ = ⟨ U P ψ_{t}, U ψ_{t} ⟩ = \int_{R} p ∣ U ψ_{t} (p) ∣^{2} d p = p_{0}$ , a real number, and $Δ P : = ∥ (P - b_{t}) ψ_{t} ∥_{2} = ∥ U (P - b_{t}) ψ_{t} ∥_{2} = (\int_{R} (p - p_{0})^{2} ∣ U ψ_{t} (p) ∣^{2} d p)^{1 / 2} = ℏ / (2 σ)$ by step 4.2, while $a_{t} : = ⟨ Q ψ_{t}, ψ_{t} ⟩ = \int_{R} x ∣ ψ_{t} (x) ∣^{2} d x = c_{t}$ and $Δ Q : = ∥ (Q - a_{t}) ψ_{t} ∥_{2} = (\int_{R} (x - c_{t})^{2} ∣ ψ_{t} (x) ∣^{2} d x)^{1 / 2} = σ \sqrt{1 + τ^{2}}$ by step 3.2; hence $Δ Q Δ P = \frac{ℏ}{2} \sqrt{1 + τ^{2}}$ .

8.1step 7.1F7algebra∎

[Robertson comparison and scope] By [F7] with $c = ℏ$ , applied to the unit Schwartz vector $ψ_{t}$ of step 7.1, one has $Δ Q Δ P \geq ℏ / 2$ ; the exact value of step 7.1 is $\frac{ℏ}{2} \sqrt{1 + τ^{2}}$ , which equals $ℏ / 2$ exactly when $τ = 0$ , i.e. exactly at $t = 0$ since $τ = ℏ t / (2 m σ^{2})$ , and is strictly larger otherwise; thus clause (d) holds, and clause (e) records that all of the above are computations inside the free model defined by $H_{0}$ , with no measurement interpretation.

Exact free Gaussian propagation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources