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The Gaussian attains equality in the Heisenberg inequality

Example

Assume countable choice. Let $n \geq 1$ and $a > 0$ and $f (x) : = e^{- π a ∣ x ∣^{2}}$ . Then $∥ f ∥_{2}^{2} = (2 a)^{- n / 2}$ , the means of $∣ f ∣^{2}$ and of $∣ \hat{f} ∣^{2}$ are $0$ , and $V_{x} (f) = \frac{n}{4 π a}, V_{ξ} (f) = \frac{n a}{4 π},$ so $V_{x} (f) V_{ξ} (f) = \frac{n^{2}}{16 π^{2}}$ and $\sqrt{V_{x} (f)} \sqrt{V_{ξ} (f)} = \frac{n}{4 π}$ ; hence $f$ attains equality in The n-dimensional Heisenberg uncertainty inequality, and FA-23's equality family $c \exp (- λ ∣ x - x_{0} ∣^{2} / 2) \exp (2 π i b_{0} \cdot x)$ is realised with $c = 1$ , $λ = 2 π a$ , and $x_{0} = b_{0} = 0$ .

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), an integer $n \geq 1$ , $a > 0$ , and $f (x) = e^{- π a ∣ x ∣^{2}}$ , whose membership in $L^{1} \cap L^{2}$ , finite moments, and zero means are verified below; the variances use Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments.

[F1]

Countable choice is assumed; it is the hypothesis carried by the Gaussian transform identity, the parameter differentiation, the reflection substitution and Plancherel below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

For every $t > 0$ the Gaussian $e^{- π t ∣ x ∣^{2}}$ is absolutely integrable with $\int_{R^{n}} e^{- π t ∣ x ∣^{2}} d x = t^{- n / 2}$ and $L^{1}$ transform $t^{- n / 2} e^{- π ∣ ξ ∣^{2} / t}$ ; every polynomial times a positive real Gaussian is absolutely integrable (Euclidean Gaussian transform with the 2π normalization); the one-dimensional Gaussian integral is $\int_{R} e^{- s^{2}} d s = \sqrt{π}$ (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ ).

[F3]

Differentiation under the integral sign: if $f (x, t)$ is integrable in $x$ for every $t$ in an open interval, differentiable in $t$ for almost every $x$ , and the $t$ -derivative is measurable in $x$ with $∣ \partial_{t} f (x, t) ∣ \leq g (x)$ for an integrable $g$ and all $t$ , then $t \mapsto \int f (x, t) d x$ is differentiable with derivative $\int \partial_{t} f d x$ (Differentiation under the integral sign).

[F4]

Complex $L^{1}$ change of variables for the reflection $T (x) = - x$ ( $∣ \det D T ∣ = 1$ ): $\int h (- x) d x = \int h (x) d x$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions). Complex $L^{2}$ carries $∥ \cdot ∥_{2}$ and Cauchy–Schwarz (Complex completeness, density, and inner product: the consumer interface), and for $g \in L^{1} \cap L^{2}$ the integral transform represents its Plancherel transform (Agreement of the integral and L2 transforms), so $∥ \hat{g} ∥_{2} = ∥ g ∥_{2}$ (Plancherel theorem).

[F5]

The power function $t \mapsto t^{α}$ on $(0, \infty)$ is differentiable with derivative $α t^{α - 1}$ (Continuity and derivatives of positive-base real powers, Real powers for positive bases, with the zero-base positive-exponent convention); in particular $\frac{d}{d c} (2 c)^{- n / 2} = - n (2 c)^{- n / 2 - 1}$ .

[F6]

The Fourier characterization identifies the $L^{2}$ classes with $⟨ ξ ⟩ \hat{f} \in L^{2}$ as $H^{1} = W^{1, 2}$ under the regular-distribution embedding (Integer-order W^{k,2} and H^k agree with equivalent norms, Statement 1 and Proof step 1.3 at $k = 1$ ).

Verification

technique · direct

1.1F1F2F4given

Norm, transform, evenness and vanishing means. By [F2] with $t = 2 a$ , $∥ f ∥_{2}^{2} = \int e^{- 2 π a ∣ x ∣^{2}} d x = (2 a)^{- n / 2} < \infty$ , and $\hat{f} (ξ) = a^{- n / 2} e^{- π ∣ ξ ∣^{2} / a}$ , so $∥ \hat{f} ∥_{2}^{2} = a^{- n} \int e^{- 2 π ∣ ξ ∣^{2} / a} d ξ = (2 a)^{- n / 2}$ by [F2] with $t = 2 / a$ and [F4]. Both $f$ and $\hat{f}$ are even and strictly positive. The functions $x \mapsto x_{j} ∣ f (x) ∣^{2}$ and $ξ \mapsto ξ_{j} ∣ \hat{f} (ξ) ∣^{2}$ are odd in their $j$ -th coordinate, so by [F4] (reflection) each integral equals its own negative; since they are integrable by [F2] and the remark above, both vanish. Hence both means are $0$ , and the variances are the uncentred second moments divided by $(2 a)^{- n / 2}$ .

2.1F2F3F5step 1.1

Second moments and variances. Differentiating the identity $\int_{R^{n}} e^{- 2 π c ∣ x ∣^{2}} d x = (2 c)^{- n / 2}$ in the parameter $c > 0$ ([F2] with $t = 2 c$ , [F5]) is legitimate by [F3]: on an open neighbourhood with closure contained in $(0, \infty)$ the derivative $- 2 π ∣ x ∣^{2} e^{- 2 π c ∣ x ∣^{2}}$ is dominated by $2 π ∣ x ∣^{2} e^{- 2 π c_{0} ∣ x ∣^{2}}$ for a positive lower bound $c_{0}$ of that interval, which is integrable by [F2]. Hence for every $c > 0$ $\int_{R^{n}} ∣ x ∣^{2} e^{- 2 π c ∣ x ∣^{2}} d x = \frac{1}{2 π} n (2 c)^{- n / 2 - 1} = \frac{n}{4 π c} (2 c)^{- n / 2} .$ With $c = a$ and step 1.1 this gives $\int ∣ x ∣^{2} ∣ f ∣^{2} = \frac{n}{4 π a} ∥ f ∥_{2}^{2}$ , hence $V_{x} (f) = \frac{\int ∣ x ∣^{2} ∣ f ∣^{2}}{∥ f ∥_{2}^{2}} = \frac{n}{4 π a} .$ Next, $∣ \hat{f} (ξ) ∣^{2} = a^{- n} e^{- 2 π ∣ ξ ∣^{2} / a} = a^{- n} e^{- 2 π c ∣ ξ ∣^{2}}$ with $c = 1 / a$ , so the same identity gives $\int ∣ ξ ∣^{2} ∣ \hat{f} ∣^{2} d ξ = a^{- n} \frac{n a}{4 π} (2 / a)^{- n / 2} = \frac{n a}{4 π} (2 a)^{- n / 2} = \frac{n a}{4 π} ∥ \hat{f} ∥_{2}^{2}$ by step 1.1, and therefore $V_{ξ} (f) = \frac{\int ∣ ξ ∣^{2} ∣ \hat{f} ∣^{2}}{∥ \hat{f} ∥_{2}^{2}} = \frac{n a}{4 π} .$

3.1F4F6step 1.1step 2.1∎

Equality. By step 2.1, $\int (1 + ∣ ξ ∣^{2}) ∣ \hat{f} ∣^{2} < \infty$ , so [F4, F6] give $f \in H^{1}$ ; the finite spatial moment is also verified there. Hence the domain of the cited Heisenberg corollary is satisfied. Steps 1.1 and 2.1 give $∥ ∣ x ∣ f ∥_{2}^{2} = \frac{n}{4 π a} ∥ f ∥_{2}^{2}$ and $∥ ∣ ξ ∣ \hat{f} ∥_{2}^{2} = \frac{n a}{4 π} ∥ f ∥_{2}^{2}$ , hence $∥ ∣ x ∣ f ∥_{2} ∥ ∣ ξ ∣ \hat{f} ∥_{2} = ∥ f ∥_{2}^{2} \sqrt{\frac{n}{4 π a}} \sqrt{\frac{n a}{4 π}} = \frac{n}{4 π} ∥ f ∥_{2}^{2},$ so the inequality of The n-dimensional Heisenberg uncertainty inequality is an equality. The product of variances is $\frac{n}{4 π a} \cdot \frac{n a}{4 π} = \frac{n^{2}}{16 π^{2}}$ and its square root is $\frac{n}{4 π}$ . Finally the entire family $c \exp (- λ ∣ x - a_{0} ∣^{2} / 2) \exp (2 π i b_{0} \cdot x)$ of the published Heisenberg theorem contains $f$ at $c = 1$ , $λ = 2 π a$ , $a_{0} = b_{0} = 0$ .

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Sources

Calder Sheagren, Uncertainty Principles with Fourier Analysis (University of Chicago REU 2017, author PDF) (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (arXiv:0903.3845) (standard reference, not scraped)