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SHINTARO FUSHIDA-HARDY

1. Heisenberg Suppose p : R → R is a density for a X. The moments of X are given by Z n n (1) mn := E(X ) = x p(x)dx. R 2 (Assuming they exist), the first moment mn is the mean of X, while m2 −m1 is the of X. The Heisenberg uncertainty principle in states that 2 (2) Var(x)Var(p) ≥ ~ 4 where x and p denote and of a , respectively. If these both have mean zero, the above equation is Z Z 2  2  2  ~ (3) t px(t)dt s pp(s)dx ≥ , R R 4 where px and pp are probability density for position and momentum. But from , the probability density for position is precisely the function of the particle under consideration. Moreover, pp is the of px with additional constants. In this document we prove a more general result from which the Heisenberg uncertainty principle follows.

2. Definitions and Theorems The Schwartz space on R are is the collection of rapidly decreasing complex valued smooth functions on R. Formally, ∞ (4) S(R) = {f ∈ C (R): kfkα,β < ∞}, where the collection of norms is defined for all α, β ∈ N by α (β) (5) kfkα,β = sup |x f (x)|. x∈R The Schwartz space is the natural space for , since the Fourier transform is a homeomorphism on this space. It also has the delicious property of being contained in p L (R) for all 1 ≤ p ≤ ∞. The Fourier transform is defined by Z (6) fˆ(ζ) = f(x)e−2πixζ dx. R 1 On S(R), the inverse is simply Z (7) f(x) = fˆ(ζ)e2πixζ dζ. R Moreover, the Parseval identity holds for all f, g ∈ S(R): Z Z (8) f(x)gˆ(x)dx = fˆ(x)g(x)dx. R R The Plancherel identity immediately follows: Z Z (9) |f(x)|2dx = |fˆ(x)|2dx. R R Lemma 2.1. Let f ∈ S(R). Then fˆ0(ζ) = 2πiζfˆ(ζ). Proof. This follows from : Z h i∞ Z 2πiζfˆ(ζ) = 2πiζ f(x)e−2πixζ dx = − f(x)e−2πixζ + f 0(x)e−2πixζ dx. −∞ R R The first term on the right hand side vanishes since f is a Schwartz function. 

3. Fourier uncertainty principle Recalling the form of the uncertainty principle in equation 2, we prove the following Fourier uncertainty principle:

Theorem 3.1. Let f ∈ S(R). Suppose f is normalised, that is, kfk2 = 1. Then Z Z  2 2  2 ˆ 2  1 (10) x |f(x)| dx ζ |f(ζ)| dζ ≥ 2 . R R 16π Proof. Let f ∈ S(R). Integration by parts gives Z h i∞ Z xf(x)f 0(x)dx = xf(x)f(x) − (xf(x))0f(x)dx −∞ R R Z Z = 0 − f(x)f(x)dx − xf 0(x)f(x)dx. R R The first term on the right side vanishes by virtue of f being a Schwartz function. Rear- ranging the above gives Z Z 1 = |f(x)|2dx = − x(f 0(x)f(x) + f(x)f 0(x))dx R R Z = −2 xReal[f(x)f 0(x)]dx. R By the Cauchy-Schwartz ,  Z 2  Z  Z  1 ≤ 2 |xReal[f(x)f 0(x)]|dx ≤ 4 |xf(x)|2dx |f 0(x)|2dx . R R R 2 Moreover, by lemma 2.1 and the Plancherel identity, Z Z Z Z (11) |f 0(x)|2dx = |fˆ0(ζ)|2dζ = |2πiζfˆ(ζ)|2dζ = 4π2 |ζfˆ(ζ)|2dζ. R R R R Combining this with the previous inequality,  Z  Z  (12) 1 ≤ 16π2 |xf(x)|2dx |ζfˆ(ζ)|2dζ , R R so Z Z  2 2  2 ˆ 2  1 (13) x |f(x)| dx ζ |f(ζ)| dζ ≥ 2 R R 16π as required. 

Department of Mathematics, The University of Auckland, Private Bag 92019, Auckland 1142, New Zealand Email address: [email protected]

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