Uncertainty principle in harmonic analysis

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Let P,MP, M be the position and momentum operators introduced in equation (11.26). These operators are not defined on all of L2(R)L^{2}(\mathbb {R}). Instead, define domains

DP={f∈L2(R):xf(x)∈L2(R)},DM={f∈L2(R):f is differentiable and f′∈L2(R)}, \begin{aligned} D_{P} & = \left\{ f \in L^{2}(\mathbb {R}): x f(x) \in L^{2}(\mathbb {R})\right\} , \\ D_{M} & = \left\{ f \in L^{2}(\mathbb {R}): f \text{ is differentiable and } f^{\prime } \in L^{2}(\mathbb {R})\right\} , \end{aligned}

which are dense subspaces of L2(R)L^{2}(\mathbb {R}). Restricted to these domains, PP maps DPD_{P} into L2(R)L^{2}(\mathbb {R}) and MM maps DMD_{M} into L2(R)L^{2}(\mathbb {R}). Show that PP and MM are unbounded even when restricted to these domains, i.e.,

sup⁡f∈DP,∥f∥L2=1∥Pf∥L2=∞=sup⁡f∈DM,∥f∥L2=1∥Mf∥L2 \sup _{\substack {f \in D_{P},\left\| f\right\| _{L^{2}} = 1}}\left\| P f\right\| _{L^{2}} = \infty = \sup _{\substack {f \in D_{M},\left\| f\right\| _{L^{2}} = 1}}\left\| M f\right\| _{L^{2}}
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The commutator of position and momentum is the operator [P,M]=PM−MP[P, M] = P M-M P. Show that [P,M]=−12πiI[P, M] = -\frac{1}{2 \pi i} I in the sense that [P,M]f=−12πif[P, M] f = -\frac{1}{2 \pi i} f for all differentiable functions ff. How does this relate to equation (11.32)?

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This exercise will give an abstract operator-theoretic version of the Uncertainty Principle.

Let SS be a subspace of a Hilbert space HH, and let A,B:S→HA, B: S \rightarrow H be linear but possibly unbounded operators. By replacing SS with the smaller space domain (AB)∩(A B) \cap domain (BA)(B A) if necessary, we may assume that A,B,ABA, B, A B, and BAB A are all defined on SS.

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(a)

Show that if A,BA, B are self-adjoint in the sense that

∀f,g∈S,⟨Af,g⟩=⟨f,Ag⟩ and ⟨Bf,g⟩=⟨f,Bg⟩, \forall f, g \in S, \quad \left\langle A f, g \right\rangle = \left\langle f, A g \right\rangle \quad \text{ and } \quad \left\langle B f, g \right\rangle = \left\langle f, B g \right\rangle ,

then

∀f∈S,∥Af∥∥Bf∥≥12∣⟨[A,B]f,f⟩∣, \forall f \in S, \quad \left\| A f\right\| \left\| B f\right\| \geq \frac{1}{2}\left| \left\langle [A, B] f, f \right\rangle \right|,

where [A,B]=AB−BA[A, B] = A B-B A is the commutator of AA and BB.

(b)

Show that equality holds in part (a) if and only if Af=icBfA f = i c B f for some c∈Rc \in \mathbb {R}.

Apply Exercises 11.29−11.3111.29-11.31 to the position and momentum operators PP and MM to derive the Classical Uncertainty Principle,

∥xg(x)∥L2∥ξg^(ξ)∥L2≥14π∥g∥L22,(11.33) \left\| x g(x)\right\| _{L^{2}}\left\| \xi \widehat{g}(\xi )\right\| _{L^{2}} \geq \frac{1}{4 \pi }\left\| g\right\| _{L^{2}}^{2}, \tag {11.33}

for g∈S(R)g \in \mathcal{S}(\mathbb {R}).

Remark An extension by density argument can be used to prove that equation (11.33) extends to all g∈L2(R)g \in L^{2}(\mathbb {R}), or integration by parts for absolutely continuous functions can be used to prove directly that equation (11.33) holds whenever ∥xg(x)∥L2∥ξg^(ξ)∥L2\left\| x g(x)\right\|_{L^{2}}\left\| \xi \widehat{g}(\xi )\right\|_{L^{2}} is finite, see [Heil].

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Modify Battle's argument to prove the Weak BLTB L T : If G(g,1,1)\mathcal{G}(g, 1,1) is a Riesz basis for L2(R)L^{2}(\mathbb {R}) then

∥xg(x)∥L2∥ξg^(ξ)∥L2∥xg~(x)∥L2∥ξg~^(ξ)∥L2=∞, \left\| x g(x)\right\| _{L^{2}}\left\| \xi \widehat{g}(\xi )\right\| _{L^{2}}\left\| x \widetilde{g}(x)\right\| _{L^{2}}\left\| \xi \widehat{\widetilde{g}}(\xi )\right\| _{L^{2}} = \infty ,

where g~\widetilde{g} is the dual system generator from Theorem 11.31(b).

Remark It requires some work, but it can be shown that the Weak BLT implies Theorem 11.35; see [DJ93] or the survey paper [BHW95].

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Suppose ff is a real, continuously differentiable function on [a, b], f(a)=f(b)=0f(a) = f(b) = 0, and

∫abf2(x)dx=1 \int _{a}^{b} f^{2}(x) d x = 1

Prove that

∫abxf(x)f′(x)dx=−12 \int _{a}^{b} x f(x) f^{\prime }(x) d x = -\frac{1}{2}

and that

∫ab[f′(x)]2dx⋅∫abx2f2(x)dx>14 \int _{a}^{b}\left[f^{\prime }(x)\right]^{2} d x \cdot \int _{a}^{b} x^{2} f^{2}(x) d x > \frac{1}{4}
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Assume ff is a measurable function on (−∞,∞)(-\infty , \infty ) such that ∣f(x)∣<e−∣x∣\left|f(x)\right| < e^{-\left|x\right|} for all xx. Prove that its Fourier transform f^\hat{f} cannot have compact support, unless f(x)=0f(x) = 0 a.e.

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In the section we claimed that

σw2=∫−∞∞w2∣f^(w)∣2dw=∫−∞∞∣Df^(w)∣2dw=∥Df∥2 \sigma _{w}^{2}=\int _{-\infty }^{\infty } w^{2}|\hat{f}(w)|^{2} d w=\int _{-\infty }^{\infty }|\widehat{D f}(w)|^{2} d w=\left\| D f\right\| ^{2}

The point of this exercise is to verify the middle equality, and hence all three equalities. Note that

Df^(w)=12π∫−∞∞f′(t)e−iwtdt. \widehat{D f}(w)=\frac{1}{\sqrt{2 \pi }} \int _{-\infty }^{\infty } f^{\prime }(t) e^{-i w t} d t.

Use integration by parts to show that this equals iwf^(w)i w \hat{f}(w), and hence that the equation holds.

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Show that (6.8) (in the text of the section) is true whenever α2+β2≤1\alpha^{2}+\beta^{2} \leq 1.

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References

[Hei11] Heil C. A Basis Theory Primer. Birkhäuser, 2011.
[Rud76] Rudin W. Principles of Mathematical Analysis. 3rd ed., McGraw-Hill, 1976.
[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.
[Sax02] Saxe K. Beginning Functional Analysis. Springer, 2002.