B Uncertainty principle
The Heisenberg-Gabor limit, is a direct result of the Heisenberg-Pauli-Weyl inequality [15].
Theorem B.1 (Heisenberg-Pauli-Weyl inequality) Let \(f\in L^2(\mathbb{R})\), then \(\forall a,b\in\mathbb{R}\) \[\begin{equation} \left(\int_\mathbb{R}(t-a)^2\left\lvert f(t)\right\rvert^2\mathrm{d}t\right)^{1/2} \left(\int_\mathbb{R}(\omega-b)^2\left\lvert\hat f(\omega)\right\rvert^2\mathrm{d}\omega\right)^{1/2} \geq \frac{\left\lVert f\right\rVert_2^2}{4\pi} \end{equation}\]
Proof. We aim to prove the uncertainty principle for self-adjoint operators on a Hilbert space \(H\) in the general case. We then introduce self-adjoint operators on a Hilbert subspace of \(L^2(\mathbb{R})\) giving rise to the uncertainty principle for realizable signals (the Heisenberg-Pauly-Weyl inequality).
Lemma B.1 Let \(A\) and \(B\) be two self-adjoint linear operators on a Hilbert space \(H\), whose commutator is defined as \[\begin{equation} [A,B]=AB-BA \end{equation}\] Then for all \(a,b\in\mathbb{R}\) and \(f\in H\). \[\begin{equation} \left\lVert(A-a)f\right\rVert\left\lVert(B-b)f\right\rVert\geq \frac{1}{2}\left\lvert\left\langle[A,B]f,f\right\rangle\right\rvert \end{equation}\]
Proof. We first notice that the commutator is invariant by translation, that is \(\forall a,b\in\mathbb{R}\) \[\begin{align*} [A-a,B-b] &= (A-a)(B-b) - (B-b)(A-a)\\ &= (AB-bA-aB+ab) - (BA-aB-bA+ab)\\ &= AB - BA\\ &= [A,B] \end{align*}\] We remind that an operator \(A\) is self-adjoint if \[\begin{equation} \forall f,g\in H, \left\langle Af,g\right\rangle = \left\langle f,Ag\right\rangle \end{equation}\]
We then deduce that \(\forall f\in H\) \[\begin{align*} \left\langle[A,B]f,f\right\rangle &= \left\langle[A-a,B-b]f,f\right\rangle \tag{translation invariance}\\ &= \left\langle((A-a)(B-b)-(B-b)(A-a))f,f\right\rangle \tag{bilinearity}\\ &= \left\langle(A-a)(B-b)f,f\right\rangle - \left\langle(B-b)(A-a)f,f\right\rangle\\ &= \left\langle(B-b)f,(A-a)f\right\rangle - \left\langle(A-a)f,(B-b)f\right\rangle \tag{self-adjointness}\\ &= \left\langle(B-b)f,(A-a)f\right\rangle - \overline{\left\langle(B-b)f,(A-a)f\right\rangle}\\ &= 2i~\mathrm{Im}\left\langle(B-b)f,(A-a)f\right\rangle \tag{$z-\bar z = 2i~\mathrm{Im}(z),\forall z\in\mathbb{C}$} \end{align*}\] Consequently, \[\begin{align*} \left\lvert\left\langle[A,B]f,f\right\rangle\right\rvert &= 2\left\lvert\mathrm{Im}\left\langle(B-b)f,(A-a)f\right\rangle\right\rvert\\ &\leq 2\left\lvert\left\langle(B-b)f,(A-a)f\right\rangle\right\rvert\\ &\leq 2\left\lVert(B-b)f\right\rVert\left\lVert(A-a)f\right\rVert \tag{Cauchy-Schwartz inequality} \end{align*}\]
We now consider the following operators \[\begin{equation} \begin{cases} Xf(t) = t\cdot f(t)\\ Pf(t) = \frac{1}{2\pi i}f'(t) \end{cases} \end{equation}\] defined over \(H=\left\{f\in L^2(\mathbb{R})\vert f', t\mapsto tf(t),t\mapsto tf'(t)\in L^2(\mathbb{R})\right\}\subset L^2(\mathbb{R})\).
We first prove that these operators are self-adjoint, let \(f,g\in H\) \[\begin{equation} \left\langle Xf,g\right\rangle =\int_\mathbb{R}tf(t)\overline{g(t)}\mathrm{d}t=\int_\mathbb{R}f(t)\overline{tg(t)}\mathrm{d}t= \left\langle f,Xg\right\rangle \end{equation}\]
Similarly, \[\begin{align*} \left\langle Pf,g\right\rangle &= \left\langle\frac{1}{2\pi i}f',g\right\rangle\\ &= \frac{1}{2\pi i}\int_\mathbb{R}f'(t)\overline{g(t)}\mathrm{d}t\\ &= -\frac{1}{2\pi i}\int_\mathbb{R}f(t)\overline{g'(t)}\mathrm{d}t\tag{integration by parts}\\ &= \int_\mathbb{R}f(t)\overline{\frac{1}{2\pi i}g'(t)}\mathrm{d}t\\ &= \left\langle f,\frac{1}{2\pi i}g'\right\rangle\\ &= \left\langle f,Pg\right\rangle \end{align*}\]
The operators are linear since multiplication and derivation are linear.
The commutator of \(X\) and \(P\) applied to \(f\in H\), \[\begin{align*} [X,P]f(t) &= XPf(t) - PXf(t)\\ &= X\left(\frac{1}{2\pi i}f'(t)\right) - P\left(tf(t)\right)\\ &= \frac{t}{2\pi i}f'(t) - \frac{1}{2\pi i}\left(f(t) + tf'(t)\right)\\ &= \frac{t}{2\pi i}f'(t) - \frac{1}{2\pi i}f(t) - \frac{t}{2\pi i}f'(t)\\ &= -\frac{1}{2\pi i}f(t) \end{align*}\]
Hence \(\forall f\in H\) \[\begin{equation} \left\langle[X,P]f,f\right\rangle = \left\langle-\frac{1}{2\pi i}f,f\right\rangle = \frac{i}{2\pi}\left\langle f,f\right\rangle = \frac{i}{2\pi}\left\lVert f\right\rVert_2^2 \end{equation}\]
Consequently \[\begin{equation}\label{eq:hilb_op} \left\lvert\left\langle[X,P]f,f\right\rangle\right\rvert = \frac{1}{2\pi}\left\lVert f\right\rVert_2^2 \end{equation}\]
As \(X\) and \(P\) are self-adjoint linear operators on the Hilbert space \(H\), then following the lemma we have \(\forall a,b\in\mathbb{R},\forall f\in H\).
\[\begin{equation} \left\lVert(X-a)f\right\rVert_2 \left\lVert(P-b)f\right\rVert_2 \geq \frac{1}{2}\left\lvert\left\langle[X,P]f,f\right\rangle\right\rvert \end{equation}\]
Moreover, from equation () we have \[\begin{equation} \left\lVert(X-a)f\right\rVert_2 \left\lVert(P-b)f\right\rVert_2 \geq \frac{1}{4\pi}\left\lVert f\right\rVert_2^2 \end{equation}\]
We develop the left-hand side \[\begin{equation}\label{eq:delta_t} \left\lVert(X-a)f\right\rVert_2 = \left(\int_\mathbb{R}\left\lvert(X-a)f(t)\right\rvert^2\mathrm{d}t\right)^{1/2} = \left(\int_\mathbb{R}(t-a)^2\left\lvert f(t)\right\rvert^2\mathrm{d}t\right)^{1/2} \end{equation}\]
For the second factor, we apply Parseval’s formula \(\left\lVert(P-b)f\right\rVert_2 = \left\lVert\mathcal{F}\left\{(P-b)f\right\}\right\rVert_2\).
\[\begin{align*} \mathcal{F}\left\{(P-b)f\right\}(\omega) &= \mathcal{F}\left\{\frac{1}{2\pi i}f' - bf\right\}(\omega)\\ &= \frac{1}{2\pi i}\hat{f'}(\omega) - b\hat{f}(\omega)\tag{linearity of $\mathcal{F}\{\cdot\}$}\\ &= \omega\hat{f}(\omega) - b\hat{f}(\omega)\tag{derivation of $\mathcal{F}\{\cdot\}$}\\ &= (\omega-b)\hat{f}(\omega) \end{align*}\]
Hence \[\begin{equation}\label{eq:delta_w} \left\lVert(P-b)f\right\rVert_2 = \left\lVert(\omega-b)\hat{f}\right\rVert_2 = \left(\int_\mathbb{R}(\omega-b)^2\left\lvert\hat{f}(\omega)\right\rvert^2\mathrm{d}\omega\right)^{1/2} \end{equation}\]
By injecting () and () in () we have \(\forall a,b\in\mathbb{R},\forall f\in H\) \[\begin{equation} \left(\int_\mathbb{R}(t-a)^2\left\lvert f(t)\right\rvert^2\mathrm{d}t\right)^{1/2} \left(\int_\mathbb{R}(\omega-b)^2\left\lvert\hat{f}(\omega)\right\rvert^2\mathrm{d}\omega\right)^{1/2} \geq \frac{1}{4\pi}\left\lVert f\right\rVert_2^2 \end{equation}\]