On the proof of Heisenberg's uncertainty principle

Question

Theorem 1 For $f \in L^2(\mathbb R)$ and $a,b \in \mathbb R$ $$ \frac{1}{2} \| f \|_2^2 \le \left( \int_{\mathbb R} (x - a)^2 | f(x) |^2 d x\right)^\frac{1}{2} \left( \int_{\mathbb R} (\xi - b)^2 | \hat{f}(\xi) |^2 d\xi \right)^\frac{1}{2}. $$ holds.

In our lecture we proved theorem 1 using the following theorem

Theorem 2 For self-adjoint (possibly unbounded) operators $S,T$ on a Hilbert space $H$ and $a,b \in \mathbb R$ $$ \| (S - a I) f \| \| (T - b I) \| \ge \frac{1}{2} | \langle [S,T] f, f \rangle | $$ holds for all $f \in \text{dom}(ST) \cap \text{dom}(TS)$, where $[S,T] := S T - T S$ is the commutator of $S$ and $T$.

Proof. Define $(S f)(x) := x f(x)$ for $f \in L^2(\mathbb R^n)$ and $(T f)(x) := i f'(x)$ for differentiable $f \in L^2(\mathbb R^n)$. For $f \in \text{dom}(ST) \cap \text{dom}(TS)$ we have \begin{align*} \tag{1} ([S,T] f)(x) & = i x f'(x) = i \frac{d}{dx} ( x \cdot f(x)) \\ & = i x f'(x) - i f(x) - i x f'(x) = - i f(x) \end{align*} and by theorem 2 $$ \tag{2} \frac{1}{2} \| f \|_2^2 = \frac{1}{2} | \langle - i f(x), f(x) \rangle | \le \| (S - a I) f \|_2 \| (T - b I) f \|_2. $$ By the Plancherel theorem we have $$ \tag{3} \| (T - b I) f \|_2 = \| \mathcal{F}((T - b I) f) \|_2 = \| (\xi - b) \hat{f} \|_2, $$ which yields the statement. $\square$

My Questions

1. Into what spaces do $S$ and $T$ map? Is it $L^2(\mathbb{R}^n)$?

2. Why can we use the Plancherel theorem? I have tried to calculate \begin{align*} \| (T - b I) f \|_2^2 & = \int_{\mathbb{R}^n} | i f'(x) - b f(x) |^2 dx \\ & = \int_{\mathbb{R}^n} | f'(x) |^2 - i b \overline{f(x)} f'(x) + i b f(x) \overline{f'(x)} + | b f(x) |^2 dx \\ & = \int_{\mathbb{R}^n} | f'(x) |^2 dx - i b \int_{\mathbb{R}^n} \frac{d}{dx} | f(x) |^2 dx + | b |^2 \| f \|_2^2 \\ & = \int_{\mathbb{R}^n} | f'(x) |^2 dx + | b |^2 \| f \|_2^2 - i b \bigg[| f(x) |^2\bigg]_{x = - \infty}^{\infty} \end{align*} Is this correct? Can we conclude this is finite?

3. How do we deal with the $\int_{\mathbb{R}} | f'(x) |^2 dx$ term?

Under the suitable assumptions (in another proof of theorem 1, where $f \in L^2(\mathbb{R})$ we used this) we can say \begin{equation*} \int_{\mathbb R^n} | f'(x) |^2 dx = \int_{\mathbb R^n} | \mathcal{F}(f')(x) |^2 dx = \int_{\mathbb R^n} x^2 | \hat{f}(x) |^2 dx, \end{equation*} but for this we would need that $f' \in L^2$. Our lecture assistant conjectured that we need to require $f' \in L^2$. How can we show this is necessary?

4. I know that $\big[| f(x) |^2\big]_{x = - \infty}^{\infty}$ only makes sense for $n = 1$. How can we generalise it? Can we conclude that it vanishes as $|f(x)| \xrightarrow{x \to \pm \infty} 0$ because $f \in L^2(\mathbb{R}^n)$?

Answer 1

@1: Yes, $S$ and $T$ are (unbounded) operators defined on suitable subspaces of $L^2$ going into $L^2$. You probably want either the Schwartz space or $H^1$ / its fourier transform respectively as domain for $T$ and $S$.

@3: If one of the integrals on the RHS of theorem 1 is infinite, there is really nothing to prove. If the other integral is zero, then $f=0$ and the inequaltiy is true. If the other integral is non-zero, then the RHS is $+\infty$ and the inequality is also true. Therefore you can assume right away and w.l.o.g. that both integrals are finite, i.e. not only $f\in L^2$, but $(x-a)f(x) \in L^2$ and $(\xi-b)\hat{f}(\xi)\in L^2$ as well so that $x f(x)\in L^2$ and $\xi \hat{f}(\xi)\in L^2$. Now since Fourier transform exchanges multiplication by $x$ with differentiation (up to some $\pm i$), this means that $f'(x)\in L^2$ as well. One has to be a bit careful here because this is only a weak derivative, but that doesn't change anything relevant.

@2: And this is the reason you can apply Plancherel's theorem: You assume wlog that all the relevant functions are in $L^2$.

Answer 2

I'm not able to answer this entirely or rigorously but I think I can share some thoughts about it

1:My guess is they go from $L^2$ to $L^2$, a priori for $T$ for example it doesn't holds but I suppose one can consider some kind of smooth space $L^2(\Omega)$ where it does

2: I don't get how the proof follows from using Plancherel but I believe it should be possible to obtain the inequality without using it.

\begin{align} \|(S-aI)f\|\|(T-bI)f\| & \overset{\text{CS}} \ge \left|\langle(S-aI)f,(T-bI)f\rangle \right|\\ & =\left[\Re(\langle(S-aI)f,(T-bI)f\rangle)^2+\Im(\langle(S-aI)f,(T-bI)f\rangle)^2\right]^{\frac{1}{2}}\\ & \ge \left|\Im(\langle(S-aI)f,(T-bI)f\rangle)\right|\\ & = \left|\frac{1}{2i}\left[ \langle(S-aI)f,(T-bI)f\rangle-\overline{\langle(S-aI)f,(T-bI)f\rangle} \right]\right|\\ & = \left|\frac{1}{2i}\left[\langle(S-aI)f,(T-bI)f\rangle-\langle(T-bI)f,(S-aI)f\rangle\right]\right|\\ & = \left|\frac{1}{2i}\left[(ST-aT-bS-abI)f^2-(TS-bS-aT-baI)f^2\right]\right|\\ & = \left|\frac{1}{2i}\left[(ST-TS)f^2\right]\right|\\ & = \left|\frac{1}{2i}\left[\langle(ST-TS)f,f\rangle\right]\right|\\ & = \left|\frac{1}{2i}\left[\langle[TS]f,f\rangle\right]\right|\\ & = \left|\frac{1}{2i}\left[\langle-if,f\rangle\right]\right|\\ & = \frac{1}{2}|\langle f,f\rangle|\\ & = \frac{1}{2}\|f\|\\ \end{align}

3: I think this question is contained in the two previous, you wrote that $\int_{\mathbb R^n} |f'(x) |^2 dx= \int_{\mathbb R^n} x^2 | \hat{f}(x) |^2 dx$ but I think you meant $\int_{\mathbb R^n} |f'(x) |^2 dx= \int_{\mathbb R^n} x^2 | \hat{f}(\xi) |^2 d\xi$ which is just what you had before $\|(T-bI)f\|_2=\|(\xi-b)\hat{f}\|_2,$ with $b=0$ from question 2 and as $Tf=f'$ then if $f'\in L^2$ should come from question 1

4: Not really sure but I believe this case wouldn't be interesting because that would mean inequality isn't bounding anything as $$ \|(S-aI)f\|\|(T-bI)f\| \ge\frac{1}{2}\|f\| =0 \quad (\text{when } f\to 0) $$

5: I don't get the question entirely but I think the case $a,b=0$ shouldn't be interesting either, if $\|(S-aI)f\|_2=0$ then $Sf=af$ that is we could think of $a$ as an eigenvalue of $S$ so in general we would be more interested in $a,b\neq 0$, also we could interpret the inequality as saying that you can't have an $a$ and $b$ being eigenvalues of $T$ and $S$ simultaneously as that would mean $$ 0=\|(S-aI)f\|\|(T-bI)f\|\ge\frac{1}{2}\|f\|\quad (\text{when } Sf=af \text{ and } Tf=bf) $$

Answer 3

FOURIER ANALYSIS

For $p\in [1,\infty)$ the space $L^p(\textbf{R}^n)$ is the space of countable functions which satisfy $$ ||f||_p:=\left(\int_{\textbf{R}^n}|f(x)|^pdx\right)^{1/p}<\infty $$ The space $L^2(\textbf{R}^n)$ have inner product $$ (f,g):=\int_{\textbf{R}^n}f(x)\overline{g(x)}dx $$ The space $L^{\infty}(\textbf{R}^n)$ is the set of all bounded in all $\textbf{R}^n$ functions $f$ (except for a posiblly set of mesure $0$) and have metric $$ ||f||_{\infty}:=\textrm{inf}\{\lambda\in\textbf{R}:\mu\{|f(x)|>\lambda\}=0\} $$ For $p\geq 1$ every $L^p$ is Banach

If $f\in L^1(\textbf{R}^n)$, we define the Fourier transform $\widehat{f}$ of $f$ as $$ \widehat{f}(\xi)=\int_{\textbf{R}^n}e^{-2\pi i (x,\xi)}f(x)dx\textrm{, }\xi\in \textbf{R}^n, $$ where $$ (x,\xi):=x_1\xi_1+x_2\xi_2+\ldots+x_{n}\xi_n $$ and $$ ||x||:=\sqrt{x_1^2+x^2+\ldots+x_n^2} $$ If $f\in L^2(\textbf{R}^n)$, then $\widehat{f}\in L^2(\textbf{R}^n)$ and $$ f(x)=\int_{\textbf{R}^n}e^{2\pi i (x,\xi)}\widehat{f}(\xi)d\xi\textrm{, }x\in\textbf{R}^n. $$ Theorem 1. If $f\in L^1(\textbf{R}^n)$, then i) $\widehat{f}$ is bounded and $$ ||\widehat{f}||_{\infty}\leq ||f||_1 $$ ii) $\widehat{f}$ is uniformly continuous.

Theorem 2. (Riemann-Lebesgue) If $f\in L^1(\textbf{R}^n)$, then $\widehat{f}(\xi)\rightarrow 0$, when $||\xi||\rightarrow\infty$.

The Schwartz space $S(\textbf{R}^n)$ is the space in which every element $f$ of it, is $C^{\infty}(\textbf{R}^n)$ (infinite times differentiatable) and every partial derivative of $f$ tends to $0$ more quickly than any polynomial i.e. for all $m_1,m_2,\ldots,m_n,N$ and $R>0$ exists positive constant $c=c(m_1,m_2,\ldots,m_n,N,R)$ such that $$ \left|\frac{\partial^{m_1+m_2+\ldots+m_n}}{\partial x_1^{m_1}\partial x_2^{m_2}\ldots\partial x_n^{m_n}}f(x)\right|\leq \frac{c}{(1+||x||^2)^N}\textrm{, }\forall x\in\textbf{R}^n. $$

$S(\textbf{R}^n)$ is dense in $L^p$, $1\leq p<\infty$

Proposition 1. If $f\in S(\textbf{R}^n)$, then $\widehat{f}\in S(\textbf{R}^n)$.

Also $$ \frac{\partial \widehat{f}}{\partial\xi_j}(\xi)=-2\pi i\int_{\textbf{R}^n}e^{-(x,\xi)}x_jf(x)dx $$ Hence with integration by parts we get $$ \widehat{\partial_jf}(\xi)=2\pi i \xi_j\widehat{f}(\xi) $$ Some usefull results are: If $f,g\in S(\textbf{R}^n)$, then $$ f(x)(-2\pi i x_j)^a\leftrightarrow\frac{\partial^a\widehat{f}}{\partial \xi^a_j} $$ and $$ \frac{\partial^af}{\partial x_j^a}\leftrightarrow (2\pi i\xi_j)^a\widehat{f}(\xi) $$ $$ \int_{\textbf{R}^n}f(x)\widehat{g}(x)dx=\int_{\textbf{R}^n}\widehat{f}(x)g(x)dx $$ $$ (f*g)(x):=\int_{\textbf{R}^n}f(x-y)g(y)dy $$ $$ (f,g)=(\widehat{f},\widehat{g}) $$ $$ \widehat{(f*g)}(\xi)=\widehat{f}(\xi)\widehat{g}(\xi) $$ Theorem.(Plancherel) Also if $f\in L^2$, then $\widehat{f}\in L^2$ and $$ ||f||_2^2=||\widehat{f}||_2^2 $$ Theorem.(Parseval) If $f,g\in L^2$, then $$ (f,g)=(\widehat{f},\widehat{g}) $$

HERMITIAN OPERATORS

Deffinition. An operator $\textbf{A}$ of a Hilbert space $H$, will called self-adjoint or Hermitian if $$ (\textbf{A}f,f)=(f,\textbf{A}f)\textrm{, }\forall f\in H $$ Definition. The characteristic values (eigenvalues) of $\textbf{A}$, are all $\lambda$ such $$ \textbf{A}f=\lambda f. $$ Characteristic elements (eigenvectors) are called all such $f$.

Theorem. A Hermitian operator have characteristic values real. The characteristic elements are orthogonal and $(\textbf{A}f,f)$ is real for every $f$. Also $$ |(\textbf{A}f,f)|\leq ||\textbf{A}||\cdot ||f||^2 $$

Theorem. If $\textbf{A}f\neq 0$ when $f\neq0$ and $\textbf{A}$ is Hermitian in a Hilbert space $H$, then the set of all $f_k$ eigenvectors are complete orthogonal base i.e. $$ (f_k,f_l)=0\textrm{, if }k\neq l $$
and every $g\in H$ have unique expansion $$ g=\sum_k(g,f_k)f_k $$ Here we omit our selfs to a descrete spectrum $\lambda_k$.

Definition. If $\textbf{A}$ is an operator of a Hilbert space $H$, then we set $$ (\Delta \textbf{A}):=\left\langle \textbf{A}^2\right\rangle-\left\langle \textbf{A}\right\rangle^2, $$ where we have set (the mean value): $$ \left\langle \textbf{A}\right\rangle f:=(\textbf{A}f,f) $$ Theorem.(Heisenberg's uncertainty principle) If $\textbf{A}$ and $\textbf{B}$ are Hermitian and $\textbf{AB}\neq \textbf{BA}$, then $$ (\Delta \textbf{A})(\Delta \textbf{B})\geq\frac{1}{2}\left|\left\langle[\textbf{A},\textbf{B}]\right\rangle\right|, $$ where $$ [\textbf{A},\textbf{B}]=\textbf{AB}-\textbf{BA}. $$

Proof. We can set in the proof that $$ \left\langle \textbf{A}\right\rangle=\left\langle \textbf{B}\right\rangle=0\tag 1 $$ If not this dont hold, then we can set $$ \widetilde{\textbf{A}}=\textbf{A}-\left\langle \textbf{A}\right\rangle $$ and $$ \widetilde{\textbf{B}}=\textbf{B}-\left\langle \textbf{B}\right\rangle. $$ Then $$ \left\langle \widetilde{\textbf{A}}\right\rangle=\left\langle \widetilde{\textbf{B}}\right\rangle=0 $$ Hence when (1) holds, then $$ (\Delta \textbf{A})^2=\left\langle \textbf{A}^2\right\rangle=(\textbf{A}^2f,f)=(\textbf{A}f,\textbf{A}f)=||\textbf{A}f||^2 $$ and $$ (\Delta \textbf{B})^2=\left\langle \textbf{B}^2\right\rangle=(\textbf{B}^2f,f)=(\textbf{B}f,\textbf{B}f)=||\textbf{B}f||^2 $$ Hence $$ \Delta \textbf{A}=||\textbf{A}f||\textrm{, }\Delta \textbf{B}=||\textbf{B}f|| $$ Hence from Schwartz inequality $$ (\Delta \textbf{A}) (\Delta \textbf{B})=||\textbf{A}f||\cdot ||\textbf{B}f||\geq |(\textbf{A}f,\textbf{B}f)|=|(f,\textbf{AB}f)|=|\left\langle\textbf{A}\textbf{B}\right\rangle|\tag 2 $$ But because $\textbf{AB}\neq \textbf{BA}$ we have $\textbf{C}=\textbf{AB}$ is not Hermitian. Hence $$ Re\left\langle \textbf{C}\right\rangle=\left\langle\frac{\textbf{C}+\textbf{C}^{+}}{2}\right\rangle\textrm{, }Im\left\langle \textbf{C}\right\rangle=\left\langle\frac{\textbf{C}-\textbf{C}^{+}}{2i}\right\rangle, $$ where $\textbf{C}^{+}=(\textbf{AB})^{+}=\textbf{B}^{+}\textbf{A}^{+}=\textbf{BA}$ is the adjoint of $\textbf{C}$. Hence $$ Re\left\langle\textbf{C}\right\rangle=\left\langle\frac{\textbf{AB}+\textbf{BA}}{2}\right\rangle\textrm{, }Im\left\langle\textbf{C}\right\rangle=\left\langle\frac{\textbf{AB}-\textbf{BA}}{2i}\right\rangle=\frac{1}{2i}\left\langle[\textbf{A},\textbf{B}]\right\rangle $$ But $$ |\left\langle \textbf{AB}\right\rangle|\geq |Im\left\langle\textbf{AB}\right\rangle|=\frac{1}{2}|\left\langle[\textbf{A},\textbf{B}]\right\rangle |\tag 3 $$ From $(2)$, $(3)$ we get the result. QED

APPLICATION. If $\textbf{S},\textbf{T}$ are Hermitian operators (with $\textbf{ST}\neq \textbf{TS}$), then we set $$ \textbf{A}f=(\textbf{S}-\textbf{I}a)f\textrm{, }\textbf{B}f=(\textbf{T}-\textbf{I}b)f\textrm{, }x\in\textbf{R}. $$ Then also $\textbf{A},\textbf{B}$ are Hermitian and $$ [\textbf{A},\textbf{B}]=(\textbf{S}-a \textbf{I})(\textbf{T}-b\textbf{I})-(\textbf{T}-b \textbf{I})(\textbf{S}-a\textbf{I})= $$ $$ =\textbf{ST}-b\textbf{S}-a\textbf{T}+ab\textbf{I}-(\textbf{TS}-a\textbf{T}-b\textbf{S}+ab\textbf{I})=[\textbf{S},\textbf{T}] $$ Hence we get from Schwartz inequality $$ ||\textbf{A}f||\cdot ||\textbf{B}f||\geq |(\textbf{A}f,\textbf{B}f)|=|(f,\textbf{AB} f)|=|\left\langle \textbf{AB}\right\rangle|\geq \frac{1}{2}|\left\langle[\textbf{A},\textbf{B}]\right\rangle|\tag 4 $$ Inequality (4) is your Theorem 2. Hence setting $\textbf{S}f(x)=xf(x)$ and $\textbf{T}=\frac{1}{2\pi i}\frac{d}{dx}$, we have from Placherel formula $$ ||\textbf{T}f-bf(x)||_2=||\frac{1}{2\pi i}f'(x)-bf(x)||_2= $$ $$ =\left\|\frac{1}{2\pi i}\int_{\textbf{R}}f'(t)e^{-2\pi i t x}dt-b\int_{\textbf{R}}f(t)e^{-2\pi i t x}dt\right\|_2= $$ $$ =\left\|-\frac{1}{2\pi i}\int_{\textbf{R}}f(t)(-2\pi i x)e^{-2\pi i t x}dt-b\int_{\textbf{R}}f(t)e^{-2\pi i t x}\right\|_2= $$ $$ =||(x-b)\widehat{f}||_2. $$ Also $$ ||\textbf{S}f(x)-af(x)||_2=||(x-a)f(x)||_2 $$ But $$ [\textbf{S},\textbf{T}]f=\textbf{ST}f(x)-\textbf{TS}f(x)=x\frac{1}{2\pi i}f'(x)-\frac{1}{2\pi i}\frac{d}{dx}(xf(x))= $$ $$ =\frac{x}{2\pi i}f'(x)-\frac{x}{2\pi i}f'(x)-\frac{1}{2\pi i}f(x)=-\frac{1}{2\pi i}f(x) $$ Hence $$ |\left\langle [\textbf{S},\textbf{T}]\right\rangle f|=\frac{1}{2\pi}||f||_2 $$ and therefore from (4) (when $f'\in L^2$): $$ \frac{1}{4\pi}||f||_2\leq ||(x-a)f(x)||_2\cdot ||(\xi-b)\widehat{f}(\xi)||_2. $$ QED

NOTES. Actualy $$ (\textbf{T}f,f)=\frac{1}{2\pi i}\left(f',f\right)=\frac{1}{2\pi i}[|f|^2]^{+\infty}_{-\infty}-\frac{1}{2\pi i}(f,f')=(f,\textbf{T}f) $$ and $\textbf{T}$ is Hermitian when $f'\in L^2$.