The Vitali Paradox of Measure

STEM fields are often interested in measuring regions. For example, an astrophysicist might be interested in measuring the volume of a star. The concept of measure is intuitive, and some of its features are self-evident. For example,

(i) if a region can be divided into countably many nonoverlapping subregions, then the measure of the region should be equal to the sum of the measures of the subregions;

(ii) if two regions are identical up to rotations, translations, and reflections, then their measures should be equal; and

(iii) the measure of a square (cube, hypercube, etc.) with side $1$ should be $1$.

We could begin to formalize this concept by attempting to measure all subsets of the set of real numbers $\mathbb{R}$, but it turns out that this is impossible, as I will now demonstrate.

Let us say that two real numbers are equivalent if their difference is rational. Construct a subset $N$ of the half-open interval $[0,1)$ in $\mathbb{R}$ such that no two distinct elements are equivalent (this is possible by the axiom of choice), and define

\begin{equation}
    N_r = \{x+r : x \in N \cap [0,1-r)\} \cup \{x+r-1 : x \in N \cap [1-r,1)\},
\end{equation}

where $r\in\mathbb Q\cap[0,1)$.

It follows that all $x \in [0,1)$ belong to exactly one $N_r$, namely $r = x-y$ or $r = x-y+1$, where $y \in N$ such that $x$ is equivalent to $y$. Otherwise, if $x \in N_r \cap N_s$, there exist $y, z \in N$ such that $x$ is equivalent to both $y$ and $z$, implying by transitivity that $y$ is equivalent to $z$, which is a contradiction. Thus, the $N_r$ are disjoint.

Then, by (i) and (ii), the measures of $N$ and $N_r$ are equal for all $r$. Moreover, by (iii) and (i),

\begin{equation}
    1=\mu\left(\left[0,1\right)\right)=\mu\left(\bigcup_rN_r\right)=\sum_r\mu\left(N_r\right)=\sum_r\mu\left(N\right),
\end{equation}

where $\mu(N)$ is the measure of $N$. However, the rightmost sum is either $0$ or $\infty$, which is a contradiction.

This problem is a consequence of our ambition to assign a measure to all of $2^{\mathbb R}$. Nevertheless, we can circumvent it by focusing on a particular subset of $2^{\mathbb R}$ rather than the whole, and this is much of the subject of measure theory.

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The Ehrenfest Theorem

While exploring the undergraduate Física Cuántica I course offered by the Pontificia Universidad Católica de Chile, I found a homework question asking to prove the momentum part of the theorem of Ehrenfest.

Although I know little to no quantum physics, I was curious whether, as a mathematician, I could understand and complete the proof.

I learned that the momentum part of the theorem of Ehrenfest relates the time derivative of the expectation value of the momentum $p$ to the expectation value of the force $F=-V'(x)$ on a massive particle moving in a scalar potential $V(x)$.

Mathematically,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\langle V'(x)\rangle.$$In simpler terms, as I understand it, this means that the average momentum of a quantum particle changes over time much like that of a classical particle under the average force from the potential (cf. second Newton law in momentum form); it connects quantum behavior with classical physics, showing that quantum averages obey familiar classical laws.

I also learned that this observation is often misused, e.g., by overlooking the assumptions on $V$, which is why I emphasize the importance of truly understanding the underlying principles rather than merely applying formulas.

I did not know how $\langle\cdot\rangle$ is defined in this context, so I looked it up: if $A$ is an operator such that $A\psi\in L^2(\mathbb R)$, where $\psi$ is the wavefunction of the particle, then$$\langle A\rangle=\underbrace{\langle \psi, A\psi \rangle}_{\text{inner product}}=\int_{\mathbb R}\psi^*(x,t)A\psi(x,t)\,\mathrm{d}x.$$In simpler terms, $\langle A \rangle$ is the average value one would get by measuring $A$ on a system described by $\psi$, weighted by the probabilities in $\psi$.

We actually study $L^p$ spaces in mathematics quite thoroughly since they are prototypical Banach spaces of Lebesgue-measurable function classes with $p$-integrable absolute value. In particular, $L^2$ is also a Hilbert space, so this quantum-mechanical setting is really nice.

By the way, a wavefunction $\psi(x,t)$ describes the state of a quantum particle: for each fixed time $t$, it is a function of position in $L^2(\mathbb R)$, meaning it is square-integrable. Moreover, physical wavefunctions are normalized so that total probability equals $1$, and we assume sufficient decay at infinity so we end up with nice equations. 😄

In order to proceed with the proof, one must know both the Schrödinger equation$$i\hbar\partial_t\psi(x,t)=\left(-\frac{\hbar^2}{2m}\partial_x^2+V(x)\right)\psi(x,t)$$and the definition of the momentum operator $p=-i\hbar\partial_x$.

I am not a physicist, so I am not concerned with how these were derived, but they seem like facts every physicist should know by heart.

Anyway, we proceed with the proof; from here I will abuse notation modestly to avoid clutter.$$\begin{align}\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle&= \frac{\mathrm{d}}{\mathrm{d}t}\int \psi^*p\psi\\&= \int \partial_t\psi^*\,p\psi + \int \psi^*p\,\partial_t\psi\\&= \frac{i}{\hbar}\int (T+V)\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*p\,(T+V)\psi\\&= \underbrace{\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi}_{\text{potential energy}}+ \underbrace{\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi}_{\text{kinetic energy}}.\end{align}$$Here, $T=-(\hbar^2/2m)\partial_x^2$. This is often called the kinetic-energy operator.

We now tackle the potential-energy term:$$\begin{align}\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi&= \frac{i}{\hbar}\int V\psi^*(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_xV\psi)\\&= \int V\psi^*\partial_x\psi - \int \psi^*\partial_xV\psi\\&= \int V\psi^*\partial_x\psi - \int \psi^*(V'\psi+V\partial_x\psi)\\&= -\int V'|\psi|^2.\end{align}$$We now tackle the kinetic-energy term:$$\begin{align}\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi&= \frac{i}{\hbar}\int \Big(-\frac{\hbar^2}{2m}\partial_x^2\psi^*\Big)(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_x)\Big(-\frac{\hbar^2}{2m}\partial_x^2\psi\Big)\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \psi^*\partial_x^3\psi\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi\\&= 0.\end{align}$$Note that we integrated the rightmost term above by parts twice, still assuming sufficient decay so that the boundary terms vanish.

Finally,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\int V'|\psi|^2+0=-\langle V'(x)\rangle,$$which is what we wanted. This result extends analogously to $\psi(\cdot,t)\in L^2(\mathbb R^n)$.

 

This exercise humbled me. The mathematics of physics is not easy: it draws on a great deal of experimentally-derived knowledge, despite being modeled with elementary, yet nontrivial, tools like finite-dimensional Hilbert spaces.

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Munkres §54: The Fundamental Group of the Circle

Munkres §53: Covering Spaces ►

Even though it has been four years since I left academia, I still miss solving math problems. They always leave my brain completely fried. 🫠 Math is fairly non-intuitive.

Topological spaces are fascinating mathematical objects. What is most fascinating is that their definition is simple: If a set $X$ and a subset $\tau$ of its power set $2^X$ are such that $\varnothing, X \in \tau$, arbitrary unions of elements of $\tau$ are elements of $\tau$, and finite intersections of elements of $\tau$ are elements of $\tau$, then $(X, \tau)$ is a topological space. A variety of other objects emerge from this definition, and one of them is the homeomorphism, which is a continuous bijection from one space to another with a continuous inverse. If two spaces are homeomorphic, then we consider them to be the same space, but finding homeomorphisms is hard. We created a new object—called the fundamental group—to represent a property of topological spaces in terms of algebraic groups, which enriches our study of them, and in this post, we take a look at the fundamental group of the circle $S^1$. However, in order to do so, we must first take a look at a handful of new definitions and lemmas.

Definition. If $p: E \to B$ is continuous and surjective, $U \subset B$ is open, and $p^{-1}(U)$ is the union of disjoint, open sets $V_\alpha \overset{p}{\simeq} U$, then $U$ is said to be evenly covered by $p$.

Definition. If $p: E \to B$ is continuous and surjective, and for all $b \in B$, there exists an open neighborhood $U$ of $b$ that is evenly covered by $p$, then $p$ is said to be a covering map.

Definition. If $p: E \to B$, and $f: X \to B$, then $\tilde{f}: X \to E$ such that $p \circ \tilde{f} = f$ is said to be a lifting of $f$.

Lemma 1. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $f$ is a path in $B$ beginning at $b_0$, then there exists a unique continuous lifting of $f$ beginning at $e_0$.

Proof. We spare the reader the pain of this intricate proof. $$\tag*{$\blacksquare$}$$

Lemma 2. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $F: I \times I \to B$ is continuous such that $F(0,0) = b_0$, then there exists a unique continuous lifting of $F$ such that $\tilde{F}(0,0) = e_0$. If $F$ is a path homotopy, then $\tilde{F}$ is a path homotopy.

Proof. We once again spare the reader the pain of this intricate proof. $$\tag*{$\blacksquare$}$$

Lemma 3. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $f$ and $g$ are path homotopic paths in $B$ beginning at $b_0$, then the unique continuous liftings of $f$ and $g$ beginning at $e_0$ are path homotopic.

Proof. If $F$ is the path homotopy between $f$ and $g$, then $F(0,0) = b_0$. Therefore—by Lemma 2—the unique continuous lifting of $F$ such that $\tilde{F}(0,0) = e_0$ is a path homotopy. Note that $\tilde{F}(x,0)$ is a lifting of $F(x,0) = f(x)$. Hence, $\tilde{F}(x,0) = \tilde{f}(x)$ since $\tilde{f}$ is unique. Similarly, $\tilde{F}(x,1) = \tilde{g}(x)$. $$\tag*{$\blacksquare$}$$

Definition. If $p: E \to B$ is a covering map, and $p(e_0) = b_0$, then $\phi: \pi_1(B, b_0) \to p^{-1}(b_0)$ defined by $[f] \mapsto \tilde{f}(1)$, where $\tilde{f}$ is the unique continuous lifting of $f$ beginning at $e_0$, is said to be the lifting correspondence derived from $p$. Note that $\phi$ depends on $e_0$ and is well-defined—by Lemmas 1 and 3.

Lemma 4. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $E$ is path connected, then the lifting correspondence $\phi$ derived from $p$ is surjective. If $E$ is simply connected, then $\phi$ is bijective.

Proof. If $E$ is path connected and $e_1 \in p^{-1}(b_0)$, then there exists a path $\tilde{f}$ in $E$ beginning at $e_0$ and ending at $e_1$. Therefore, $p \circ \tilde{f} = f$ is a loop in $B$ at $b_0$, and $\phi([f]) = \tilde{f}(1) = e_1$.

If $E$ is simply connected, then suppose that $\phi([f]) = \phi([g])$ and $[f] \neq [g]$. If $\tilde{f}$ and $\tilde{g}$ are the unique continuous liftings of $f$ and $g$ beginning at $e_0$, then $\tilde{f}(1) = \tilde{g}(1)$. Therefore, there exists a path homotopy $\tilde{F}$ between $\tilde{f}$ and $\tilde{g}$, and $p \circ \tilde{F}$ is a path homotopy between $f$ and $g$, which is a contradiction. $$\tag*{$\blacksquare$}$$

Lemma 5. $p: \mathbb{R} \to S^1$ defined by $x \mapsto (\cos 2\pi x, \sin 2\pi x)$ is a covering map.

Proof. $p$ is clearly continuous and surjective. $U = \{(x, y) \in S^1 : x > 0\}$ is open. $p^{-1}(U)$ is the union of disjoint intervals $V_n = (n - 1/4, n + 1/4)$. $p|_{\bar{V}_n}$ is bijective onto $\bar{U}$ since it is clearly surjective and $\sin 2\pi x$ is strictly monotonic. $\bar{V}_n \overset{p}{\simeq} \bar{U}$ since $\bar{V}_n$ is compact and $\bar{U}$ is Hausdorff. Therefore, $V_n \overset{p}{\simeq} U$. Repeat this argument with the remaining $90^\circ$ rotations of $U$. $$\tag*{$\blacksquare$}$$

Theorem. The fundamental group of $S^1$ is isomorphic to the additive group of integers.

Proof. If $p$ is the covering map of Lemma 5, $e_0 = 0$, and $b_0 = p(e_0)$, then $p^{-1}(b_0) = \mathbb{Z}$, and—by Lemma 4—the lifting correspondence $\phi: \pi_1(S^1, b_0) \to \mathbb{Z}$ derived from $p$ is bijective since $\mathbb{R}$ is simply connected.

If $\tilde{f}$ and $\tilde{g}$ are the unique continuous liftings of $f$ and $g$ beginning at $e_0$, $\tilde{f}(1) = n$, and $\tilde{g}(1) = m$, then $\tilde{\tilde{g}} = n + \tilde{g}$ is a lifting of $g$ beginning at $n$ since $p(n + x) = p(x)$. Moreover, $\tilde{f} * \tilde{\tilde{g}}$ is the lifting of $f * g$ beginning at $e_0$, and $\tilde{\tilde{g}}(1) = n + m$. Therefore, $\phi([f] * [g]) = n + m = \phi([f]) + \phi([g])$.$$\tag*{$\blacksquare$}$$

Claim. If $p:E\to B$ is a covering map, $\alpha$ and $\beta$ are paths in $B$ such that $\alpha(1)=\beta(0)$, $\tilde\alpha$ and $\tilde\beta$ are liftings of $\alpha$ and $\beta$ such that $\tilde\alpha(1)=\tilde\beta(0)$, then $\tilde\alpha*\tilde\beta$ is a lifting of $\alpha*\beta$.

Proof.

$$\begin{align}p\circ\tilde\alpha*\tilde\beta(x)&=\begin{cases}p\circ\tilde\alpha(2x)=\alpha(2x)&0\leq x\leq1/2\\p\circ\tilde\beta(2x-1)=\beta(2x-1)&1/2<x\leq1\end{cases}\\&=\alpha*\beta(x).\tag*{$\blacksquare$}\end{align}$$

Claim. If $p:E\to B$ is a covering map, $E$ is path connected, and $B$ is simply connected, then $p$ is a homeomorphism.

Proof. The lifting correspondence $\phi: \pi_1(B, b_0) \to p^{-1}(b_0)$ derived from $p$ is surjective—by Lemma 4—since $E$ is path connected, and its domain is trivial since $B$ is simply connected. Therefore, $p$ is injective.

If $V\subset E$ is open and $b\in p(V)$, then there exists an open neighborhood $U$ of $b$ that is evenly covered by $p$. If, $e\in V$, $p(e)=b$, and $V_\alpha$ is the slice containing $e$, then $p(V\cap V_\alpha)\subset p(V)$ is open since $p|_{V_\alpha}$ is a homeomorphism. Therefore, $p(V)$ is a union of open sets. Hence, $p$ is open.$$\tag*{$\blacksquare$}$$

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