Is there a 'certainty' principle?

Question

Heisenberg's uncertainty principle is a restriction on which probability distributions can describe the position and momentum of a quantum particle.

In mathematical terms it says that if $\psi\in L^2$ is normalized, and we define $f,g\in L^1$ by $f(x)=|\psi(x)|^2$ and $g(k)=|\hat\psi(k)|^2$ then we have $$V(f)V(g)\geq\frac14$$ where $V$ is the variance of the probability distribution with the given density function.

There are various other uncertainty principles, including the Entropic uncertainty principle and Hardy's uncertainty principle. Define $f,g\in L^1$ to be compatible if there exists $\psi\in L^2$ such that $f(x)=|\psi(x)|^2$ and $g(k)=|\hat{\psi}(k)|^2$. Then each uncertainty principle states a condition which compatible $f$ and $g$ must obey.

I noticed a curious fact, which holds true of everything I could find in the literature calling itself an 'uncertainty principle'. For fixed $f$ the restriction on $g$ is always a convex set. For example the set of $g$ satisfying $V(g)\geq\frac1{4V(f)}$ is convex because variance is a concave function on the space of probability distributions.

This does makes sense with the name 'uncertainty principle'. Intuitively, mixing probability distributions cannot produce a result that is more 'certain' than all of them.

However, playing with the Discrete Fourier Transform as a toy model, I noticed that the set of $g$ compatible with a given $f$ need not be convex.

Randomly sampled $g$ compatible with $f = (0.46,0.46,0.08)$ and $f = (0.46,0.46,0.07,0.01)$:

Note that these sets are nonconvex, and don't even contain the maximally-uncertain uniform distribution (the centre point of the simplex of possible distributions). So the uncertainty of distributions in these sets is bounded above as well as below.

In the case of distributions on $\mathbb R$, can we even find a single $f$ for which we can prove the set of compatible $g$ is not convex?

Is there a 'certainty principle' that, for $f$ within some class, puts an upper bound on the variance or entropy of compatible $g$?

Answer 1

By request, I add a comment as an answer with some additional details; but what I meant is really straightforward. The simplest realization is as follows: take any $\psi\in L^2$. Split its support into finitely many parts to obtain a representation $\psi=\sum_{k= 0}^N\psi_k$ where $\psi_0$ is small in $L^2$ (the infinite tail) and $\psi_k$ for $k>0$ are small (less than $\varepsilon$) in $L^1$ (short intervals). Now multiply each $\psi_k$ with $k>0$ by $e^{2\pi i Mkx}$ with $M$ chosen so that $\sup_{\lvert y\rvert>M,1\le k\le N}\lvert\widehat\psi_k(y)\rvert\le \frac{\varepsilon}N$ (it exists by Riemann–Lebesgue). Then the Fourier transform of the resulting function at any point $y$ will be bounded by $\lvert\widehat\psi_0(y)\rvert+3\varepsilon$. The first part doesn't influence anything because its $L^2$-norm is small and the rest is uniformly small and, therefore, spread wide.

If $\psi\in L^1\cap L^2$, then no special treatment of $\psi_0$ is needed. Also, you can get the true uniform smallness by splitting into countably many parts and choosing the phase shifts inductively instead of just using an arithmetic progression. And so on, and so forth.

Edit: Now about convexity. Take $f$ to be the characteristic function on $[0,1]$ and consider $g(k)$ where $k\in\mathbb Z$ (in this case the point values are continuous functionals). Clearly, every sequence with all zeroes and one $1$ is admissible ($\psi(x)=e^{2\pi ik_0x}$ on $[0,1]$). Thus, if the convexity had held, we would be able to construct a function on $[0,1]$ that is identically $1$ (or, at least, as close to that as we would like) such that $g(0)=g(1)=\frac 12$ and all other $g(k)=0$. However, that would be just a two-term polynomial with equal coefficients, so it would vary quite a bit in absolute value on $[0,1]$. This proves at least that sometimes convexity does not hold. I suspect that this trick can be generalized quite a bit but the details are elusive yet.

Answer 2

With $D_x=\frac{d}{i dx}$, the Heisenberg uncertainty principle in its most classical form can be deduced from the equality $$ 2\Re \langle \hbar D_x u, ix u \rangle_{L^2(\mathbb R)}= \langle \bigl[\hbar D_x, ix\bigr] u, u \rangle_{L^2(\mathbb R)}=\hbar\Vert u\Vert_{L^2(\mathbb R)}^2, $$ which implies $ \Vert \hbar D_x u\Vert_{L^2(\mathbb R)}\Vert xu\Vert_{L^2(\mathbb R)}\ge \frac\hbar 2\Vert u\Vert_{L^2(\mathbb R)}^2, $ where the constant $\hbar/2$ can be proven sharp by testing on a Gaussian function. So much for the lowerbound. Maybe a "certainty principle" would mean that we want to deal with the upperbound (?) We have $$ \Vert \hbar D_x u\Vert_{L^2(\mathbb R)}\Vert xu\Vert_{L^2(\mathbb R)}\ge \Re \langle \hbar D_x u, ix u \rangle_{L^2(\mathbb R)}= \frac\hbar 2\Vert u\Vert_{L^2(\mathbb R)}^2, $$ but it is true that the left-hand-side could be much larger than the rhs: take for instance with $\omega$ smooth, valued in $[0,1]$, equal to 1 for $\vert x\vert\ge 2$, to 0 on $\vert x\vert\le 1$, $\lambda \ge 1$, $$ u_\lambda(x)=(x^2+1)^{-1/2}\omega(x/\lambda),\quad \Vert u_\lambda\Vert_{L^2(\mathbb R)}^2 \le π, $$ $$ u'_\lambda(x)=-\underbrace{x(x^2+1)^{-3/2}\omega(x/\lambda)}_{\text{bounded in $L^2$}}+\underbrace{\frac1\lambda \omega'(x/\lambda) (x^2+1)^{-1/2}}_{\substack{ \text{with limit $0$ in $L^2$}\\\text{since $\omega'$ has support $[\lambda, 2\lambda]$} }}, $$ $$ x u_\lambda(x)=\frac{x}{\sqrt{x^2+1}} \omega(x/\lambda),\quad \Vert xu_\lambda(x)\Vert_{L^2(\mathbb R)}={+\infty}. $$ As a consequence, the upperbound is $+\infty$.

Answer 3

You might find the von Neumann-Koopman mechanics of interest. Here, classical mechanics is formulated in the same formal language of Diracs transformational theory which superseded both the Wave Mechanics of Schrodinger and the Matrix Mechanics of Heisenberg.

Observables, as in Quantum Mechanics, are represented by self-adjoint operators on the Hilbert space of KvN wave functions. However, unlike quantum mechanics, these operators commute and so are simultaneously measurable. This means that the uncertainty principle of Heisenberg disappears to be replaced by the usual deterministic laws of classical Newtonian mechanics - aka, a 'certainty principle'.