Abstract

We show that the classical dynamic system x’=x^3 has a non-deterministic time-evolution and we argue that for most real-world cases the so-called general solution with a singularity is not valid. We also show that time-independent quantum Hamiltonians are expressive enough, even in the presence of singularities and time-dependent differential equations at the cost of choosing a larger sample space.

General picture

Consider a sample space defined by the real line. Since we can always define a wave-function by taking the square-root of the probability density function, the Koopman-von Neumann version of classical statistical mechanics[1][2][3][4] defines classical statistical mechanics as a particular case of quantum mechanics where the algebra of observable operators is necessarily commutative (because the time-evolution is deterministic).

However, it is well known that for two unbounded self-adjoint operators, the fact that they commute does not imply that their corresponding spectral measures commute[5], then we would say that they strongly commute.

Thus in principle, it is possible to have a classical dynamical system defined by an ordinary differential equation where the unitary time-evolution is non-deterministic, that is, the algebra of observable operators is non-commutative.

Since the real line is a simply connected group (for addition), then by Bargmann’s theorem any strongly continuous projective representation of the real line is induced by some strongly continuous representation of the real line on the corresponding Hilbert space[6][7]. Then, Stone’s theorem implies that such representation on the Hilbert space is the exponential of a time-independent Hamiltonian[6][7]. Thus, if there is a solution in a standard probability space of a differential equation without singularities for all times (including negative times), then there is a quantum Hamiltonian corresponding to such solution which is time-independent. Moreover, if such solution is deterministic then the fact that there are more Hamiltonians that generate the same differential equation does not change the unique solution, up to initial conditions: suppose that $x(t)=U(t)x(0)U^\dagger(t)$ is a deterministic solution to the differential equation, then $x(t)=f(x(0),t)$ is a measurable function of $x(0)$ and it can be rewritten as $c(t)=U^\dagger_2(t)x(t)U_2(t)=U^\dagger_2(t)f(x(0),t)U_2(t)$. It verifies $\dot{c}=U^\dagger_2(t)[H-H_2,f(x(0),t)]U_2(t)=0$, because the difference between Hamiltonians must preserve the differential equation. Thus $x_2(t)=U_2(t)x(0)U^\dagger_2(t)=U_2(t)c(t)U^\dagger_2(t)=x(t)$. Then, $x(t)=U(t)x(0)U^\dagger(t)$ is the unique solution, up to initial conditions.

On the other hand, if there is no deterministic solution without singularities for all times (including negative times), then a solution without singularities must be defined by a self-adjoint Hamiltonian (by Bargmann’s and Stone’s theorems) and it must be non-deterministic. Moreover, since any solution in a standard probability space is given by a time-dependent self-adjoint Hamiltonian in a larger probability space, then the singularities in a standard probability space are consequence of a time-dependent Hamiltonian which is non-integrable. But since any time-dependent Hamiltonian can be converted into a time-independent Hamiltonian in a even larger sample space, then we can always resolve the singularities in a larger sample space which better defines the dynamical system. Thus, solutions without singularities for all times (including negative times) defined by time-independent Hamiltonians are expressive enough, even to deal with singularities at the cost of choosing a larger sample space.

Example

Consider the differential equation[8]: $\dot{x}=x^3$. The general deterministic solution is:

With the plus or minus sign chosen such that the solution is continuous at $t=0$. The general deterministic solution has a singularity at $t=\frac{1}{2(x(0))^2}$. There are no deterministic solutions without singularities[8].

On the other hand, the corresponding quantum Hamiltonian operator $H=\frac{1}{2}(p x^3+x^3 p)$ (such that the commutator $[x,p]=i$ and the domain is the dense set of smooth functions with compact support), can be proved to be essentially self-adjoint[9][10] (using $H^2$ as a positive auxiliary operator in Corollary 1.1 of [9]), meaning that the unitary time-evolution $U(t)$ is uniquely defined. Even if $x(t)=U(t)x(0)U^\dagger(t)$ would commute with $x(0)$, these are unbounded self-adjoint operators so the relevant question is whether or not they strongly commute for all times. If they would strongly commute for all times, then they could be simultaneously diagonalized, thus there would be a deterministic solution without singularities. This would be in contradiction with the fact that there are no deterministic solutions without singularities[8].

If we now assume that the deterministic solution Equation 1 exists for some $t>0$, then there is an isometry $T(t)$ such that $x(t)=T^\dagger(t)x(0)T(t)$ is a real number for all eigenvalues of $x(t)$, since the total probability must be conserved and the solution is deterministic. The eigenvalues such that $(x(0))^2>\frac{1}{2t}$ have no corresponding $x(t)$ which is a real number given by the isometry in agreement with the deterministic solution (ref?), because the deterministic solution becomes ill-defined for $t\geq\frac{1}{2 (x(0))^2}$. We can repeat the same reasoning for $t\geq 0$. Thus, we conclude that the deterministic solution Equation 1 is implemented by a non-unitary isometry when working with probability measures, for all $t\neq 0$. This is a time-dependent restriction on the allowed initial probability measures, for instance any initial Gaussian distribution is not allowed for any $t\neq 0$. The deterministic solution can only be implemented by a unitary operator at the cost of choosing a larger sample space, which we will do in the following.

We could think that a unitary time-evolution is somehow too demanding, that only with restrictions to the allowed initial probability measures we can find all solutions to the differential equation defined above. But as a matter of principle that should not be correct since we always work with such restrictions under the assumption that we will somehow be able to recover a corresponding result in the space of probability measures, otherwise such result cannot be tested experimentally or even in most numerical simulations (due to approximation uncertainties). Indeed, if we consider the space $L^2(\mathbb{R}^2)$ and the complex number $z(t)=x(t)+iy(t)$ satisfying $\dot{z}(t)=z^3(t)$ then the singularity condition $t\to\frac{1}{2 (x(0))^2}$ cannot be satisfied except in a set of null measure where $y(0)=0$, thus there is no singularity in $L^2(\mathbb{R}^2)$. But $y(0)$ can be as concentrated around zero as we want, so this unitary solution in a larger sample space recovers the corresponding non-unitary solution in the original sample space.

Now we can compare both unitary solutions in different sample spaces and choose the one we need.

For most real-world problems, we need $y(0)=0$ strictly. But this is not possible in $L^2(\mathbb{R}^2)$, so the general deterministic solution is not valid for most real-world problems and there are no singularities as expected.

Conjecture

We can easily see that for instance, all time-independent polynomial quantum Hamiltonians such that the domain is the dense set of smooth functions with compact support, can be proved to be essentially self-adjoint[9][10], using $H^2$ as a positive auxiliary operator in Corollary 1.1 of [9]. Thus, the conclusions apply to many more differential equations with would-be singularities and we can find non-deterministic solutions instead.