Can we quantize Aristotelian physics?

Aristotelian physics, shorn of whatever the historical Aristotle actually believed, is pretty similar to Newtonian physics. Instead of "An object in motion stays in motion unless acted on by an unbalanced force", we have "An object at rest stays at rest unless acted on by an unbalanced momentum." Newton's $$F~=~ma,$$ which is a second order differential equation, becomes $$p~=~mv,$$ which is a first order differential equation. Otherwise, we have business as usual.

My question is: Can we quantize this theory? Instead of constructing Hilbert space operators using the representation theory of the full Galilei group, we just use the representation theory of the Galilei group excluding Galilean transformations, i.e. just consisting of spatial translations, spatial rotations, and time translations. What would such a quantum theory look like? I can tell right off the bat there will probably be fewer conserved quantities, but not much more.

Any help would be greatly appreciated.

• Didn't you mean $F=mv$?
– user154997
Jul 5, 2017 at 7:20

I'm rewriting this answer after enjoyable discussion in chat with user23660. The following is what I took away from our discussion, and doesn't necessarily represent user23660's interpretation. The tl;dr answer is that I don't think there is any way of quantizing this system that offers nontrivial results while preserving standard QM -- we end up with either violations of the basic axioms of QM, or a trivial system with no dynamics and no measurement processes.

I'm only going to analyze the one-dimensional case. I think the important insight is gained by considering the zeroes of the function $p$, i.e., the points $x_0$ where $p(x_0)=0$. These are the places where particles can exist in equilibrium. They are also possible locations for collisions, if $p$ crosses from positive to negative there.

We have two logical possibilities: (1) The function $p(x)$ is always constrained to be a smooth function, and when it crosses zero, it never has a vanishing derivative. (2) No such constraint applies to $p$.

(1) In this case, we have $p\approx\alpha (x-x_0)$ for $x$ close enough to $x_0$, with $\alpha\ne0$. The interesting case, where $p$ crosses from positive to negative, is $\alpha<0$. The classical motion is an asymptotic approach to $x_0$ from either side. This means that the universe is split into two parts, each of which is unobservable from the other side. It is not possible for particles approaching $x_0$ from the two sides to collide with each other, since they would take infinite time to reach the point of collision. Since collisions are impossible, measurement processes are also impossible -- you can't get your measuring apparatus to touch the object that you're trying to observe. There is also no measurable notion of time, since the nonexistence of periodic motion makes it impossible to build a clock. Since clocks and rulers are impossible, we can arbitrarily rescale the coordinates in any way we wish. Since collisions are impossible, there are no dynamics at all.

As a way of simplifying the system, consider that we can do any smooth, one-to-one change of coordinates $x\rightarrow u(x)$, $p\rightarrow p^*=(du/dx)p$. Since there are no clocks and no observables, this change of coordinates is just a relabeling, and there is no way to tell whether $x$ or $u$ was the more natural coordinate system. In general, if we have any continuous function $p$, we can take any region where $p\ne0$ and define a "special" coordinate $u=\int dx/p$. The problem is transformed into one in which particles always move to the right with velocity 1.

User23660's answer analyzes how to quantize this system. The quantized system has no possibility of scattering, absorption, or emission, so there is no notion of measurement. Although I think user23660's analysis is correct, it seems to me that a theory of quantum mechanics without time, observables, or dynamics is not very interesting. As a shortcut in quantizing this system, we can switch to the special coordinates. Since in these coordinates the only possible velocity is 1, the solutions to the wave equation must be simply $\Psi_u=f(u-t)$, where $f$ is an arbitrary function. If we transform back to the $x$ coordinates, then $\Psi_u\rightarrow\Psi_x$ has to pick up an additional time-dependent normalization factor if want to preserve conservation of probability.

I originally oversimplified by saying that motion in case #1 was dissipative in the sense of Liouville's theorem. This is not quite right, since we can only express these notions in a space where we have measurement. Because the time and space coordinates are unmeasurable, we can always patch up conservation of probability simply by throwing in arbitrary time-dependent normalization factors.

(2) If there is no constraint on $p$, then in the classical system it's possible to have equilibrium points that can be reached by particles in finite time. We can conceivably have collisions and nontrivial dynamics. The motion may be dissipative in the sense that if a particle reaches an equilibrium point, all information about its past motion is lost. It should also be possible to have acausal solutions a la Norton's dome. Explicitly, if we let $p=|x|^{3/4}$, then the solution is either $x=0$ or anything of the form $x=\pm (4^{-4})(t-t_0)^4$, so for example, a particle could come in from the left, then sit at the origin for a while, and then at some undetermined time take off again to the right. So in general, particles might be able to collide, stay merged for some unpredictable amount of time, and then reemerge from the collision.

On quantization, we expect this nonconservation of information to show up as a violation of unitarity. This does seem to show up in user23660's analysis, in which the quantized system has real-valued wavefunctions. If the wavefunctions are real-valued, and we can have collisions, then we have to give up at least one of the following: unitarity, linearity, or the Born rule. For otherwise, a positive pulse colliding with a negative pulse would superpose and suffer destructive interference, violating unitarity. But anyway it's not clear that user23660's analysis really tells us anything about case #2, since the analysis makes assumptions about the behavior of $p$.

• Can you clarify what "loss of information" means, and why it implies nonunitarity? Oct 13, 2013 at 0:06
• @KeshavSrinivasan: Try these links: en.wikipedia.org/wiki/Unitarity_%28physics%29 motls.blogspot.com/2008/06/black-hole-information-puzzle.html
– user4552
Oct 13, 2013 at 0:45
• Unfortunately, those links don't really help. Here's my basic confusion: in ordinary quantumm mechanics, if a particle enters a region where there's no force acting on it, then it's quantum state will have no information about its prior acceleration. How is that different from a particle's quantum state having no information about its prior velocity if it enters a region where no momentum is acting on it? Oct 13, 2013 at 1:49
• @BenCrowell In Newtonian mechanics, we have a second-order differential equation, so current position and velocity suffice to tell you $x(t)$ for all past times $t$. In Aristotelian mechanics, we would be dealing with a first-order differential equation, so current position suffices to tell you $x(t)$ for all past times $t$. So what's the difference? Oct 13, 2013 at 4:09
• Going to higher dimensions/multiple particles relieves some of this triviality. For example, take $H=\omega\left(p_{1}q_{2}-p_{2}q_{1}\right)$. This generates simple harmonic motion for $q_{1,2}$ classically. Quantum mechanically, introducing polar coords $r^2=q_1^2+q_2^2,\tan(\phi)=q_2/q_1$, the general solution to the Schrodinger equation is $\psi(r,\phi,t)=f(r)g(\phi+\omega t)$ which is obviously periodic in time. Obviously I'm just taking the usual Hamiltonian system and reinterpreting the $p$s as $q$s, which is a bit of a cheat. A criticism may be that the force is nonlocal in the $q$s. Oct 18, 2013 at 8:28

We can quantize the aristotelian mechanics, and upon discussion Ben Crowell also seems to agree on this point.

Short summary: to quantize the system governed by Aristotelian mechanics we extend its configuration space with canonically conjugate momenta and then perform ordinary canonical quantization. The resulting quantum system exhibits somewhat boring behavior and depending on the choice of force field could has some serious issues.

One example of Aristotelian mechanics is the motion of object in very viscous fluid where we can neglect inertial forces: Stokes (or creeping) flow. Since the viscous drag force $\mathbf{F}_\text{drag}$ is proportional to the velocity of an object, the application of external force $\mathbf{F}_\text{ext}$ would rapidly (on a timescale determined by inertial forces: the smaller they are, the faster it happens) result in a motion where $\mathbf{F}_\text{ext}= - \mathbf{F}_\text{drag}$. This means that the resulting equation of motion is $$\frac {d\, \mathbf{x}}{dt} = \alpha\, \mathbf{F}_\text{ext}(\mathbf{x},t), \tag{1}$$ which is a first order differential equation with some positive constant $\alpha$.

This model, by the way, also establishes the relevance of aristotelian mechanics to such fields as micro- and nanofluidics.

So, the quantization of aristotelian mechanics could be done from within the framework of quantization of dissipative systems, which is a well established theory.

For instance, this paper

Tarasov, Vasily E. "Quantization of non-Hamiltonian and dissipative systems." Phys. Lett. A 288.3 (2001): 173-182. (arxiv:quant-ph/0311159)

gives the following summary of approaches to the problem:

We can divide the most frequent methods of quantization of dissipative and non-Hamiltonian systems into two groups. The first method uses a procedure of doubling of phase-space dimension [6]-[8]. The second method consists in using an explicitly time-dependent Hamiltonian [9]-[16].

Bateman has shown [6] that in order to use the usual canonical quantization methods a procedure of doubling of phase-space dimension is required. To apply the usual canonical quantization scheme to dissipative and non-Hamiltonian systems, one can double the numbers of degrees of freedom, so as to deal with an effective isolated system. The new degrees of freedom may be assumed to represent by collective degrees of freedom of the bath with absorb the energy dissipated by the dissipative system [7, 8].

The cited Bateman's 1931 paper

Bateman, Harry. "On dissipative systems and related variational principles." Physical Review 38.4 (1931): 815. (doi:10.1103/PhysRev.38.815)

also states:

In a recent paper P. S. Bauer stated that a linear dissipative set of differential equations with constant coefficients cannot be derived from a variational principle. This is only true if the variational principle is required to give no additional equations. Now a dissipative system is physically incomplete and so additional equations are to be expected when an attempt is made to derive the defining equations from a variational principle. We must look, then, for a complementary set of equations.

Since the equation (1) is of a first order, we could try to look for minimal extension of configuration space with a set of momenta variables. With the mind toward canonical quantization we try to find a Hamiltonian which would produce the equation (1) as one of the canonical equations.

It is surprisingly easy: the Hamiltonian required is: $$H(\mathbf{p},\mathbf{x}) = \alpha\, \mathbf{p} \cdot \mathbf{F}_\text{ext}(\mathbf{x}), \tag{2}$$ which would give equation (1) along with additional set of canonical equations for momenta.

So we could simply consider Hamiltonian (2) as corresponding to aristotelean mechanics and use a canonical quantization, which in the coordinate representation gives us the Schrödinger equation: $$i \hbar\frac{ \partial }{\partial t} \Psi = \mathrm{Sym}\,[H(-i\hbar\vec{\nabla}, \mathbf{x})] \Psi ,$$ Where $\mathrm{Sym}$ means that we must provide appropriate symmetrization to ensure that Hamiltonian is hermitian. Note, that since the Schrödinger equation contains no more than 1st derivatives in $x$ and $t$ it could be integrated through method of characteristics. Additionally, Schrödinger equation could be made fully real, so real function at initial moment $\Psi(\mathbf{x},0)$ will stay real at all times. Thus, the dynamics of any system would be much simpler than ordinary QM systems. Once we have solutions to Schrödinger equation we can compute various matrix elements for observables, bearing in mind that momenta here are actually auxiliary variables.

To illustrate, let us consider a couple of special cases of force field for 1D systems:

1. Constant force $F = b$. This gives us $\Psi = f(x-\alpha\, b\, t),$ that is, impulse propagating with constant speed in the $x$-direction. Note the absence of dispersion.

2. Elastic spring $F = - x$. Here, general solution would be $\Psi = f(y+\alpha t) \cdot \exp(\frac{\alpha \,t}2)$, where $y = \ln x$.

Even from these two simple examples we could conclude that the dynamics of these quantized systems is less interesting than ordinary QM. In particular, we see that there is no quantum interference: the double-slit experiment in aristotelian quantum mechanics would be very boring. This is, of course, determined by the underlying classical system: since there we have a simple flow on the configuration space, quantization won't add much.

Addition. Recently the question 'A “Hermitian” operator with imaginary eigenvalues' (and answer by Emilio Pisanty) highlighted the type of problems we could encounter in the quantum aristotelian mechanics. Specifically, the 'problem' hamiltonian there is of the type defined by equation (2). The problems were also mentioned in Ben Crowell's answer to this question, the most serious of which is non-conserved probability. Indeed, for some rather simple force fields we could not construct self-adjoint hamiltonian operator. But as the paper

Classical symptoms of quantum illnesses. Chengjun Zhu and John R. Klauder. Am. J. Phys. 61 no. 7, 605 (1993) doi:10.1119/1.17221.

showed, these problems have roots in the classical dynamics. So, in order to obtain reasonable quantum system, we must ensure good behavior of the underlying classical one. In particular, we might require that classical solutions exist for all values of time and no singularities are encountered at finite time. This might restrict allowable fields but results in correct quantum version.

• In your example of the elastic spring, you give the solution $\Psi = f(y+\alpha t) \cdot \exp(\alpha t/2)$. Since most choices of $f$ and $\alpha$ give $(d/dt)\int \Psi^2 \ne 0$, this is not compatible with the Born rule and conservation of probability. You've lost unitarity. You wrote in a comment: That we would need some generalization of ordinary QM was obvious from the start. So I think we're actually in agreement here. You need to generalize QM in some way so that nonunitarity no longer applies.
– user4552
Oct 13, 2013 at 14:04
• @BenCrowell: Wrong. We have to integrate over $x$ not $y$ (which is $\ln x$). So $\int\Psi^2dx=\text{const}$ at all times, which is as it should be, since it is solution of Schrödinger equation with hermitean Hamiltonian $H=\frac 12 (xp+px)$. Oct 13, 2013 at 14:14
• Suppose I take $f$ to be defined as $f(u)=1$ for $0 \le u \le 1$, $f(u)=0$ everywhere else. Let $\alpha=1$. Then I get $\int \Psi^2 dx=e^t(e^{1/t}-1)$, which isn't constant.
– user4552
Oct 13, 2013 at 14:38
• @BenCrowell: In your example function $\Psi$ is nonzero on the interval $x\in [\exp(-t),\exp(-t+1)]$ so you have some mistake there Oct 13, 2013 at 14:51
• I don't understand your most recent comment. Isn't $f$ meant to be an arbitrary function? I just made a choice of $f$. Actually there's a much simpler way to see that you don't have conservation of probability. You have real-valued wavefunctions, and therefore you can't have both linearity and unitarity. If you collide a positive pulse with a negative pulse, the total probability will go down when they superpose.
– user4552
Oct 13, 2013 at 14:57

Eugene Wigner discussed the quantization of Aristotelian physics in the very short paper Conservation Laws in Classical and Quantum Physics. You can quantize the theory, but there's no obvious way in which the quantized version is at all physically distinct from the classical version.