The (not so) paradoxical asymmetry between position and momentum

post by Erik Jenner (ejenner) · 2021-03-28T13:31:08.785Z · LW · GW · 10 comments

Contents

  False assumptions?
  Hamiltonian part I: Locality
  Hamiltonian part II: "Weak" locality
  Speculations
None
10 comments

(Cross-posted from my blog with light edits)

Related to: Which Basis is More Fundamental? [LW · GW]

Epistemic status: Not an expert on physics, please point out any mistakes.

In physics, there appears to be a deep duality between position and momentum, in the sense that one can switch between position-based and momentum-based views of the same system. In classical mechanics, exchanging position and momentum via $x\mapsto p$ and $p\mapsto -x$ is a canonical transformation, meaning that it leaves the dynamics unchanged. In quantum mechanics, the roles of position and momentum can be similarly switched by the Fourier transform.

So mathematically speaking, it would appear that there is nothing special about either position or momentum, both yield similar and equally good descriptions. And yet, human cognition treats position and momentum very differently, they don't feel like dual descriptions of reality. To us, there is a big difference between a car that is close to us and moving with a high relative velocity (distant in momentum space) and one that is far away and more or less stationary with respect to us.

But human cognition runs on brains, which run on physics, which seems to treat position and momentum equivalently. So how can this be? How does the cognitive asymmetry arise from what seems to be symmetry on the fundamental physical level?

The motivation for this post is mostly to point out the question. Below, I'll give my best guesses for the answer, though I'm not sure that those are correct.

False assumptions?

It could be that minds don't actually "run on physics", or that they exploit unknown physics where the symmetry between position and momentum breaks in new ways. But I think most readers here will find that unlikely, at least as long as we can find a simpler explanation, so I'm not going to discuss it further.

It's also possible that the symmetry just vanishes completely in a relativistic setting (this is where actually knowing QFT would come in handy). But even then, I would also expect an explanation on a classical level because I have the intuition that you could have minds in a classical universe that perceive position and momentum differently.

Hamiltonian part I: Locality

This explanation is specific to quantum mechanics. So if it turns out to be the reason for the asymmetry between position and momentum, this would mean that this feature of our cognition is inherently quantum mechanical and would not appear in a classical universe. As mentioned, this seems intuitively unlikely to me, but I think the explanation is still interesting.

The Schrödinger equation, which determines the time evolution of a system, can be written in terms of position as follows: $$i\hslash \frac{\partial }{\partial t}\psi (x,t)=(-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}+V(x\left)\right)\psi (x,t).$$ This time evolution is local in the following sense: to calculate $\frac{\partial }{\partial t}\psi (x,t)$, we only need to know the wave function $\psi$ in an arbitrarily small neighborhood of $x$ (so that we can calculate its second spatial derivative).

We can also write the Schrödinger equation in terms of momentum: $$i\hslash \frac{\partial }{\partial t}\psi (p,t)=(\frac{p^2}{2m}+V\left(i\hslash \frac{\partial }{\partial p}\right))\psi (p,t).$$ The term $V\left(i\hslash \frac{\partial }{\partial p}\right)$ probably deserves some explanation: we're assuming that $V$ is given by a power series, then $V\left(i\hslash \frac{\partial }{\partial p}\right)$ is defined by plugging in $i\hslash \frac{\partial }{\partial p}$ into that power series.

If $V$ happens to be a polynomial, this is just a sum of normal differential operators, and the time evolution is local in exactly the same sense as for position. But in general, $V$ can be an infinite power series, and we will take arbitrarily high derivatives of $\psi$. This means that locality can be violated -- this power series of derivatives may depend on points that are far away in momentum space (I saw this point made in this comment [LW(p) · GW(p)]). The most famous example for a power series of differential operators being non-local is probably the fact that $\exp \left(a\frac{\partial }{\partial x}\right)f\left(x\right)=f(x-a)$ (see e.g. this StackExchange post). $f(x-a)$ depends on the value of $f$ outside a small enough neighborhood (if $a\ne 0$), so in such cases, the time evolution in terms of position is not local in the sense described above.

This raises the question: where does the asymmetry between these two formulations of the Schrödinger equation come from? The answer is that it's all the Hamiltonian's fault. The Schrödinger equation can be written in basis independent form as $$i\hslash \frac{\partial }{\partial t}\psi =\hat{H}\psi ,$$ where $\hat{H}$ is the Hamiltonian operator. This Hamiltonian usually has the form $$\hat{H}=\frac{{\hat{p}}^2}{2m}+V\left(\hat{x}\right).$$ So the asymmetry on the level of the Hamiltonian is that the momentum operator appears as a second power, whereas the position operator is plugged into the potential, which may be an infinite power series.

In the position basis, $\hat{p}$ turns into a derivative whereas in the momentum basis, $\hat{x}$ becomes a derivative. This leads to our observation that time evolution is local in the position formulation in a sense that does not hold for momentum.

Hamiltonian part II: "Weak" locality

In the previous section, we considered only a single particle (though the same asymmetry applies to multiple particles -- having only a single particle is the weaker assumption). If we have multiple interacting particles, we get a different sense of locality that doesn't require QM anymore.

In the beginning, I mentioned the difference in our cognition between a distant stationary car and a nearby car that's moving fast. It's very reasonable that we think about these situations differently: if a car is very far away, it can't interact with us, i.e. hit us. The same is not true for momentum: if a car is moving very fast, it can still hit us, even though it is far away in momentum space.

We might call the fact that spatially distant objects tend to interact less "weak locality". "Weak" because they can still interact, just typically not as much. So position satisfies weak locality while momentum apparently doesn't.

The reason for that can again be found in the Hamiltonian. For multiple particles $i=1,\dots ,n$, the Hamiltonian usually has the form $$H=\sum _{i=1}^n{H}_i(x_i,p_i)+\sum _{i\ne j}V(|x_i-x_j|).$$ Here, $x_i,p_i$ are the position and momentum of particle $i$. $H_i$ is the Hamiltonian for a single particle, which only depends on the position and momentum of that particle. This includes the kinetic energy and any potentials that are not caused by particle interactions.

The second sum in the Hamiltonian describes the interactions between particles. The way I wrote it, it can model any pairwise interaction that depends only on the distance between particles. It so happens that for the forces that actually occur in our universe, the interaction potential $V$ diminishes as the distance between the interacting particles increases. This is what leads to the weak locality in position space. Since the interaction does not depend on the momenta of the particles, there is no analogous weak locality for momentum.

As in the previous section, the asymmetry again boils down to the Hamiltonian being asymmetric in position and momentum. This fits rather well with my own intuition. For example, the Hamiltonian of a harmonic oscillator is completely symmetric with respect to position and momentum, and they really do seem much "more equivalent" there than in other systems.

Speculations

All of this raises the question of why the Hamiltonian has such an asymmetric form. Classical mechanics or QM themselves don't have an answer; after all, symmetric Hamiltonians such as the harmonic oscillator work completely fine in principle, it's just that our universe isn't a harmonic oscillator.

I don't know whether QFT can shed light on this question, otherwise, maybe theories of quantum gravity can. This would likely mean a more fundamental difference between position and momentum, which looks very different from what I've described, but which leads to the asymmetry in the Hamiltonian (in the non-relativistic limit).

Another approach is to say that most possible Hamiltonians aren't symmetric in position and momentum, so it's not surprising at all that ours isn't. This doesn't feel quite as satisfying and whether you buy into that argument at all depends on how you think about the "probability" of physical laws being a certain way. In a similar vein, one could appeal to the anthropic principle: we can only observe Hamiltonians that permit observers to exist in the universe they describe. A harmonic oscillator is presumably too simple for that and maybe the same is true for any Hamiltonian that treats position and momentum exactly equivalently.

Feedback appreciated! (including regarding style, grammar, etc.)

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