# The (not so) paradoxical asymmetry between position and momentum

_(Cross-posted from [my blog](https://ejnnr.github.io/posts/position-momentum-asymmetry/) with light edits)_

Related to: [Which Basis is More Fundamental?](https://www.lesswrong.com/posts/XDkeuJTFjM9Y2x6v6/which-basis-is-more-fundamental)

_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\hbar \frac{\partial}{\partial t}\psi(x, t) = \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\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\hbar \frac{\partial}{\partial t}\psi(p, t) = \left(\frac{p^2}{2m} + V\left(i\hbar\frac{\partial}{\partial p}\right)\right)\psi(p, t)\,.\\]
The term \\(V\left(i\hbar\frac{\partial}{\partial p}\right)\\) probably deserves some explanation: we're assuming that \\(V\\) is given by a power series, then \\(V\left(i\hbar \frac{\partial}{\partial p}\right)\\)
is defined by plugging in \\(i \hbar \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](https://www.lesswrong.com/posts/XDkeuJTFjM9Y2x6v6/which-basis-is-more-fundamental?commentId=A5Pux22d5QKj58fXi)). 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(x) = f(x - a)\\) (see e.g. [this StackExchange post](https://math.stackexchange.com/questions/1341495/exponential-of-powers-of-the-derivative-operator/1495596)). \\(f(x - a)\\) depends
on the value of \\(f\\) outside a small enough neighborhood (if \\(a \neq 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 \hbar \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(\hat{x})\,.\\]
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, \ldots, n\\),
the Hamiltonian usually has the form
\\[H = \sum_{i = 1}^n H_i(x_i, p_i) + \sum_{i \neq 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.)_