# Quantum mechanics on Cantor set?

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Has quantum mechanics been studied on highly singular and/or discrete spaces? The particular space that I have in mind is (usual) Cantor set. What is the right way to formulate QM of a particle on a Cantor set?

I can only guess that:

i) There will be no momentum operator.

ii) Hilbert space will be spanned by position vectors corresponding to points of Cantor set.

iii) Corresponding to any two points $x$ and $y$ in Cantor set there will be a unitary operator $P_{xy}$ (analog of the exponential of momentum operator) such that $P_{xy}|x>=|y>$; $P_{xy}=P_{yx}^{-1}$; and $P_{yz}P_{xy}=P_{xz}$ for all $x,y,z$.

But I am not able to see what would be correct notion of free particle, and what would be corresponding Hamiltonian.

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user user10001
retagged Mar 30, 2015

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You can define quantum mechanics on a cantor set, but in order for it to be nontrivial, it needs to be a Levy quantum mechanics, not a Gaussian quantum mechanics, in that it will be the quantum analog of a Levy process, not a Brownian motion, as the ordinary Schrodinger equation is.

To define Schrodinger quantum mechanics, you take the continuum limit of a nearest neighbor amplitude random walk. To do this, I will first remind you of the standard imaginary time map between random walks and quantum mechanical systems. When you have a stochastic process on a discrete space in discrete time, you have a transition operator:

$$\rho_j(t+1) = \sum_i \rho_i(t) K_{i\rightarrow j}$$

where $K_{i\rightarrow j} = K_{ij}$ is a stochastic matrix:

$$\sum_j K_{ij} = 1$$

These stochastic matrices generically have a stationary distribution, which I will call $\rho^0$. I will assume that this stationary distribution obeys detailed balance, or in mathemtical jargon, that it is the "reversing measure" for K:

$$\rho^0_i K_{ij} = \rho^0_j K_{ji}$$

This says that the transitions between states i and j balance in equilibrium separately from any other transitions. The stationary distribution for random walk on a graph obeys detailed balance and it is ${1\over D(i)}$ where D is the degree of the vertex.

When you take the continuous time limit, you make K equal to the identity plus an infinitesimal transition rate, and the stochastic equation becomes:

$${d\over dt} \rho_j = \sum_i \rho_i R_{ij}$$

And you still have a stationary distribution $\rho0$ for the continuous time case. Now you can define a symmetric H from the continuous time random process:

$$H_{ij} = {1\over \sqrt{\rho^0_i}} R_{ij} \sqrt{\rho^0_j}$$

and the detailed balance condition gives you symmetry of H. You then can define the imaginary time continuation as a standard quantum mechanical unitary time evolution, generated by this Hamiltonian. This is the most abstract form of Wick continuation.

If you do this process on a random walk whose limit is a Brownian motion, you get ordinary Schrodinger quantum mechanics. If you do the same process on a random walk which takes steps of size s according to a distribution:

$$P(s) \propto {1\over s^{1+\alpha}}$$

Where $0<\alpha<2$, you get Levy quantum mechanics.

So to define quantum mechanics on a Cantor set, all you need is an appropriate stochastic motion. The ordinary Brownian motion fails to have a limit, it just stays still on the Cantor set--- it ends up fully localized. But the Levy process generalizes just fine.

The Cantor set can be defined as all base 3 numbers with digits which are all 0 or 2. A discrete approximation is truncating this at N digits. Define a random walk on this graph by toggling a digit between 0 to 2 at digit position k with a rate which goes as:

$$e^{-ak}$$

where $a>0$. If you take the limit of continuous time, timesteps of size $\epsilon$, and $a= {A\over \epsilon}$, you get a hop which is a power law in size (since it is an exponetial distribution on exponentially shrinking sizes, and this is a powerlaw in the size), and the continuum limit is Levy quantum mechanics restricted to the Cantor set.

This is related to the question of localizing Dirac Fermions, since the |k| dispersion relation is Levy. You don't localize Levy particles with a local potential, unlike normal Schrodinger particles. This was the subject of this question: How to localize the massless fermions in Dirac materials? .

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user Ron Maimon
answered Aug 25, 2012 by (7,720 points)
Hi Ron, thanks for your answer. But i don't understand why usual QM would be trivial in this case. I was thinking that since each point of Cantor set can be represented as an infinite sequence of (say) -1's and 1's; so position states are of the form $|i_1,i_2,...>$ where each $i_k$ is either -1 or 1. Now these states can be mapped to states of a half infinite chain of quantum spins. So i think problem of QM of a particle on Cantor set can equivalently be looked at as QM of an infinite chain of quantum spins. Is that correct ?

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user user10001
@dushya: yes, but the normal Schrodinger quantum dynamics does not allow the spins to flip, since the distances are not uniformly small--- flipping each successive spin is a factor of 3 larger distance jump. So if you want a nontrivial dynamics, you need powerlaw jumps, hence the Levy flights.

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user Ron Maimon
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It is difficult to devise any classical system on the Cantor set (as a target space). For example, the set is totally disconnected, so any continuous map from a connected space into it is constant. Therefore, there can be no propagating objects (even just particles).

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user Ryan Thorngren
answered Aug 25, 2012 by (1,925 points)
Right, but there's no problem doing quantum mechanics on finite, discrete Hilbert spaces, so the lack of an ability to consider a classical particle needn't necessarily be a problem.

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user Nathaniel
That's not the problem dushya is considering, though.

This post imported from StackExchange Physics at 2015-03-30 13:26 (UTC), posted by SE-user Ryan Thorngren

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