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Quantum mechanics on Cantor set?

+ 7 like - 0 dislike
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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
asked Aug 25, 2012 in Theoretical Physics by user10001 (635 points) [ no revision ]
retagged Mar 30, 2015

2 Answers

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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 Ron Maimon (7,535 points) [ no revision ]
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
+ 1 like - 0 dislike

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 Ryan Thorngren (1,605 points) [ no revision ]
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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