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I am looking at doing something cool with these rings of LEDs sold by adafruit.

https://www.adafruit.com/products/1463

Initially, I thought that a cool thing to do would be to simulate three objects with different masses bouncing around the ring. The idea is that each mass has a position and momentum, if they get "close" they bounce off each other. After each delta of motion you apply a correction to keep the total momentum and energy constant. This way they should bounce realistically.

But it occurred to me that you could treat each of the 16 (or 24) LEDs as a quantum state. I imagine that each of my three objects would be modelled as a vector of 16 complex numbers, and we'd let the system of 48 numbers evolve over time - again, applying some sort of correction after each delta to keep the values reasonable.

As to how it would look, various things might be possible. For instance, for two of the masses we'd display a single bright white spot at the modal led, and for the other we'd display the modulus and argument of it's amplitude at each led as brightness and hue. So you'd see how the position of the particle gets smeared out over the 16 possible states.

Anyway - does this make any sense at all? My question comes down to - I have these three arrays of complex numbers for the 16 amplitudes for position of these objects, (maybe I need another set of amplitudes for the momenta) - what arithmetic do I do to make these bounce around in a way that's analagous to macroscopic objects bouncing off one another? In a ring?

Obviously - I'm expecting references to the information, not for someone to tell me the actual equations. Although to be frank, that would be nice.

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So you may have heard that quantum physics is computationally intractable. This post will (hopefully) explain why. Lets say that you have a single particle and it is in the state:

$|\psi\rangle = a|0\rangle_1+b|1\rangle_1$

where $aa^{*}+bb^{*}=1$.

For two particles the most general state is not:

$|\psi\rangle = (a_1|0\rangle_1+b_1|1\rangle_1)(a_2|0\rangle_2+b_2|1\rangle_2)$

it is:

$|\psi\rangle = a|0\rangle_2 |0\rangle_1 +b|0\rangle_2 |1\rangle_1 + c|1\rangle_2 |0\rangle_1 + d|1\rangle_2 |1\rangle_1$

with:

$aa^{*}+bb^{*} +cc^{*} +dd^{*}=1$.

for three particles you have:

$|\psi\rangle = a|0\rangle_3|0\rangle_2 |0\rangle_1 +b|0\rangle_3|0\rangle_2 |1\rangle_1 + c|0\rangle_3|1\rangle_2 |0\rangle_1 + d|0\rangle_3|1\rangle_2 |1\rangle_1 + e|1\rangle_3|0\rangle_2 |0\rangle_1 + f|1\rangle_3|0\rangle_2 |1\rangle_1 + g|1\rangle_3|1\rangle_2 |0\rangle_1 + h|1\rangle_3|1\rangle_2 |1\rangle_1$

and so on. For each particle you add, the number of states in yours system increases not by the number of states in that subsystem, but as the product of the number of states in the previous system and the number of states in the subsystem. This means that you would need $2^{16}$ states to describe $16$ quantum particles or $2^{24}$ to describe $24$. On top of this, this is just to set up the state! to actually evolve it forwards in time you would need to look at the time evolution generated by the Hamiltonian of the system. For a $16$ particle system, the Hamiltonian is $2^{1}6\times2^{16}=2^{32}$. So for only $16$ particles you are looking at $64$ million states and $16$ gigs of ram to store the Hamiltonian (Since it will in general be dense even for only nearest neighbor couplings).

So imagine that you are trying to compute wave-functions in 3D (like for a hydrogen atom in a non uniform magnetic field for example). You will need somewhere in the vicinity of a $100\times100\times100$ grid to adequately represent the wave-function of the electron. If you put in another electron you don't need 2 million grid points, you need 1 trillion. This is why perturbation theory is so important to quantum mechanics as opposed to fields like general relativity where you can just brute force the equations.

What would be possible to do though would be to do something "quantumy" like the 1D Ising model of magnetism. It wouldn't be terribly difficult to modify that into a real time display of annealing.

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  • $\begingroup$ Well, consider me a little deflated, but it's kind of what I was expecting I'll go back to plan A. :) $\endgroup$ Commented Dec 16, 2015 at 9:03

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