cosmology - How many bits are needed to simulate the universe?

This is not the same as: How many bytes can the observable universe store?

The Bekenstein bound tells us how many bits of data can be stored in a space. Using this value, we can determine the number of unique states this space can be in.

Imagine now, we are to simulate this space by enumerating each state along with which states it can transition to with a probability for each transition.

How much information is needed to encode the number of legal transitions and the probabilities? Does this question even make sense? If so, is there any indication that any of these probabilities are or are not Computable numbers?

Edit

Here's a thought experiment.

1. Select your piece of space and start recording all the different states you see.


2. If the Bekenstein bound tells us we can store n bits in our space, wait until you see 2^n different states. Now we've seen all the states our space can be in (otherwise we can violate the Bekenstein bound).


3. For any state, record any other state that the space can legally transition to without violating any physical laws.

To simulate this portion of space, take it's state and transition it to a legal state. Repeat.

We have only used a finite number of bits and we have modeled a section of space.

Answer

There's a huge difference between the number of bits you can store in a given space and the number of bits you need to describe that space.

Take a single atom of iron with its 26 electrons. For a complete description, you need the many-particle wavefunction $\psi(\vec{x}_1, \vec{x}_2, \vec{x}_3, \dots, \vec{x}_{26})$ (ignoring spin for the moment). Imagine you want to sample it in a given region of space with a very crude grid of 10 points for each direction, so you have $1000$ points in total.

This means you need $1000^{26} = 10^{3\cdot 26} = 10^{78}$ numbers to store it. For decent precision, you want to use at least $16$ bits, so you end up with approximately $10^{79}$ to $10^{80}$ bits. This is more than (or of the same order as) there are atoms in the entire universe.

Now taking it up from here, for a super-exact simulation of the universe you need the complete wavefunction of the entire universe, so replace the $26$ from the example above with something much higher, and of course you want it to be more precise, so replace the 1000 with something much higher, and then note that due to quantum field theory, the number of particles isn't even fixed, so a simple wavefunction isn't even enough... In a black body, for example, you can have an infinite number of photons. Although the probability for this decays exponentially, you'd still have to include it in an exact simulation...