Quantum physics today tells us that we may not predict outcomes with certainty. We perceive events as random.

So is this old philosophical fantasy of an all knowing computer possible? Meaning that all particules at a time T would have a determined state.

If it is, and assuming that there is going to be some randomization to go from T to T+1, how precise of an estimation could we get of the future?

  • can a weather forecast for the next day be 100% reliable?
  • can we answer questions like "if hitler had died as a kid" with 100% reliability?
  • how much further could we predict from T, with good accuracy, knowing that there is randomness?
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    $\begingroup$ Sorry, I cannot agree with your first sentence. QT does not say actual randomness exists, only what we may not predict outcomes with certainty. We perceive events as random. Further, mathematics tells us that an all knowing computer is impossible. A computational universe evolves from state to state by applying its computational rules and like all rules in our science, these rules are time-symmetric. This means it can go from state T to T+1 or conversely from state T+1 to state T. $\endgroup$ – Nick Oct 26 '14 at 20:47
  • $\begingroup$ @NickR edited the question. $\endgroup$ – Sheraff Oct 26 '14 at 20:49
  • $\begingroup$ Thanks. I didn't want to sound to nit-picky. +1 since no one has done so thus far and it is a good question. You may wish to note that it is Godel's Incompleteness Theorem that says no computer can be all knowing. $\endgroup$ – Nick Oct 26 '14 at 20:56
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    $\begingroup$ I can't help thinking the all-knowing computer would end up in an infinite loop. Let's say the computer knows the state of the universe right now, it would have to factor its own state into the calculation, but of course that would change its own state, so now it has to factor the changed state of itself into the calculation which would change its state again, etc. $\endgroup$ – colmde Jul 26 '18 at 8:58
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    $\begingroup$ First of all, any such a computer would be required to be outside of our universe, otherwise it would need to compute itself computing itself computing itself etc. Second, it can produce a complete wave vector solution, but that would only be providing exact probabilities of all the possible outcomes. No outcome is ever 100% guaranteed. $\endgroup$ – Alice Jul 26 '18 at 13:24

10 Answers 10


Maybe. Maybe not. People disagree about determinism.

However, you don't need quantum mechanics to see that such a computer is not going to work. Suppose you've constructed a computer that can store $n$ bits of memory. Then you'll need those $n$ bits to track their own state. You've got no memory left to track the rest of the universe. If you use any of the bits to store the state of the universe, you become unaware of your own state.

By choosing a suitable approximation you might end up having reasonably good information about everything. At this point, chaos theory comes into the picture. That's a big topic with some interesting results. I don't recall the answers to your specific questions off the top of my head, but the general answer is that even without randomness, there are things you're just not going to be able to predict given finite information.

A little more detail about the compression issues.

Suppose you live in a very technologically advanced world and you build computers for a living. At some point, a physicist comes to a great breakthrough: the world can only be in one of $2^n$ possible states. That's the number of ways the atoms (or quarks, or fundamental Lego pieces...) in the world can be arranged.

You immediately decide to build a computer that can store which of these $2^n$ states you are in and then compute the next state you'll end up in. To distinguish between $2^n$ states you need $n$ bits of information. Any less, and you'll end up being unsure which state you are in. Thus, knowing exactly how much memory you need, you go ahead and build it.

Suppose you've managed this task. At least, you've managed to build your $n$ bits of memory and they work: you've checked you can store all the necessary values. That means that your computer's memory alone can take on $2^n$ states. If the physicist was right and the amount of memory you made was enough, the state of your computer's memory completely determines the state of the rest of the universe. You're not going to be able to calculate much with that.)

(Interestingly, if this computer exists, it must exist in all $2^n$ states, meaning it already existed when you started building it.)

  • $\begingroup$ your first paragraph seems to ignore compression. Can't a raw data of n fit into m (m*<*n) if it has redundancy? And if all bits describe the universe, then it's 2 times the exact same data :) $\endgroup$ – Sheraff Oct 26 '14 at 19:12
  • $\begingroup$ Note however that even with deterministic interpretations of quantum mechanics, the full information is not available to us, not for practical, but for fundamental reasons. So we could not predict a quantum event even if it were 100% determined. $\endgroup$ – celtschk Oct 26 '14 at 19:19
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    $\begingroup$ @Sheraff: It wouldn't work. If you can represent the state of the universe in $n$ bits, then the universe has at most $2^n$ states. If you can construct a computer with $n$ bits of memory, the universe has at least $2^n$ states, plus however much is necessary to perform computation. Those boundaries are contradictory. $\endgroup$ – Anton Golov Oct 26 '14 at 19:19
  • $\begingroup$ I feel like I disagree but I have no argument right now. Do you have some reference? $\endgroup$ – Sheraff Oct 26 '14 at 19:26
  • $\begingroup$ @Sheraff: See en.wikipedia.org/wiki/Lossless_compression#Limitations for some limitations of compression. I'll edit my answer to be more thorough, though. $\endgroup$ – Anton Golov Oct 26 '14 at 19:44

Given a sufficiently detailed model of the universe, and a sufficiently fine-grained time step, you could get pretty close from a macroscopic point-of-view view. However, once you look at small enough scales you're going to have a problem...

The Heisenberg Uncertainty Principle forbids knowing the exact state of the universe at any given time. This is not simply because observing a particle causes it to change (although that is part of the problem). Instead, it is a fundamental property that arises from the wave-like nature of matter under quantum physics. The most well-known (but not the only) formulation of the uncertainty principle says that the product of the uncertainty in position and the uncertainty in momentum must be greater than a certain threshold. The more precisely one is known, the less precisely the other is known.

It may seem like these small changes are insignificant, but chaos theory demonstrates that seemingly insignificant differences can potentially lead to vastly different results. For example, a small change in the velocity of an Earth-bound spacecraft could lead to it skipping off the atmosphere to be lost in space, or burning up in the same.

  • $\begingroup$ Was hoping to see a Uncertainty principal answer for this question +1 $\endgroup$ – Twelfth Oct 27 '14 at 18:51
  • $\begingroup$ Good reason for why we can never measure the state of T but not for forbidding the knowing of it. If it was compact enough to know (on say a sheet of paper) the Uncertainty Principle doesn't prevent you from knowing it, just measuring it (and hence validating with 100% accuracy). Good overview of the fundamental problems though +1. $\endgroup$ – Black Nov 3 '14 at 19:04
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    $\begingroup$ As I understand it, the UP is not about measuring state (this is a common misconception). It says that there is, fundamentally, no determinate state to measure. States are inherently indeterminate, or "fuzzy". It has been experimentally shown that this is true, even when the effect of the measurement falls below the uncertainty threshold: scientificamerican.com/article/… $\endgroup$ – Caleb Hines Nov 3 '14 at 21:23
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  • $\begingroup$ The Heisenberg uncertainty principle, in combination with the mass and electrical interactions of particles (especially electrons) determines the volume taken up by those particles. Without the uncertainty, our atoms' volumes cease to exist. $\endgroup$ – Jasper Jul 26 '18 at 14:35

So is this old philosophical fantasy of an all knowing computer possible?


Information theory dictates that this is not possible. If you accumulate up all of the bits of information that define the universe at a given instant of time, all of the states of those particles you get some giant number N.

How are you going to store that information?

Information theory says (in short) that to store a universe's worth of particle state, you need a universe worth of particles to store it. A particle already represents the entirety of its information (position, composition, velocity, angular momentum, etc, etc). The only way to store that same information in less storage is to lose information via encoding (assuming that the nuclear arrangement of atoms are identical to others of the same element for example).

So ignoring the whole 'can you predict state based on existing state' question - you can't even get the whole existing state in the first place.

  • $\begingroup$ I'd like to point out that while theory says you can't store the universe's particle state without at least as many bits that theory isn't Information Theory. In fact that particular discipline might hold out the olive branch of compression to store your universe worths of data in reasonable space (if there's enough information redundancy). (Example being antimatter twins of regular matter, they're always exactly opposite). $\endgroup$ – Black Nov 3 '14 at 19:09

Yes and No

Quantum mechanics is full of probability. You simply can't get around it. You can never know for sure if a particle at time $t$ in the future will be at point $A$ or point $B$. Quantum mechanics forbids it. Sure, the odds could be essentially 100-to-1 against that it will be at point $B$, but there's still a chance that it could be there. So there will always be randomization.

However, there's a scale where this doesn't matter a lot, and that's on the macroscopic scale. Yes, there's a chance that a baseball hurtling towards you will move an inch to the left for a split second, but that would require every single particle in the ball to move one inch to the left. Once you combine all those wavefunctions, you find that that is really improbable.

So if you want to make a large scale simulation, you can sort of ignore that. That's one reason we didn't really think about probability being extremely important in physics until we could work with tiny particles, because the effects didn't show up on a macroscopic scale. So while you can't predict the minute details of each particle's motion, you can model the ball fairly well.

  • $\begingroup$ I think you certainly cannot always ignore quantum mechanics at macroscopic scale. There are systems that exponentially amplify arbitrarily small errors until they become large. It is called a butterfly effect and it is exhibited for example in weather forecasting. Sure, some systems are quite deterministic (falling apples), but definitely not almost all of them. $\endgroup$ – Irigi Nov 3 '14 at 16:40
  • $\begingroup$ @Irigi True, but the butterfly effect generally doesn't apply to single particles. The consequences arise from a small pseudo-chaotic effect, such as a small breeze. That breeze is an area of different pressure going through an area of many particles. You'd need a lot of quantum effects to account for that. $\endgroup$ – HDE 226868 Nov 3 '14 at 16:42
  • $\begingroup$ I am not so sure. For example, I think that the positions of gas molecules in a closed bottle are governed by a very chaotic equations. Even the tiniest disturbances lead to completely different positions of molecules after a while. You do not need Navier-Stokes equations and whole atmosphere to go chaotic, I think that even one "misplaced" molecule counts. And this is exactly what quantum mechanics gives you. $\endgroup$ – Irigi Nov 3 '14 at 17:29
  • $\begingroup$ And even if the odds are 100 for point $A$ to 1 for point $B$, what's to say the odds are non-zero for points $A'$ and $B'$? My guess would be that both $P(A') \gt 0$ and $P(B') \gt 0$ for some values of $A'$ and $B'$. $\endgroup$ – a CVn Jan 3 '17 at 10:42

Sort of.

Quantum particles can exist in multiple states simultaneously. When they interact with other particles they then collapse into one of their possible states, with there being no way to know for certain which they will collapse into in advance. If you're looking only at quantum particles you can't predict even one interaction into the future with absolute reliability.

However, this clearly doesn't give the whole story. A baseball contains an uncountable multitude of quantum particles, each interacting with other particles and constantly collapsing into various states, yet any child can predict the path of its movement in spite of that. The explanation for this is that as you add more quantum particles the system they describe becomes more and more deterministic as the probability of some significant portion of the particles collapsing into a possible but unlikely state approaches zero. We humans are made of so very, very many quantum particles that the effect of quantum randomness on our behavior is negligible, we are for all intents and purposes entirely deterministic.

The weather tomorrow could be predicted with near-perfect accuracy provided one knew everything there is to know about the current state of the Earth's atmosphere and any nearby objects. The current state of the human world if Hitler had been killed as a baby could be predicted with near-perfect accuracy provided that one knew everything there is to know about the state of the world prior to Hitler's birth and all ensuing events. However, either of those predictions is well beyond our computing capabilities for the foreseeable future, much less our data-gathering capabilities.


Disclaimer: My science levels are all over the place so this may not be 100% accurate. I'm specifically concerned about my first paragraph which is a alternative interpretation of Gödel's Incompleteness Theorems. It takes a different view on the matter, one that I feel is correct but am not certain of. I added a second concrete example to illustrate the thought process. Slightly unrelated is Turing's Proof which rankles me and may be related to decoding the universe. See Note2 for an argument on Turing's Proof that I way more certain of than I am of my argument on Gödel.

Our current math theorems only tells us that we can never have a complete system in the sense that we can never know with certainty that our axioms and such are complete, etc. But this rides on the assumption that the parts our math doesn't allow us to grasp aren't the ones that override this assumption. (Our current axiom set roughly says "You can't have all the rules because you can't logically get to some of them, no matter what set of axioms you pick" but there may be a rule that we have to hit by chance or take on faith, since we can't get to it logically by our current axiom set per Gödel, that translates out to roughly: "Disregard all other rules. The last axiom of math is the 5th item on your shopping list on every billionth day. Just take my word for it."). Like all science we may discover that we need to rewrite the axioms when we hit new areas of math. This could open the way to us discovering the true set of axioms that explains all (if they exist). That is, by virtue of logic failing we could potential be forced to have our minds wander until we hit the right answer by chance (which sounds a lot like the current state of affairs for our current search for a ToE).

A good example for the possible last axiom (well, maybe for a moral system and not a mathematical one) would be the rule: "Everything is a grey-area of some extent, there are no black and white rules." which seems decent if you can apply it to just about everything. But when you get around applying that rule to itself it becomes a statement of "Sometimes things are black and white, because this rule is a grey-area itself." So do you edit it to "Everything is grey except for this rule." or do you take it as it stands or do you toss it completely. The right course of action seems difficult. Assuming that this is one of the true axioms of the universe the two framings have different outcomes: strict rules with a loophole written in (that may or may not be exploited by the universe at any given moment), or with exceptions only for themselves. Either way its status as an axiom relegates it to the "take it on faith" principle which in the case of this self-paradoxical rule is a little harder to swallow that 1+1=2.

So even if the universe is deterministic to a fault we would most likely need a lucky guess or a change in mathematics to get all the rules right.

It could be that even without perfect rules we could move towards a snapshot of the universe at time T. Our certainty would never reach 100% but we could get very very close. If your predicting the future accurately left and right though, then you pretty much don't care if you're 100% right, you might just assume you are spot-on which is basically what Bayesian statistics would tell you. Sure you missed that thing-a-ma-what's-it axiom, but that only causes a simulation glitch once every billion years... Do you care?

If the universe is boot-strappable from a small set of data and has implicit T states, then we could operate on all of this data and recreate a time-location-state(which is what you're asking, so this is the caveat). The simulation could even reduce as far as a random seed number. This may be a number with only finite possibilities. We could test the parameters until we reached one that modeled our current T value correctly. This would get us to the point of modeling the universe without violating the Heisenberg Uncertainty Principle or any other building blocks.

If the universe fails any of these assumptions (is not deterministic, etc.), then our math can give us possible futures with error measures. You could extract information as long as your signal-to-noise ratio for the universe was relatively decent. Non-implicit universes would have diminishing returns on the signal over time and increased noise (which would model our heat as information loss perfectly!).

Note: Figured I'd add that our current science makes the needed assumptions when it assumes to know the effects of the constants on the universe. "Our Constants are so finely tuned it must be the work of an intelligent design". What gets more in the way is the problem of storing a given snapshot of the universe so that we can compute on it. If there isn't a small boot-strappable data set then it can't be done. There are various physical effects that operate as a whole and not additively, so you would need an atom to record all the information about that atom, etc. resulting in the universe. Otherwise you'd need a frequent snapshot of the area of interest to work off of. If you could acquire a frequent snapshot of an area there becomes little need to forecast very far ahead which reduces the error we can have in T+1 allowing more inaccurate rules. (Our weather forecasting is an example. Snapshots of poor detail are acquired and a simulation plots out 7 days. Days closer to the start of the simulation are more accurate. Increasing resolution, simulation frequency, or computing power increases accuracy overall).

Note2: So this isn't directly related except for the fact that if I'm right all of a sudden we can compute Chaitin's Constant which will probably give us a decent platform to create our boot-strappable data set (it also flirts with violating Gödel's Incompleteness Theorems)... So Turing's Proof has always rankled me because it amounts to saying "If you give me a program H that computes Halt() I can give you a program B that always invalidates H" the problem is that his formulation involves B containing H which amounts to saying "I can construct a wrapper function/super-set to H that invalidates it" so the implicit assumption is B is always bigger than H. You can in fact (I wrote a proof) construct an H that uses B to expand itself recursively. This makes sure H is always able to get the correct solution for Halt() as long as H is at-least 1 bit larger than B. If you extend this out to infinity then technically Turing's Proof is right because both H and B will end up at the same infinite size and we can't decide when their the same size (at-least my function can't, although it would be possible if you made a quantum extension of my algorithm (but that may give B new tricks of subverting H as written)) so we end up with an infinitesimal fraction of all programs that are undecidable. But that just means your constant has an infinitesimal error and so is still usable for our purposes.

Edit2: Oh yeah your questions...

  • How precise of an estimation could we get of the future? It depends, see the other answers.
  • Can a weather forecast for the next day be 100% reliable? I'm pretty sure we can already do this, we just don't run them every day. It's run the minimum amount of times for tolerable accuracy because weather "people"/stations pool money to run a simulation on a supercomputer (or they did anyway...)
  • Can we answer questions like "if hitler had died as a kid" with 100% reliability? If we achieve exactly 99.999% accuracy in our rule set for time-steps of 1 yr (noting we'll never reach 100% unless we just take it on faith and it turns out to actually be our magical rule set) and get a perfect snapshot today (2014) then your looking at 99.874% accuracy on our Hitler predictions. So no, not 100% at that point but pretty damn close.
  • How much further could we predict from T, with good accuracy, knowing that there is randomness? Well if you compute Chaitin's Constant for your universe you can actually predict your randomness given enough bits of information (this is exactly the same as predicting the seed of a random number generator given x bits where Chaitin's is your seed and the rule set is the generator. You tweak things snipped long explanation until your correct and then randomness doesn't matter). If we take the other track and assume you can't then you have the same situation as for the Hitler situation. The perfection of your rule set determines how close you can get. You multiply the accuracy every time-step (so starting with .999999999999 you lose 1 off the least significant digit each step and there are Z steps each second which is at least the speed of light divided by the number of planck lengths in a meter in magnitude (so 1.855492(18) x10^43) which means for an accuracy to the 12 decimal place you lose at least (.999999999999 accuracy to the 2 x10^43 ticks/sec to the 31600800 sec/yr) .2% accuracy each year (I think I did that right...)). So if we place our limit for good accuracy at 90% then we have 50 years.
  • $\begingroup$ From half-awake original post to clarity-induced megapost. Your welcome. Or not, the votes will tell. As an aside, if someone wants to play with my Turing "anti-proof" let me know. $\endgroup$ – Black Nov 3 '14 at 16:30

Is the predictor located inside the universe? If so, no.

Thomas Breuer has proven that universally-valid deterministic or probabilistic theories are impossible.

What does it mean? It means that an observer cannot measure his own quantum state (wave function, or probability). Maximum what he can measure about himself is possibility which is more lose thing. Thus he cannot predict even probabilities for the future.

If you predict future possibilistically, you can say very little about it, only that some things are possible and other are impossible.

In other words:

If the predictor is inside of universe:

  • Deterministic prediction is impossible

  • Probabilistic prediction is impossible

  • Possibilistic prediction is possible

If the predictor is outside the universe:

  • Deterministic prediction is impossible

  • Probabilistic prediction is possible

  • Possibilistic prediction is possible (because probabiblistic is possible)



It is theoretically possible to completely observe the state of a universe at time T, and predict its state perfectly at time T+1. Because our models of the universe are models, they usually assume there is some source of perfect state to be had.

Issue #1: Current QM theory suggests that it is not possible to measure all of the perfect state of our universe at time=T. There must be information missing (see Heisenburg Uncertainty). However, nothing exists that says QM is the final model of the universe... there might be a way for all we know.

Issue #2: If you try to predict the universe while still inside it, you run into funny issues. In a very long story short, the fact that your model of the universe has to include a model of your computer, your universe is at least strong enough to do integer arithmetic. This instantly puts your universe (and its model) under the control of Kurt Goodel's incompleteness theorem. Handwaving the math terms, it means you will have to accept at least one undesirable trait to make your computer a reality:

  • The model must be incomplete, failing to model at least SOME part of the universe
  • The model must be inaccurate, getting the state of part of the universe wrong
  • The model must be illogical, failing to follow the rules of First Order Logic
  • The model must be non-enumerable, unable to ever actually be contained on a sheet of paper (or on the hard drive of a computer)
  • The model must be unprovable, so you THINK you have the right answer but have no possible way of proving it

Unfortunately, by the rules of integer arithmetic and First Order Logic, that's the breaks. It is only feasible to predict the universe using a computer outside the universe because of this issue.


No. Any system with three or more independent variables will be chaotic. That does not mean not repeatable, but it does mean that the only way to determine the outcome is to run the process to its end condition.

See "Period Three Implies Chaos", by Tien-Yien Li and James A. Yorke, published in The American Mathematical Monthly Vol. 82, No. 10 (Dec., 1975), pp. 985-992

Link: https://www.jstor.org/stable/2318254?seq=1#page_scan_tab_contents

  • $\begingroup$ While a general analytical solution of a chaotic system is not possible, numerical solutions can be calculated to arbitrarily high precision. $\endgroup$ – timuzhti Jul 26 '18 at 12:56


I agree with the information-theoretic and Heisenberg answers outlined above, but haven't seen anyone bring up the Halting Problem objection. Here's the Halting Problem: Is it possible to write a computer program ORACLE that will look at another program FOO and the inputs BAR that program will receive, such that ORACLE will say with 100% accuracy whether FOO will terminate if given BAR?

The answer is no. This is not possible. The reason why it's not possible is applicable in other contexts (including here) so I'll give the proof.

In short, if I have such an ORACLE, I can write a program S (for smartass) that calls into the ORACLE, and then does whatever the opposite of what ORACLE claims will happen.

Predicting the future runs into this exact problem: Unless it is literally impossible for someone to build another, then other people will have predictors. If they're good at using them, now they know what you think they're going to do. It won't be hard for them to find clever ways to screw around with you by deviating from those predictions.


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