Why MWI cannot explain the Born rule

[Muppetmaster]

So, what you are saying is "Mathematics is not enough. There must be also a way to translate Birds view into Frogs view". Is this correct?

Yes String Theory is philosophy and barely even a hypothesis let alone a theory. MWI is the latest in a long line of totally unprovable fairy tales.

Maths is just a model. Shut up and calculate.

 

[dmtr]

Are you working on anything relating quantum and BBP? Do you have any papers on this topic? Read 0908.4348 to see why I ask.

Sorry. I'm not a physicist. I've studied the 'basic modern minimum' - CM, SR/CFT, SM, GR, QM, but that's pretty much it. This QFT interpretation paper is interesting, but it is also a way over my head.

Still, I really like the idea of how they interpret the "probability amplitudes" as "symmetry amplitudes". It's just change of wording, but it makes a lot of sense. I couldn't get if they use the MWI approach (many subjective histories) or some kind of superdeterminism (one history). And how they use BBP I couldn't understand at all.

 

[Muppetmaster]

Sorry. I'm not a physicist. I've studied the 'basic modern minimum' - CM, SR/CFT, SM, GR, QM, but that's pretty much it. This QFT interpretation paper is interesting, but it is also a way over my head.

Still, I really like the idea of how they interpret the "probability amplitudes" as "symmetry amplitudes". It's just change of wording, but it makes a lot of sense. I couldn't get if they use the MWI approach (many subjective histories) or some kind of superdeterminism (one history). And how they use BBP I couldn't understand at all.

The equations for both would be identical if you think about it, the only difference being one is imaginary and one is real, if you see what I mean.

The usual Dirac equation would create 2 versions: Dirac=MWI+CI

ie.

\left(\beta mc^2 + \sum_{k = 1}^3 \alpha_k p_k \, c\right) \psi (\mathbf{x},t) = i \hbar \frac{\partial\psi(\mathbf{x},t) }{\partial t}

1 version would be all possible words ie MW_{wave}=all real states in different dimensions/realities ie \int_a^{b} x_p=\infty the other would be superposition CI_{wave}^{i}=imaginary solutions of the wave function and one possible "real" state probabilistically.

x_p=probability of x where x = variable of wave function.

 

[Dmitry67]

I think the additional assumption is:

"We are on the NORMAL branch"

Otherwise Born rule can not be proven even experimentally. No matter what results you get one must deny them saying "we are just on one of the unprobable branches"

 

[RUTA]

Still, I really like the idea of how they interpret the "probability amplitudes" as "symmetry amplitudes". It's just change of wording, but it makes a lot of sense.

The different wording is to emphasize the fact that the amplitude Z is independent of the field Q, so the fundamental mathematical objects being "evaluated" by the path integral are the discrete differential operator K and source J. That is, K and J are the mathematical objects describing the experimental arrangement and Q is merely an integration variable or "tool" to analyze K and J.

I couldn't get if they use the MWI approach (many subjective histories) or some kind of superdeterminism (one history). And how they use BBP I couldn't understand at all.

There is only one configuration being analyzed, that described by K and J. No Many Worlds. So, the question is, What constrains K and J in Z? The answer is BBP, which gives Kx = J where x is the vector of plaquettes, links or nodes, depending on whether you're doing tensor, vector or scalar field theory, respectively. So, you start off with Kx = J at the graphical level and compute Z for the graph. But, what good is that? You need to recover classical field theory (CFT) at some point which all about Q and your mathematical object Z is totally void of Q. Answer: You know Z is a partition function in this case (Kx = J gives a Euclidean action), so you use Z to compute the probability of measuring some particular value of Q at a particular graphical location, e.g., the k^th node, i.e., the probability that Q_k=Q_o. That's easy enough to compute since Z is a partition function we know that the probability simply equals Z(Q_k=Q_o)/Z. When you're finding such probabilities you're doing QFT (actually, a discrete counterpart). You recover the discrete counterpart to CFT when you attempt to find the most probable value of Q_o at Q_k. You obtain KQ_o = J, which you recognize as CFT and which you know conforms to BBP. So, you assume BBP at the fundamental graphical level and find that it leads to both CFT in accord with BBP (as required), as well as a new interpretation of QFT that resolves all its foundational issues (as explained in the paper).

Anyway, that's why I was wondering how you used BBP to explain quantum physics. I wanted to compare your method with that of this paper.

 

[Demystifier]

Demystifier, is the argument about the Born rule similar to what I am saying here https://www.physicsforums.com/showthread.php?t=230032&highlight=many+worlds ?

Yes it is.

 

[dmtr]

Interesting paper on the subject: http://www.ensmp.fr/aflb/AFLB-333/aflb333m533.pdf

RUTA: I'll try to understand/answer the "K and J constrains/BBP" part and if I like it or not, but that will need time.

 

[pellman]

Yes it is.

Very cool.

Isn't the failure of the MWI to predict the statistics a rather glaring problem? I would think some MWI proponent has covered this. Surely we're not covering new ground here?

 

[Demystifier]

Isn't the failure of the MWI to predict the statistics a rather glaring problem? I would think some MWI proponent has covered this. Surely we're not covering new ground here?

You are right. However, most literature on that issue is not very easy to read. My intention is to present and discuss this issue in a simpler way, so that all physicists can easily understand it.

 

[hamster143]

Very cool.

Isn't the failure of the MWI to predict the statistics a rather glaring problem? I would think some MWI proponent has covered this. Surely we're not covering new ground here?

See my comments earlier in the thread, I believe I've addressed that specific issue.

If you repeat an experiment that branches the wavefunction as

<br /> |\psi\rangle = \sqrt{\frac{1}{1000}}|A\rangle + \sqrt{\frac{999}{1000}}|B\rangle<br />And you throw in some decoherence and an outcome counter, after 1000 runs you'll have 2^1000 branches of the whole universe, but only 1001 branches of the counter, all with varying amplitudes. And the branch of the counter with the largest amplitude will be the one that counted 1 outcome A and 999 outcomes B.

Your approach seems to be a straw man argument. Sure, if you take MWI and add an axiom that all observers are equally likely in any branching, you can arrive at all sorts of contradictions (including the most obvious one - you can't have 2^1000 equally likely histories and 1001 equally likely counter outcomes at the same time). If you allow that every branch has an amplitude and the "likelihood" is in some way monotonically related to the absolute value of the amplitude, you can naturally deduce that the relationship has to take form of the Born rule, P_i = |c_i|^2 / \sum_k |c_k|^2.

 

[Demystifier]

If you allow that every branch has an amplitude and the "likelihood" is in some way monotonically related to the absolute value of the amplitude, you can naturally deduce that the relationship has to take form of the Born rule, P_i = |c_i|^2 / \sum_k |c_k|^2.

The question is: WHY would likelihood be monotonically related to the absolute value of the amplitude?

Let me use a classical analogy. Assume that we have two houses: a big one with 100 apartments, and a small one with only 2 apartments. These two houses are analogs of two worlds.

Now we ask the question: Which house is more probable, the big one or the small one? It is not even clear what this question means. Nevertheless, one meaningfull answer is that both houses are equally probable. On the other hand, the Born rule corresponds to the claim that the big house is 50 times more probable than the small one. But if the two houses are ALL we have, then it is difficult to justify such a claim.

Now consider a modified question: What is the probability that Jack lives in the big house? Now it is reasonable to assume that each APARTMENT has equal a priori probability to be the Jack's home, which implies that probability that Jack lives in the big house is 50 times larger than probability that Jack lives in the small house. In other words, by introducing an additional "hidden" variable (called Jack) that lives in one of the worlds, the Born rule attains a natural explanation. Needless to say, Jack is an analog of the pointlike particle in the Bohmian interpretation.

 

[pellman]

You are right. However, most literature on that issue is not very easy to read. My intention is to present and discuss this issue in a simpler way, so that all physicists can easily understand it.

Looking forward to reading it. If you do write that paper, shoot me a PM or email when its done.

 

[Demystifier]

Looking forward to reading it. If you do write that paper, shoot me a PM or email when its done.

Well, I didn't say that I will write a paper. But I didn't say that I will not write it either.

By the way, what do you think about my analogy in #71? In my view, this IS the simplest way (though certainly not the most rigorous one) to explain my opinion on that stuff.

 

[Hurkyl]

Let me use another classical analogy.

Suppose we have an ordinary six-sided die. I ask the question "are the sides equally likely?" What does that question really mean?

Of course, I'm sure you have a ready answer for that, based on the classical frequentist interpretation of probabilities -- mutter something about propensities, repeated experiments, argue that, for each positive epsilon, the propensity of "The proportion of times '1' comes up is in (1/6 - \epsilon, 1/6 + \epsilon)" converges to 1 as the number of experiments goes to infinity, and then we go and do controlled empiricial trials until we're satisfied we're close close enough to infinity to believe the answer.


I offer the following theorem:

Let \psi be a state a Hilbert space H. Let \psi^{(n)} := \psi \otimes \psi \otimes \cdots \otimes \psi (n times).

Let M_{n, p, \epsilon} be a "toy frequency counting operator" on H^{(n)} \otimes \mathbf{C}^2. (C² is the space of pure qubit states) -- for some basis {e_m} of H, M is given by

• If the proportion of the a_k equal to 1 is in (p - \epsilon, p + \epsilon), then
  M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m \oplus 1\rangle
• Otherwise,
  M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle

We can construct a function f_{p, \epsilon, \psi} that maps each positive integer n to the state space of a qubit given by taking the partial trace of M \left(\psi^{(n)} \otimes |0\rangle) -- heuristically, this function tells us the (probably mixed!) quantum state of the result of toy measurement if we do n trials.

Finally, define P(e_n ; \psi) = p if and only if, for every positive epsilon, we have \lim_{n \to +\infty} f_{p, \epsilon, \psi}(n) is the pure state |1\rangle.

Theorem: P as defined above is given by the Born rule.


Is the above mildly complex? Sure -- but if you aren't going to handwave over everything, isn't the corresponding argument in classical mechanics similarly complex?

 

[pellman]

By the way, what do you think about my analogy in #71?

Not entirely sure I follow it. Is the analogy such that it corresponds to an observation/experiment in which result BIG HOUSE is 50/2 times more likely than result LITTLE HOUSE?

In other words, by introducing an additional "hidden" variable (called Jack) that lives in one of the worlds, the Born rule attains a natural explanation. Needless to say, Jack is an analog of the pointlike particle in the Bohmian interpretation.

Are we going for the Many-worlds-Bohmian interpretation now? Ladies and gentlemen, the captain has now turned on the "fasten seat belt" sign. Passengers should return to their seats.

 

[Fredrik]

Let me use another classical analogy.

Suppose we have an ordinary six-sided die. I ask the question "are the sides equally likely?" What does that question really mean?

Of course, I'm sure you have a ready answer for that, based on the classical frequentist interpretation of probabilities -- mutter something about propensities, repeated experiments, argue that, for each positive epsilon, the propensity of "The proportion of times '1' comes up is in (1/6 - \epsilon, 1/6 + \epsilon)" converges to 1 as the number of experiments goes to infinity, and then we go and do controlled empiricial trials until we're satisfied we're close close enough to infinity to believe the answer.

I do have an answer, but it's not that one. I think both the question and that answer are unscientific at best, and nonsense at worst. The question is asking for the correct definition of something undefined. That can't ever make sense. And that answer asserts the existence of a "limit" which is "defined" by a description of an act that can't be performed. That's just more nonsense.

This is the short version of my answer: Probability in mathematics is a number assigned by a probability measure. If we try to associate such a number with something in the real world, we're no longer doing pure mathematics. We're either doing science or pseudoscience.

A scientific theory associates numbers assigned by a probability measure with possible results of experiments. The definition of science requires that we test theories by comparing those numbers to relative frequencies in the results of a large but finite number of almost identical experiments.

If someone claims that probability "is" something more than this, they're making a statement about the real world, not about mathematics, and that means that the statement must satisfy the definition of a theory to be worthy of further consideration.

See also #11 in this thread, and my posts in the thread I linked to there.

I offer the following theorem:

Let \psi be a state a Hilbert space H. Let \psi^{(n)} := \psi \otimes \psi \otimes \cdots \otimes \psi (n times).

Let M_{n, p, \epsilon} be a "toy frequency counting operator" on H^{(n)} \otimes \mathbf{C}^2. (C² is the space of pure qubit states) -- for some basis {e_m} of H, M is given by

• If the proportion of the a_k equal to 1 is in (p - \epsilon, p + \epsilon), then
  M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m \oplus 1\rangle
• Otherwise,
  M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle

We can construct a function f_{p, \epsilon, \psi} that maps each positive integer n to the state space of a qubit given by taking the partial trace of M \left(\psi^{(n)} \otimes |0\rangle) -- heuristically, this function tells us the (probably mixed!) quantum state of the result of toy measurement if we do n trials.

Finally, define P(e_n ; \psi) = p if and only if, for every positive epsilon, we have \lim_{n \to +\infty} f_{p, \epsilon, \psi}(n) is the pure state |1\rangle.

Theorem: P as defined above is given by the Born rule.

I don't understand what you're doing here, but see #13 in the thread I linked to above for a comment about the 1968 article by Jim Hartle that (I think) introduced frequency operators.

 

[Count Iblis]

The argument by Hartle allows you to replace the Born rule by the weaker rule that says that measuring an observable of a system if the system is in an eigenstate of that observable, will yield the corresponding eigenvalue with certainty.

 

[Fredrik]

The argument by Hartle allows you to replace the Born rule by the weaker rule that says that measuring an observable of a system if the system is in an eigenstate of that observable, will yield the corresponding eigenvalue with certainty.

I don't think that's correct. This is what some Wikipedia article said, right? I suspect that whoever came to that conclusion believed that every system is always in an eigenstate. (An idea which as you know has been thoroughly disproved by Bell inequality violations). It seems to me that all Hartle did was to prove that for each eigenstate |k>, there's an operator that has every |s\rangle\otimes|s\rangle\otimes\cdots (where |s> is an arbitrary state) as an eigenvector with eigenvalue |\langle k|s\rangle|^2.

It's natural to interpret this operator as a frequency operator when it acts on a tensor product of eigenstates, but it doesn't seem to have a natural interpretation when it's acting on |s\rangle\otimes|s\rangle\otimes\cdots with |s> arbitrary. The only way I can justify the interpretation as a frequency operator is to use the Born rule, i.e. the rule he's trying to prove.

I would also say that Hartle used the Born rule before that, when he assumed that the Hilbert space of a system is the tensor product of the Hilbert spaces of its subsystems. I don't think that assumption can be justified without appealing to the Born rule.

So my opinion is that Hartle's article is completely useless. If I'm wrong, I'd like to know it, so feel free to try to convince me. Post #18 in the other thread (linked to above) has a direct link to the article. My comments are in #13.

 

[Demystifier]

Not entirely sure I follow it. Is the analogy such that it corresponds to an observation/experiment in which result BIG HOUSE is 50/2 times more likely than result LITTLE HOUSE?

No, but 100/2=50.

Are we going for the Many-worlds-Bohmian interpretation now?

Not necessarily, but why not to mention it if is related to MWI and to the Born rule.

 

[pellman]

No, but 100/2=50.

Right. That's what I meant.

Not necessarily, but why not to mention it if is related to MWI and to the Born rule.

No reason. I was kidding. I lean towards it myself.

 

[pellman]

Let \psi be a state a Hilbert space H. Let \psi^{(n)} := \psi \otimes \psi \otimes \cdots \otimes \psi (n times).

Let M_{n, p, \epsilon} be a "toy frequency counting operator" on H^{(n)} \otimes \mathbf{C}^2. (C² is the space of pure qubit states) -- for some basis {e_m} of H, M is given by

• If the proportion of the a_k equal to 1 is in (p - \epsilon, p + \epsilon), then
  M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m \oplus 1\rangle
• Otherwise,
  M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle

We can construct a function f_{p, \epsilon, \psi} that maps each positive integer n to the state space of a qubit given by taking the partial trace of M \left(\psi^{(n)} \otimes |0\rangle) -- heuristically, this function tells us the (probably mixed!) quantum state of the result of toy measurement if we do n trials.

Finally, define P(e_n ; \psi) = p if and only if, for every positive epsilon, we have \lim_{n \to +\infty} f_{p, \epsilon, \psi}(n) is the pure state |1\rangle.

Theorem: P as defined above is given by the Born rule.

I might be able to understand this with a little help. What are the a_k?

 

[Count Iblis]

I don't think that's correct. This is what some Wikipedia article said, right? I suspect that whoever came to that conclusion believed that every system is always in an eigenstate. (An idea which as you know has been thoroughly disproved by Bell inequality violations). It seems to me that all Hartle did was to prove that for each eigenstate |k>, there's an operator that has every |s\rangle\otimes|s\rangle\otimes\cdots (where |s> is an arbitrary state) as an eigenvector with eigenvalue |\langle k|s\rangle|^2.

It's natural to interpret this operator as a frequency operator when it acts on a tensor product of eigenstates, but it doesn't seem to have a natural interpretation when it's acting on |s\rangle\otimes|s\rangle\otimes\cdots with |s> arbitrary. The only way I can justify the interpretation as a frequency operator is to use the Born rule, i.e. the rule he's trying to prove.

I would also say that Hartle used the Born rule before that, when he assumed that the Hilbert space of a system is the tensor product of the Hilbert spaces of its subsystems. I don't think that assumption can be justified without appealing to the Born rule.

So my opinion is that Hartle's article is completely useless. If I'm wrong, I'd like to know it, so feel free to try to convince me. Post #18 in the other thread (linked to above) has a direct link to the article. My comments are in #13.

There is another discussion of Hartle's arguments http://arxiv.org/abs/hep-th/0606062" where it is suggested that you need to introduce a discrete Hilbert space to fix the derivation. The problem they address is the fact that the number of factors in the tensor product is always finite in practice, so the a-typical branches don't exactly get a zero norm.

About the assumptions, I don't think one assumes the Born rule, only the L^2 norm. The notion of a probability is eliminated from the postulates. The MWI is supposed to be a deterministic theory, so talking about probabilities in the postulates is unnatural. Instread, you can attempt to replace them by statements about certainties, like that measuring a system in an eigentate will yield a certain outcome with certainty.
This combined with cutting away zero norm sectors of Hilbert space gives you the Born rule.

 

[Fredrik]

There is another discussion of Hartle's arguments http://arxiv.org/abs/hep-th/0606062" where it is suggested that you need to introduce a discrete Hilbert space to fix the derivation.

They don't address the issues I mentioned, so they're probably not aware of them. If they're right about the need for a discrete state space, they've just found another problem with an article that was already useless. It also seems to me that this article is much worse than Hartle's. Their summary of Hartle's argument looks like complete crackpot nonsense to me.

This isn't meant as criticism against you. I'm getting a bit upset about the fact that this article got published, but I don't think you had anything to do with that. I actually hope I'm wrong about this article, because it's quite sad if articles as bad as I think this one is can get published.

About the assumptions, I don't think one assumes the Born rule, only the L^2 norm.

How does that justify the use of the tensor product, or the interpretation of Hartle's operator as a frequency operator?

The MWI is supposed to be a deterministic theory, so talking about probabilities in the postulates is unnatural.

You get a deterministic model if you just remove the Born rule, but it's not a theory since we don't have a way to interpret the mathematics as predictions about results of experiments. It also doesn't include anything about "worlds", since the Born rule is what tells us that there's something in the mathematics that we can think of as worlds. It doesn't even include a description in terms of subsystems, since the Born rule is what justifies the use of the tensor product.

Instread, you can attempt to replace them by statements about certainties, like that measuring a system in an eigentate will yield a certain outcome with certainty.

I haven't seen any reason to believe that. I haven't even seen an attempt to prove it. I just have a vague memory of seeing that claim in a Wikipedia article that I wasn't able to find when I tried yesterday, and that used Hartle's article as a reference. But I don't think Hartle's results could be described that way even if his argument had been valid.

This combined with cutting away zero norm sectors of Hilbert space gives you the Born rule.

There are already no such sectors in the individual system Hilbert space, so they're clearly talking about the tensor product of infinitely many copies of that space. The removal of the zero norm "vectors" that get included by accident when we take the tensor product of infinitely many copies of a Hilbert space is necessary to ensure that the result is a Hilbert space. It doesn't have anything to do with the Born rule.

 

[PTM19]

The minimal set of assumptions defining MWI is:
1. Psi is a solution of a linear deterministic equation.
2. Psi represents an objectively real entity.

This is of course not enough to define MWI, for that one has to redefine what "objectively real entity" means. Normal definition clearly doesn't include parallel universes (If I sold you an "objectively real" car and later claimed it's in a parallel universe you would likely be upset). Without parallel universes the interpretation based on above assumptions is simply ruled out by experiment. To avoid being ruled out MWI requires a lot of additional postulates which add parallel unobservable worlds and so on.

The point being that MWI is in no sense a minimal interpretation - on the contrary it is the most extremely baroque of interpretations, as it necessitates belief in an immense number of unobservable universes, what's more it cannot even be falsified making it unscientific. MWI is the most extreme violation of Ockham's razor principle one can imagine.

The minimal interpretation is an ensemble interpretation (without PIV, described here for example (thx for link Fredrik) https://www.physicsforums.com/showthread.php?t=360268&page=1) since it doesn't require one to believe in anything beside what can be experimentally verified and agrees with all experimental evidence.

 

[RUTA]

The problem I have with MWI is that you can’t know whether your physics is legit. Suppose the “real” experimental distribution is 50-50, someone occupies a universe that is 100-0 and someone else 0-100 (with others at all possible distributions in between). So, theoretical physicists in those universes (and most others in between) who predict the “real” 50-50 distribution will not get tenure in physics, but will have to try for tenure in philosophy or religious studies. Physicists in those universes who predict the empirical, but “wrong,” results are heralded and rewarded. I don’t understand why anyone in physics would subscribe to the MWI ontology.

 

[Hurkyl]

To avoid being ruled out MWI requires a lot of additional postulates

 

[Fredrik]

The point being that MWI is in no sense a minimal interpretation - on the contrary it is the most extremely baroque of interpretations, as it necessitates belief in an immense number of unobservable universes,

I addressed those claims in my reply to you in the thread in the philosophy forum (and also earlier in this thread).

what's more it cannot even be falsified making it unscientific.

True, but the same thing holds for every claim that some theory describes what actually happens. This is because experiments can't tell us anything except how accurate a theory's predictions are.

MWI is the most extreme violation of Ockham's razor principle one can imagine.

If there's anything I'm sure about here, it's that this claim is completely wrong.

The minimal interpretation is an ensemble interpretation...

See my comment in the philosophy forum.

 

[PTM19]

For example it has to postulate the existence of parallel universes and that those universes are unobservable. The concept and existence of parallel universes certainly does not follow from those two axioms, it has to be postulated in their interpretation.

 

[Fredrik]

For example it has to postulate the existence of parallel universes and that those universes are unobservable. The concept and existence of parallel universes certainly does not follow from those two axioms, it has to be postulated in their interpretation.

Those two axioms don't even say anything about one world, because they don't make any predictions about (probabilities of) results of experiments.

 

[dmtr]

The problem I have with MWI is that you can’t know whether your physics is legit. Suppose the “real” experimental distribution is 50-50, someone occupies a universe that is 100-0 and someone else 0-100 (with others at all possible distributions in between). So, theoretical physicists in those universes (and most others in between) who predict the “real” 50-50 distribution will not get tenure in physics, but will have to try for tenure in philosophy or religious studies. Physicists in those universes who predict the empirical, but “wrong,” results are heralded and rewarded. I don’t understand why anyone in physics would subscribe to the MWI ontology.

I don't see any problem, as long as the symmetric apparatus gives you symmetric probabilities (probability amplitudes to be precise). Let me illustrate this with this simple example: I'll go to the website generating random numbers from the physical origin and generate two groups of numbers from 0 to 1. After that I'll sum them. Now 0 would correspond to 0:100; 1 would correspond to 50:50 and 2 would correspond to 100:0.

Now, you wouldn't be surprised to see some zeros, and twos? Right?
Here are the numbers:
GR1: 1 0 0 1 0 0 0 1 1 1
GR2: 0 1 1 0 0 0 0 0 1 0
SUM: 1 1 1 1 0 0 0 1 2 1