3.5 The distribution of universes
The last two figures of the previous Step 3.4 (Discrete universes) provide useful insight into to how the multiverse of discrete universes must be structured. If you know that you are in one of the eight block universes in the semi-circular array in these figures, how likely is it that, in your universe, both Alice and Bob have detected spin-down? Not very likely, as it happens, because only one of the eight universes has that combination of outcomes. So you would say that the probability of finding both outcomes spin-down is ⅛. This is an example of the self-location uncertainty that we mentioned in Step 3.2 (The Many Worlds of Hugh Everett III). However, the probability that you will find both outcomes spin-up is ⅜, because three of the eight universes have that combination of outcomes.
We calculated these fractional probabilities from the product of the Born probability of Alice detecting a particular outcome and the Born probability of Bob detecting a particular probability. We can keep track of which probability refers to which combination of the two outcomes by writing the sum of the probabilities for Alice’s outcomes and the sum for Bob’s in the following way, and multiplying them together:
(¾ Au + ¼ Ad) ⨯ (½ Bu + ½ Bd) = ⅜ AuBu + ⅜ AuBd + ⅛ AdBu + ⅛ AdBd.
So, if we have a total of eight universes, we can find the number of universes for each combination of outcomes by multiplying the above result by 8, giving
3 AuBu + 3 AuBd + 1 AdBu + 1 AdBd.
This is just the distribution of discrete universes that you see in the semi-circular array in the last two figures of Step 3.4 (Discrete universes). Nothing subtle is going on here; essentially, all we are saying is that the number of universes containing a particular outcome of a given event is proportional to the (Born) probability of that outcome. (Later, we shall come to the question of what to do with Born probabilities that can’t be expressed as such nice round fractions like ⅛.)
Now, although I said that the probability of finding, say, both outcomes spin-up is ⅜, if you are in one of the eight universes, you can’t deduce such a probability from a single observation, regardless of whether you see two spin-up outcomes or something else. Instead, we ask a simple question of the multiverse structure: if we perform a series of quantum experiments (say Alice’s 60º experiment) many times in succession, and then aggregate all of the measured outcomes of every experiment in the series, do they converge towards the result predicted by the Born rule?
Suppose Alice performs her 60º spin-filter experiment just twice in succession with two different, non-entangled electrons. Then, according to our scheme, the Born probabilities for the distribution of universes containing the possible outcomes of the two experiments are
(¾Au + ¼Ad) × (¾Au + ¼Ad) = (¾Au + ¼Ad)2 = (1/16) {9(Au)2 + 6AuAd + (Ad)2}
This diagram shows the distribution of universes containing the outcomes of two 60º experiments which each version of Alice performs sequentially in each universe that she inhabits. The relative distribution of universes containing both outcomes can be found simply by squaring the expression for the distribution of universes in the first experiment, which, for the 16 universes, is 16 × (¾ Au + ¼ Ad)2 = 9 (Au)2 + 6 AuAd + (Ad)2.
This is illustrated in the figure, where I have shown the experiments taking place in 16 universes. I have again drawn each universe as a (pale yellow) box, although, this time, the outcomes of the two experiments are stacked one on top of the other, because they are carried out at the same place, with the earlier outcome being the lower of the two in the box. When the trunk of 16 universes splits into 12 and 4, the upper of the two outcomes in each box is still a question mark, since the experiment hasn’t been performed yet, at that time. I chose to use 16 universes because it neatly cancels the factor of 1/16 that you can see in the equation after you multiply the probabilities. However, I could equally have chosen 32 universes or, for that matter, 16 million (in that case, with each rectangle in the figure representing a million identical universes). In fact, the number of universes in our own multiverse will turn out to be unimaginably large.
If each version of Alice performs her experiment not twice but four times in succession with non-entangled electrons, then the Born probabilities will be (¾ Au + ¼ Ad) multiplied by itself four times:
(¾Au + ¼Ad)4 = (1/42) {81(Au)4 + 108(Au)3Ad +54(Au)2(Ad)2 + 12Au(Ad)3 + (Ad)4}.
Since 44 is 4 ⨯ 4 ⨯ 4 ⨯ 4 = 256, then the probability that the result of every one of her four measurements will be spin-up, i.e., the outcome represented by (Au)4, is 81/256. The most likely scenario is for three of her results to be spin-up and one spin-down (since 108 is the largest of the coefficients) with a probability of 108/256. The total of all the probabilities is
(81 + 108 + 54 + 12 + 1)/256 = 256/256 = 1
as expected, because the probability of Alice finding one of the four possible outcomes, AuBu, AuBd, AdBu and AdBd, is just unity – it is certain to happen!
Out of a total of 256 universes, this is the number of universes in which 0, 1, 2, 3 and 4 outcomes of spin-up (or, equally, 0%, 25%, 50%, 75% and 100%) are observed in a sequence of four of Alices 60º experiments.
If there were a total of 256 universes, each containing a sequence of Alice’s four experiments, then we can plot the above expression as a histogram. Even with just four experiments, we see that the most popular universe is one of those in which Alice detects three spin-ups (75%). So, if I could reach into this multiverse of 256 universes and pick out one universe at random and ask Alice about her outcomes, it is more likely that she will attest to having seen three spin-up outcomes than two or one or none or even four.
Of course, alongside Alice’s experiment, there is a myriad other interactions going on quite independently of Alice’s activities, and each of these also needs to be similarly represented by multiplications of universes with Alice in them, so that the number of universes containing Alice’s experiments is increased tremendously. If we had used 512 universes instead of 256, the histogram would have looked no different, except that the number above each histogram bar would be doubled, as would the values of the graduation marks on the y-axis. Importantly, the relative sizes of the five histogram bars would be the same. No matter how large is the number of universes, the shape of the histogram doesn’t alter.
As for the previous figure, but, in addition to just 4 experiments per sequence, showing 100, 1,000 and 10,000 experiments per sequence. Note that the sharpness of the curve has nothing to do with the total number of universe in the multiverse: it depends entirely on the number of experiments conducted in a sequence in the individual universe.
In the next histogram, in addition to sequences of just four experiments, I have added three more sequences of experiments: sequences each containing 100 experiments, sequences each containing 1,000 experiments and sequences each containing 10,000 experiments. The number of universes required for the 10,000 experiments would, instead of 44, be 410,000. However, what affects the shape of the histogram is not the number of universes, but the number of experiments within each universe, i.e., 4 or 100 or 1,000 or 10,000. So let us see what histograms for these extra three sequences would look like with, say, 10 to the power of 10 to the power of 10 universes! In the figure, all four sequences of experiments have been plotted.
For the histogram showing sequences of four experiments in a universe I have just copied the previous one, with the difference that, instead of 108 universes containing three spin-up outcomes, there are now (108/256) × the total number of universes. The histogram for the 100 experiments is the same as that in Step 1.11 (Our absolutely indeterministic universe). Again, of course, I have made the number of universes very much greater than the 100,000 sequences, or universes, which I used for the first histogram, but that doesn’t alter the shape of the histogram, which depends upon the number of experiments in a universe, not the number of universes.
As the number of experiments in each universe increases through 1,000 to 10,000, you can see that the peak becomes successively narrower, converging on the fraction, 75%, predicted by the Born rule for an indefinitely large number of such experiments. In other words, using this scheme, since nearly all of the universes are clustered around the 75% result when the number of successive experiments is very high, then it is extremely likely that you are in a universe where you will find that the proportion of spin-up outcomes is exactly, or very close to, 75%.
The conclusion is that, if the parallel universes are distributed in the multiverse in proportion to the product of the outcome probabilities of each independent interaction, then the probabilities of these outcomes determined within these universes will be in accordance with quantum mechanics.
Notice that, in order to draw the histogram where Alice repeat her 60º experiment four times, we multiplied the Born probabilities by 256 so that we could get whole numbers of universes. Indeed, we only just got there by the skin of our teeth there, because, for the case where the outcome for each of the four experiments was spin-down, there was only one universe out of the 256, shown at the 0% point on the x-axis. That was because the probability for that point was 1/256. But suppose the Born probability for a given result wasn’t a nice round figure, but was instead something involving an irrational number, like the square root of 2. For instance, the algebra of quantum mechanics shows us that, if the Stern-Gerlach filter in Alice’s first experiment had been angled at 45º rather than 60º, then the corresponding spin-up and spin-down probabilities would be [½ +/- 1/(2√2)] , or 0.853553390… and 0.146446609… . We can write the sum of these probabilities as (853,553,390 Au + 146,446,609)/(1,000,000,000) so that, if we have a billion universes in which Alice performs just one experiment, then, in our scheme, Alice will find a spin-up result in 853,553,390 universes and spin-down in 146,446,609 universes. But that’s not quite right, because the digits after the decimal point in an irrational number go on forever, so that the number of universes would have to be indefinitely large to give the correct probabilities. So, if the number of universes in the multiverse is finite, then quantum mechanics will turn out to be only an ideal approximation (no matter how accurate) to the actual structure of the multiverse. In fact, in practice, the number of universes in the multiverse will be so large that it would be impossible to show any difference experimentally between the quantum ideal and the actual structure. We shall come back to the question of the number of universes in the multiverse in Step 4.4 (The multiverse and its universes are all finite).
