## Thursday, June 30, 2022

### Predictions and Everett

Imagine this unfortunate sequence of events will certainly befall you in a classical universe:

1. You will be made to fall asleep.

2. Upon waking up, you will be shown a red square.

3. You will be made to fall asleep again.

4. While asleep, your memory will be reset to that which you had in step (1).

5. Upon waking up, you will be shown a green triangle.

6. You will be made to fall asleep for a third time.

7. While asleep, your memory will be reset again to that which you had in step (1).

8. Upon waking up, you will be shown a green circle.

9. You will then be permanently annihilated.

Questions:

1. How likely is it that you will be shown a green shape?

2. How likely is it that you will be shown a red shape?

The answers to these questions are obviously: one and one. You will be shown a green shape twice and a red shape one, and that’s certain.

Now consider a variant story where personal identity is not maintained in sleep. Perhaps each time in sleep the person who fell asleep will be annihilated and replaced by something that is in fact an exact duplicate, but that isn’t identical with the original according to the correct metaphysics of diachronic personal identity. (We can make this work on pretty much any metaphysics of diachronic personal identity. For example, we can make it work on a materialist memory theory as follows. We just suppose that before step (1), you happen to have three exact duplicates alive, who are not you. Then during the nth sleep cycle, the sleeper is annihilated, and a fresh brain is prepared and memories will be copied into it from your nth doppelganger. Since these memories don’t come from you, the resulting brain isn’t yours.)

And in the variant story, let’s ask the questions (10) and (11) again. What will the answers be? Again, it’s easy and obvious: zero and zero. You won’t be shown any shapes, because you will be annihilated in your sleep before any shapes are shown.

Now consider Everettian branching quantum mechanics. Suppose there is a quantum process that will result in your going to sleep in an equal superposition of states between having a red square, a green triangle and a green circle in front of your head, so that upon waking up an observation of the shape will be made. Now ask questions (10) and (11) again.

I contend that this is just as easy as in my classical universe story. Either the branching preserves personal identity or not. If it preserves personal identity, the answer to the questions is one and one. If it fails to preserve personal identity, the answer to the questions is zero and zero. The only relevant ontological difference between the quantum and classical stories is that in the quantum stories the wakeups might count as simultaneous while in the classical story the wakeups are sequential. And that really makes no difference.

In none of the four cases—the classical story with or without personal identity and the branching story with or without personal identity—are the answers to the questions 2/3 and 1/3. But those are in fact the right answers in the quantum case, contrary to the Everett model.

Now, one might object that we care more about decisions than predictions. Suppose that you have a choice between playing a game with one of two three-sided fair quantum dice:

• Die A is marked: red square, green triangle, green circle.

• Die B is marked: green square, red triangle, red circle.

And suppose pain will be induced if and only if the die comes up red. Which die should you prudentially choose for playing the game? Again, it depends on whether personal identity is preserved. If not, it makes no difference. If yes, clearly you should go for die A on the Everett model—and that is indeed the intuitively correct answer. But the reason for going for die A on the Everett model is different from the reason for going for it on a non-branching quantum mechanics. On the Everett model, the reason for going for die A is that it’s better to get pain once (die A) rather than twice (die B).

So far so good. But now suppose that you’ve additionally been told that if you go for die A, then before you roll A, an irrelevant twenty-sided die will be rolled. (This is a variant of an example Peter van Inwagen sent me years ago, which was due to a student of his.) Then, intuitively, if you go for die A, there will be twenty red branches and forty green branches on Everett. So on die A, you get pain twenty times if personal identity is preserved, and on die B you get pain only twice. And so you should surely go for die B, which is absurd.

One might reasonably object that there are in fact infinitely many branches no matter what. But then on the no-identity version, the choice is still irrelevant to you prudentially, while on the identity version, no matter what you do, you get pain infinitely many times no matter what you choose. And that doesn’t work, either. And if there is no fact about how many branches there will be, then the answer is just that there is no fact about which option is preferable on the identity version, and on the no-identity version, indifference still follows.

This is all basically well-known stuff. But I like the above way of making it vivid by thinking about classically sequentializing the story.

Nagy Zsolt said...
This comment has been removed by the author.
Nagy Zsolt said...

One might reasonably object that there are in fact infinitely many branches no matter what. But then on the no-identity version, the choice is still irrelevant to you prudentially, while on the identity version, no matter what you do, you get pain infinitely many times no matter what you choose. And that doesn’t work, either.

And how is that so exactly?!?
I think in the infinitely many branches case, I would rather play with and prefer the die where the ratio of branches with pain is less to branches without pain: pain/not-pain<1
Sure, one might calculate that with one metric like cardinality given there to be a one-to-one correspondence between branches with pain and branches without pain (Measurable cardinal). But how is that exactly the correct metric here or what we want to measure?!?
If there is a statistical distribution of branches with pain and branches without pain, then the correct metric and measurement of the previous ratio pain/not-pain is to be calculated with the Lebesgue measure obviously.

Nagy Zsolt said...
This comment has been removed by the author.
Nagy Zsolt said...

Because of "personal identity" or to really say, that because of "subjective reality"?!?
What? I don't get it.
I mean, if I were to be asking (M10) "How likely is it that you will be shown a green shape?" and (M11) "How likely is it that you will be shown a red shape?"
and that person from that scenario might response and answer with:
(Y12) "The answers to these questions are obviously: one and one. I will be shown a green shape twice and a red shape one, and that’s certain."
Then my response would be:
(M13) "First of all it is quite remarkable, how you can answer my questions, when you have been permanently annihilated given (9). Or "will" that be the case? Ups. Spoiler Alert!
Second of all, "One and one." are not even considered answers to my questions. So I reiterate my questions here:
(M10) "How LIKELY is it that you will be shown a green shape?"
(M11) "How LIKELY is it that you will be shown a red shape?"
Third of all, how can you know, that "[You] will be shown a green shape twice and a red shape one, and that’s certain.", when your memory has been actually reseted two times given (4) and (7)? Or again those cases "will" be the case? Ups. Another Spoiler Alerts!"

That person from that scenario again:
(Y14) "To your second point: Ahh. Now I get it. It will depend on the distribution of shown green and non-green shapes as it will depend on the distribution of shown red and non-red shapes, how LIKELY it will be, that I will be shown a green shape and/or a red shape. Of course. How could I ever forget being a frequentist. Duh.
To your first and third point: What?!? And I don't know, maybe?!? I'm just a character and imagination of a person, maybe?!? Maybe ask the narrator, why is he so inconsistent with his narration? Again, I don't know."

I have so many questions. I guess the most important one from me is the following:
What exactly is "personal identity"? And how exactly does that relate to an objective reality?
Your example doesn't explain, what exactly "personal identity" is supposed to be.
I guess, that wasn't your intent. Fair enough! You can assume for your given example, that the reader knows exactly, what "personal identity" is supposed to be. But it appears to me, that you think, that your given example somehow explains or shows, how "personal identity" can relate or is related to an objective reality.
Am I correct with this impression of mine? Is that given example/scenario of yours explaining or demonstrating that?!?
If so, then I'm so confused.
How is "One and one. You will be shown a green shape twice and a red shape one, and that’s certain." such an "obvious" and trivial or rational answer to anything here?!?

Nagy Zsolt said...

Imagine the following unfortunate events happening to Fred simultaneously at the current moment:

(1) Fred has never married before. Hence, Fred is currently a bachelor.
(2) Fred is marrying at the current moment. Hence, Fred is married currently.

So obviously Fred identifies himself to be a "married bachelor" at this current moment, because one green shape and one red shape equals to be two shown green shapes and one shown red shape.
I mean, that's the only rational thing, which can be thought of here.

Alexander R Pruss said...

Personal identity is easy. There is personal identity between x and y iff x is a person, y is a person and x is the same entity as y. :-)

I don't see any reason to measure with Lebesgue measure here. Take a non-quantum example. Suppose you're a two-dimensional being and are about to fission into continuum many two-dimensional future selves, stacked along the third dimension between z=0 and z=1 (each future self to one z-value).

Option 1: The future selves between z=0 and z=0.6 are in pain and the future selves between z=0.6 and z=1 are OK.

Option 2: The future selves between z=0 and z=0.6 are OK and the future selves between z=0.6 and z=1 are in pain.

By Lebesgue measure, you should prefer option 2, since the measure of those that are OK in Option 2 is 0.6 and the measure of those that are OK in Option 1 is 0.4. But why should the _position_ of the selves matter at all? I can see how _counting_ them matters, but then on both options equal cardinalities are OK and in pain.

Now imagine this. As soon as the pains are induced, the selves are moved, so that the self at coordinate z gets moved to z^2. After the movement:

Option 1: Pain is had by those now at coordinates between z=0 and z=0.36.

Option 2: Pain is had by those now at coordinates between z=0.36 and z=1.

Now it looks like Option 1 is better by Lebesgue measure. But simply moving the slices around doesn't actually change anything important! So the Lebesgue measure doesn't matter.

Zsolt Nagy said...

Ah, so that's "personal identity".
And how does that amount or relate to an objective perspective of the actual world?!? ;-)

Did you know, that once something happened in the past, then that's fixed.
Sure. One might be able to move points on the space axes from left to right and from right to left or from z to z². But how does that exactly supposed to work on the time axes?!?
Besides that yes, there are reasons to use Lebesgue measure - to be more precise there is a century, the last century amount of physics and confirmed hypothesis of quantum mechanics relying on the Lebesgue measure - a lot of reasons for using Lebesgue measure.
I know, you are more of an a priori knowledge or prime face guy. You know, I'm more of an a posteriori knowledge guy - I can not simply ignore such theories resulting in confirmed results and predictions physically measured to such a high precision and accuracy - to the most precision and accuracy of science history to this current date. That's not my thing.

Zsolt Nagy said...

Besides quantum mechanics is all about LINEAR functions AND the Lebesgue measure is an invariant measure, the invariant measure, for linear transformations.
I very much think so, that this is a very good reason to consider the Lebesgue measure especially for quantum mechanics. Maybe not for your non-linear example and transformations, not analogous to quantum mechanics in any given way but for quantum mechanics the Lebesgue measure should do the trick.
Wouldn't you also think that, Alexander?