Imagine this unfortunate sequence of events will certainly befall you
in a classical universe:
You will be made to fall asleep.
Upon waking up, you will be shown a red square.
You will be made to fall asleep again.
While asleep, your memory will be reset to that which you had in
step (1).
Upon waking up, you will be shown a green triangle.
You will be made to fall asleep for a third time.
While asleep, your memory will be reset again to that which you
had in step (1).
Upon waking up, you will be shown a green circle.
You will then be permanently annihilated.
Questions:
How likely is it that you will be shown a green shape?
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.