On 28-04-2022 07:51, Bruce Kellett wrote:

On Thu, Apr 28, 2022 at 3:24 PM Brent Meeker <meekerbr...@gmail.com> wrote:On 4/26/2022 5:32 PM, smitra wrote:On 27-04-2022 01:37, Bruce Kellett wrote:Changing the weights of the components in the superposition does not change the conclusion of most observers that the actual probabilities are 0.5 for each result. This is simple mathematics, and I am amazed that even after all these years, and all the times I have spelled this out, you still seek to deny the obvious result. Your logical and mathematical skill are on a par with those of John Clark. It's indeed simple mathematics. You apply that to branch counting to arrive at the result of equal probabilities.I have not used branch counting. Please stop accusing me of that.

`You are considering each branch to have an equal probability when there`

`is no logical reason to do so, and when that's also being contradicted`

`by QM.`

'

So, the conclusion has to be that one should not do branch counting. The question is then if this disproves the MWI. If by MWI we mean QM minus collapse then clearly not. Because in that case we use the Born rule to compute the probability of outcomes and assume that after a measurement we have different sectors for observers who have observed the different outcomes with the probabilities as given by the Born rule.In which case the Born rule is just an additional arbitrary assumption: it is not part of the Schrodinger equation. Your theory of QM minus collapse is not well-defined. You simply take whatever you want from text-book quantum mechanics, with no regard to the consistency of your model.

`QM includes the Born rule. QM minus collapse is just that: QM minus`

`collapse. It's not QM minus collapse minus the Born rule.`

You then want to argue against that by claiming that your argument applies generally and would not allow one to give different sectors unequal probabilities. But that's nonsense, because you make the hidden assumption of equal probabilities right from the start.I simply assume the Schrodinger equation. Then, following Everett, we take it to be deterministic, so that all branches occur on every trial. Since it is deterministic, there is no concept of probability inherent in the Schrodinger equation, and I do not assume any definition of probability. So the branches occur as they occur, there is no assumption of equal probability. It is just that the construction means that all 2^N branches occur on the same basis and necessarily count equally in the overall branching picture.

`Why do they necessarily count equally? What is the meaning of the`

`wavefunction? Why don't the amplitudes matter?`

There is nothing in QM that says that branches must count equally, and the lottery example I gave makes it clear that you can have branching with unequal probabilities in classical physics.As I have said, there is no classical analogue of an interaction in which all outcomes necessarily occur. So your lottery example is useless. There is no concept of probability involved in any of this.

`The lottery example I gave clearly is a classical example in which all`

`outcomes necessarily occur. Your reasoning does not involve any QM at`

`all, you just apply it to the MWI. Your argument goes through also in`

`case of the lottery example, in which case it leads to an obviopusly`

`wrong conclusion. So, it's your reasoning that's at fault not the MWI`

`taken to be QM minus collapse.`

Saibal

BruceYes, there's nothing in QM that says the branches must count equally. But there's also nothing in the evolution of Schroedingers equation that they must count as _a^2_ and _b^2_. Of course IF they are probabilities then it follows from Gleason's theorem that they follow the Born rule. But in that case you have reintroduced almost all the philosophical problems of the Copenhagen interpretation. When exactly does this splitting occur? Can the split be into irrational numbers of branches? A splitting is in some particular basis and not in other bases. What determines the pointer basis? Brent-- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLTBdJpDkw_duZDuvMvArLte-3OoxJcs8-3vXjroKSti8g%40mail.gmail.com [1]. Links: ------ [1] https://groups.google.com/d/msgid/everything-list/CAFxXSLTBdJpDkw_duZDuvMvArLte-3OoxJcs8-3vXjroKSti8g%40mail.gmail.com?utm_medium=email&utm_source=footer

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