# Why are there no probability amplitudes in MWI?

• I
• entropy1
In summary: If there is an appropriate interaction Hamiltonian between the two systems (the thing whose basis states are ##A## and ##B##, and the thing whose basis states are the ##E## states), yes. And this is true regardless of what complex amplitudes you put in front of the ##A## and ##B## terms: they just get carried through the process since the process is unitary.
entropy1
If I consider the MWI, one of the notions for what happens during measurement is that the initial wavefunction, if I use Dirac notation and two dimensions, ##|A\rangle+|B\rangle## undergoes the transformation ##(|A\rangle+|B\rangle)|E_{before}\rangle \rightarrow |A\rangle|E_{after}\rangle+|B\rangle|E_{after}\rangle##. So would it then also be correct that we can include the probability amplitudes, to get for instance this?: ##(a|A\rangle+b|B\rangle)|E_{before}\rangle \rightarrow a|A\rangle|E_{after}\rangle+b|B\rangle|E_{after}\rangle##? It seems to me the latter is not correct, because in MWI there are no amplitudes used. Why is that?

entropy1 said:
in MWI there are no amplitudes used
Yes, they are. MWI uses the same math as all QM interpretations, and that math has amplitudes.

Paul Colby and entropy1
PeterDonis said:
Yes, they are. MWI uses the same math as all QM interpretations, and that math has amplitudes.
Do my example formulas in #1 agree for the large part with the math you are referring to, or are they far off?

entropy1 said:
Do my example formulas in #1 agree for the large part with the math you are referring to
I'm not sure what you are trying to describe with the math in the OP. But the basic math of QM is described in any QM textbook.

entropy1 said:
##|A\rangle+|B\rangle## undergoes the transformation ##(|A\rangle+|B\rangle)|E_{before}\rangle \rightarrow |A\rangle|E_{after}\rangle+|B\rangle|E_{after}\rangle##.
Did you intend the two ##|E_{after}\rangle##'s to be different? If they're the same, what you're writing does not represent any interaction at all.

Why are there no probability amplitudes in MWI?

I notice that you often start threads with a statement that isn't true. I can't believe this is the most effective way of learning. As Peter said, MWI uses the same math as all QM interpretations.

PeterDonis said:
Did you intend the two ##|E_{after}\rangle##'s to be different? If they're the same, what you're writing does not represent any interaction at all.
Yes, sorry, that is an error. It should be _afterA and _afterB, or _measuredA and _measuredB. Is the formula correct if these errors are corrected?

I guess I have to start with that textbook or go study something. Thanks so far.

entropy1 said:
Is the formula correct if these errors are corrected?
If there is an appropriate interaction Hamiltonian between the two systems (the thing whose basis states are ##A## and ##B##, and the thing whose basis states are the ##E## states), yes. And this is true regardless of what complex amplitudes you put in front of the ##A## and ##B## terms: they just get carried through the process since the process is unitary.

## 1. Why are there no probability amplitudes in MWI?

The Many-Worlds Interpretation (MWI) of quantum mechanics does not assign probabilities to different outcomes because it assumes that all possible outcomes of a quantum event actually occur in different parallel universes. Therefore, there is no need for probabilities to determine the likelihood of a specific outcome in a single universe.

## 2. How does MWI explain the collapse of the wave function without using probability amplitudes?

In MWI, the collapse of the wave function is seen as an illusion caused by the observer's limited perspective. Instead of the wave function collapsing to a single outcome, it branches off into multiple parallel universes, each containing a different outcome. This eliminates the need for probability amplitudes to explain the collapse of the wave function.

## 3. Can MWI be tested or proven without the use of probability amplitudes?

MWI is a philosophical interpretation of quantum mechanics and cannot be proven or disproven through experimentation. However, it is a mathematically consistent interpretation that does not require the use of probability amplitudes, making it a valid alternative to other interpretations.

## 4. Why do some scientists reject MWI due to the lack of probability amplitudes?

Some scientists reject MWI because it goes against the traditional interpretation of quantum mechanics and lacks empirical evidence. Additionally, the lack of probability amplitudes makes it difficult to make predictions and test the theory, which goes against the scientific method.

## 5. Are there any potential consequences of accepting MWI without probability amplitudes?

One potential consequence of accepting MWI without probability amplitudes is that it challenges our understanding of causality and the concept of free will. It also raises questions about the nature of reality and the existence of multiple parallel universes. However, these consequences are philosophical in nature and do not affect the validity of the theory itself.

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