Undergrad Why are there no probability amplitudes in MWI?

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SUMMARY

The discussion centers on the role of probability amplitudes in the Many-Worlds Interpretation (MWI) of quantum mechanics. Participants clarify that MWI employs the same mathematical framework as other quantum mechanics interpretations, which includes probability amplitudes. The transformation of wavefunctions, represented in Dirac notation, is confirmed to be valid as long as the interaction Hamiltonian is appropriately defined. Errors in the initial formulas regarding the states after measurement were acknowledged and corrected for clarity.

PREREQUISITES
  • Understanding of Dirac notation in quantum mechanics
  • Familiarity with quantum mechanics (QM) textbooks
  • Knowledge of interaction Hamiltonians in quantum systems
  • Basic concepts of unitary transformations in quantum mechanics
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  • Study the mathematical framework of quantum mechanics, focusing on probability amplitudes
  • Learn about the Many-Worlds Interpretation and its implications in quantum mechanics
  • Explore the role of interaction Hamiltonians in quantum state transformations
  • Investigate unitary processes and their significance in quantum mechanics
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Quantum physicists, students of quantum mechanics, and anyone interested in the mathematical foundations of the Many-Worlds Interpretation.

entropy1
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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?
 
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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.
 
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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. :wink:
 
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.
 

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