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B What is the difference between MWI and CI?

  1. Jul 7, 2016 #1
    Suppose we measure a normalized state ##|\Psi \rangle = \alpha _0 | \lambda _0 \rangle + \alpha _1 | \lambda _1 \rangle + \alpha _2 | \lambda _2 \rangle + ...## with ##| \lambda _i \rangle## the eigenvalues of the measured observable. Is it true that, in the CI, the wavefunction collapses into an arbitrary eigenvector ##| \lambda _i \rangle## with probability ##\alpha _i^* \alpha _i##, while in MWI, all possible 'collapses' ##| \lambda _i \rangle## are realized simultaneously in separate branches?

    Which leads me to the question: what happens to the probability ##\alpha _i^* \alpha _i## in MWI?
     
    Last edited: Jul 7, 2016
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  3. Jul 7, 2016 #2

    Nugatory

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    Informally, it's the probability that you end up in the branch corresponding to ##|\lambda_i\rangle##. That's why it is impossible to tell the difference between CI and MWI - either way, you find yourself with one of the possible outcomes according to the Born rule probabilities.

    (That's "informally" - there are some unpleasant mathematical difficulties in applying probabilities in this way, but this is not an "A" level thread so we don't have to go there).
     
  4. Jul 8, 2016 #3
    Suppose we measure an observable with N eigenstates ##\lambda _{0..N-1}##. In MWI, all possible eigenstates ##|\lambda _i \rangle## become realized, right? So, suppose we conduct M measurements. In MWI, if we get as a first outcome ##\lambda _0## in some branch, then, if the probability to get ##\lambda _0## is fixed (determined by the measurement setup), then yielding ##\lambda _0## must decrease the probability (the number of times) of getting it again in that tree of branches, right?
     
  5. Jul 8, 2016 #4

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    No, that is completely wrong.

    Our first measurement causes a branching. We end up in the branch in which the wave function is ##|\lambda_0\rangle## (Everyone watching.... Please remember that this is a B-level thread and refrain from expressing your altogether justifiable disdain for this description) because our measurement yielded the result ##\lambda_0##. There are ##N-1## other branches as well, but we aren't in them so they are irrelevant to any future measurements that we make - we have no way of even knowing whether they exist. So now we go to make our second measurement. We're in the ##|\lambda_0\rangle## branch, so that's what the wave function is in our universe, and our second measurement will yield exactly the same results as if the wave function had collapsed to ##|\lambda_0\rangle## and there were no other universes.
     
    Last edited: Jul 8, 2016
  6. Jul 8, 2016 #5
    So, is it then the relative number of universes realized with outcome ##\lambda _0## as compared to the number of the other ones that reflects the probability ##\alpha^*_0 \alpha_0## of getting ##\lambda _0##?
     
    Last edited: Jul 8, 2016
  7. Jul 8, 2016 #6
    Suppose we prepare a particle in state ##\alpha_0 | \lambda_0 \rangle + \alpha_1 | \lambda_1 \rangle## with ##| \lambda_{0,1} \rangle## eigenvectors of observable O. Suppose ##\alpha^*_0 \alpha_0## is close to 1, and we measure observable O. Then the probability of measuring ##\lambda_0## is close to 100% and that of measuring ##\lambda_1## is close to 0%, right?

    So if, in MWI, if we repeat this experiment repeatedly with exactly the same preparation, each time both ##| \lambda_0 \rangle## as well as ##| \lambda_1 \rangle## become realized. So, doesn't this lead to the conclusion that ##| \lambda_0 \rangle## and ##| \lambda_1 \rangle## both have the same probability of occuring, namely 100%? It seems as if the probability got lost. How does it fit the theory?
     
  8. Jul 8, 2016 #7
    Re post #5. Wouldn't that require an infinite number of universes?
     
  9. Jul 8, 2016 #8
    I guess that depends on whether the number of measurements done in the lifespan of the universe is finite of infinite.
     
  10. Jul 8, 2016 #9

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    Collapse interpretations assume, as a postulate, that the probability of collapsing to ##|\psi_i\rangle## is ##\alpha_i\alpha_i^*##.
    The equivalent assumption in MWI is that the probability of ending up in the universe corresponding to ##|\psi_i\rangle## is ##\alpha_i\alpha_i^*##.

    (But please remember the cautionary note in #2 of this thread).
     
  11. Jul 8, 2016 #10

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    No, for the reason that you give in #6. But please remember the cautionary note in #2 of this thread.
     
  12. Jul 8, 2016 #11
    I don't get it. In both branches, everything exists, right? (with the caveat of the difference in measurement outcome) So everything ends up in both branches, right?
     
  13. Jul 8, 2016 #12

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    Not right.

    We start with a wave function that looks like ##\alpha_0|\psi_0\rangle+\alpha_1|\psi_1\rangle## at the moment of measurement. Collapse interpretations say that it collapses to one of the two possibilities and that's the only world there is. MWI says that it describes two worlds, completely disconnected and unable to interact, one for the ##|\psi_0\rangle## case and one for the ##|\psi_1\rangle## case.

    Meanwhile, the math of quantum mechanics says that there's no particular reason to take either picture seriously. You can accept either, both, or neither... and no matter what, you will get one result with probability ##\alpha_0\alpha_0^*## and the other result with probability ##\alpha_1\alpha_1^*##. There aren't any worlds or collapses here, just a simple statement of the probability of getting a particular result.
     
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