Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Collapse Wavefunction

  1. May 1, 2014 #1
    After a measurement a wavefunction collapses.

    You measure the position of a particle, the particles assumes a definite position, let's say point C.

    The coefficient of C is let's say c, so the probability that it takes that value is c²
    Wavefunction collapses at C (actually in the vicinty of C ?)
    A second measurement immediately after the first, you would have to yield C again. Because a physical experiment has to be reproducible.
    So, immediately right after your collapse, you get C with a probability of 100% because it just collapsed.

    My question is:

    If you don't measure right immediately after your first measurement, the wavefunctions spreads out again, according to the original Schrodinger equation right?

    If you perform the same measurement again on your particle, would it yield C again with probability 100% or c²

    Please make your answer as simple as possible. I don't want to be bothered with Quantum Decoherence or Bell's paradox or Dirac Brakets at this point right now.

    First chapter Griffiths and I got some basic notions of QM, but not that much. :)
  2. jcsd
  3. May 1, 2014 #2


    User Avatar

    Staff: Mentor

    For position, and anything else that doesn't commute with the Hamiltonian, the answer is "neither". The wave function evolves according to Schrodinger's equation starting from the initial post-measurement state, and there's no particular reason why that evolution should return it to the pre-measurement state in which the probability of getting C was ##|c^2|##.

    For things that do commute with the Hamiltonian, you get C with probability 100%.

    (Be aware that wave function collapse isn't necessarily something that really happens. It's an interpretation, one way of thinking about what the math is telling us. It's not the only way, and it's not everyone's favorite way. It does work really well for visualizing this particular situation though).
    Last edited: May 1, 2014
  4. May 1, 2014 #3


    User Avatar
    2017 Award

    Staff: Mentor

    That depends on the interpretation.
    At whatever your measurement gives. No position measurement is exact.

    Without delay? Then 100%. With delay? Then it depends on the delay, the system and everything else - it depends on how the wavefunction evolves.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook