5 ms·

> if it interacted with a force and no observation was made, the wave function doesn’t collapse, does it? Any two systems interacting will cause the collapse.

• strogonoff 6y ago
  > Any two systems interacting will cause the collapse. I suppose that means if a photon, say, is reflected by a mirror, that should collapse its wave function and any measurements after that should not have any effect on it? Maybe it should be qualified what kind of interaction collapses wave function? > I'm a software developer, not a quantum physicist. Great, I’m not a quantum physicist either—yet here we are, talking about quantum physics!
  • pault 6y ago
    > I suppose that means if a photon, say, is reflected by a mirror, that should collapse its waveform and any measurements after that should not have any effect on it? I'm not sure if that example is the right one to use, but yes, that's roughly my understanding.
    • strogonoff 6y ago
      Okay. I actually thought the paradox was that wave collapses only if the photon is influenced by measurement activity but I’s been a while since I was reading up on this topic so I may need a refresher. I suppose if we can calculate exactly how a given force would influence a photon, that would be essentially the same as “measuring” it.
      • sdenton4 6y ago
        Turns out you need to run your quantum computer in near-zero-kelvin conditions to avoid accidental 'observations' from collapsing the precious waves you're using to factorize the number 6.
• simonh 6y ago
  That’s not what Schroedinger’s equation says though, it can combine the wave functions of two quantum systems, say a particle and a force, and show how they evolve with no problem, in fact that literally what it does. Quantum mechanics simply doesn’t have a complete account of collapse to a specific state.