What happens to the wave function if...

  • #1
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The question is;

In an experimental small universe, a photon is released from a source. It continues its path as a probabilistic wave function. if it interacted with mass, we could say the wave function collapsed and observe a particle photon hitting an object.

But what happens when the photon never hits with any mass(object) until it reaches the Schwarzschild radius of the source and can no longer transmit information to the source.

The photon never existed in our experimental universe then? A superposition that never actualized?
 

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  • #2
PeterDonis
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what happens when the photon never hits with any mass(object) until it reaches the Schwarzschild radius of the source
If the source is a black hole, it can't emit a photon to begin with. If the source is an ordinary object like a star, it has no horizon, so "reaches the Schwarzschild radius" is meaningless. (If you calculate the Schwarzschild radius of, say, the Sun, you will find that it is much, much smaller than the actual radius of the Sun--which is just another way of saying that the Sun has no horizon.) So your scenario is meaningless and your questions about it can't be answered because they aren't well-defined to begin with.

You might want to take a step back and think about what your actual issue is, instead of trying to construct scenarios that might not be well-defined.
 
  • #3
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But what happens when the photon never hits with any mass(object) until it reaches the Schwarzschild radius(observable universe) of the source and can no longer transmit information to the source.

It has not yet been confirmed by the scientific community that the Schwarzschild radius also gives the radius of the observable universe at a given universe when you take *your* observable universe's total mass. I am just rushing ahead in my head, sorry. But I doubt it was that important to dwell on.
 
  • #4
PeterDonis
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what happens when the photon never hits with any mass(object) until it reaches the Schwarzschild radius(observable universe) of the source and can no longer transmit information to the source
The term "observable universe" is better; better still would be "cosmological horizon", since that is the actual boundary between the region that can send light signals to us and the region that can't. The observable universe is the region that has sent light signals in the past that have reached us up to now, which is not the same.

The answer to your question depends on what interpretation of QM you adopt. Some interpretations say that the wave function (note that, strictly speaking, we should be using quantum field theory to model photons, in which the term "wave function" is somewhat problematic, but I'll ignore that here) is physically real, so we can use it to describe a quantum system even if it never interacts with anything else. Other interpretations say the wave function only models the probabilities of results of measurements that could be made on the system, so there is no point in even assigning a wave function to a system that never gets observed or measured. We have no way of deciding which interpretation is correct at our current state of knowledge, since they all make the same predictions for all observations.

It has not yet been confirmed by the scientific community that the Schwarzschild radius also gives the radius of the observable universe
No, it is well known in the scientific community that the universe as a whole does not have a Schwarzschild radius because it is not an isolated system surrounded by vacuum, which is the only kind of system to which the term "Schwarzschild radius" even applies. The spacetime geometry of the universe is very different, and the observable universe is not the same as a black hole horizon or a Schwarzschild radius.
 
  • #5
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Can you share the name of the interpretations so that I can check them out in detail?

Cosmological horizon yes, that's the correct word. The use of the Schwarzschild radius is a bad definition like you have put. I had that problem and now consider it like a disability when I am trying to say something and use a word to define it, I assume everybody understands what I am thinking in my head. Of course, I am just trying to mean the equation instead of the definition of the term.

I have given up a few years ago and just accepted I am not good at communicating, Instead of getting frustrated and stressed lol.
 

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