3 ms·

Yes there are many perspectives to take. The basic problem with your suggestion is that it does not explain how anything actually "happens". Unitary evolution

Yes there are many perspectives to take. The basic problem with your suggestion is that it does not explain how anything actually "happens". Unitary evolution is a theory where nothing changes. People seem to think that they can evoke decoherence, or similar, and these issues disappear. But they do not. Anyway, I'm not positing any solution, I am just saying that it irks me that all these ideas are dismissed whenever it comes to the problem of the asymmetry of time. It just seems bizarre to me.

antidesitter 8y ago
What do you mean by “unitary evolution is a theory where nothing changes”?
- mathgenius 8y ago
  For any two state vectors, |u> and |v>, unitary evolution preserves the amplitude <u|v>: <u|U^\dag U|v> = <u|v>.
  - monktastic1 8y ago
    Sure, if U operates on both |u> and |v>, their inner product doesn't change. But why does this mean that "nothing ever changes?" |u> can evolve independently of |v>, after all. OTOH, I can understand the argument that nothing ever happens. One reasonable requirement for something to happen is that it should not be able to "unhappen," which is why some are looking for the possibility of irreversible decoherence.
    - mathgenius 8y ago
      " |u> can evolve independently of |v> " This is not unitary. Quantum theory has two parts: unitary evolution, and measurement processes. That's it.
      - monktastic1 8y ago
        Eh? If U is a unitary operator that can act on |u>, then U @ I is a unitary operator on the state |u> @ |v>, producing U|u> @ |v>. Can you explain what you mean?
  - antidesitter 8y ago
    Doesn’t that assume |v> evolves the same way |u> does (according to U)?
    - mathgenius 8y ago
      Anything else is not unitary evolution.
      - antidesitter 8y ago
        What do you mean? Let u and v be qubits. Apply a Hadamard gate to u while leaving v unchanged. Then they have evolved independently (but unitarily): u according to H, and v according to I.

antidesitter 8y ago

What do you mean by “unitary evolution is a theory where nothing changes”?

mathgenius 8y ago
For any two state vectors, |u> and |v>, unitary evolution preserves the amplitude <u|v>: <u|U^\dag U|v> = <u|v>.
- monktastic1 8y ago
  Sure, if U operates on both |u> and |v>, their inner product doesn't change. But why does this mean that "nothing ever changes?" |u> can evolve independently of |v>, after all. OTOH, I can understand the argument that nothing ever happens. One reasonable requirement for something to happen is that it should not be able to "unhappen," which is why some are looking for the possibility of irreversible decoherence.
  - mathgenius 8y ago
    " |u> can evolve independently of |v> " This is not unitary. Quantum theory has two parts: unitary evolution, and measurement processes. That's it.
    - monktastic1 8y ago
      Eh? If U is a unitary operator that can act on |u>, then U @ I is a unitary operator on the state |u> @ |v>, producing U|u> @ |v>. Can you explain what you mean?
- antidesitter 8y ago
  Doesn’t that assume |v> evolves the same way |u> does (according to U)?
  - mathgenius 8y ago
    Anything else is not unitary evolution.
    - antidesitter 8y ago
      What do you mean? Let u and v be qubits. Apply a Hadamard gate to u while leaving v unchanged. Then they have evolved independently (but unitarily): u according to H, and v according to I.