Wavefunction Collapse

  • Thread starter jewbinson
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  • #1
jewbinson
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Simple question.

So the energy of a particle is observed to be E_1 (for example) at time t=0.

At time t=0 the wavefunction psi(x) collapses to phi(x)exp(-i(E_1)t/h). At time t>0 the wavefunction is also in this state (right?). Is it in this state until it interacts with another particle or what? Or does it go into the state phi(x)exp(-i(E_1)t/h) when you observe it but then after it goes back into the state psi(x)? As far as I am aware it collapses into the state phi(x)exp(-i(E_1)t/h) for t=0 and stays in that state (but until when?) for t>0.
 

Answers and Replies

  • #2
cbetanco
133
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Assuming no other interactions occur with the outside world, the state will state in the energy eigenstate you measured it in (up to an unimportant phase). This is because once you measure its energy, you force it into an energy eigenstate, which only change their phase when evolved with shrodingers equation
 
  • #3
jewbinson
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Thanks.

"(up to an unimportant phase)"... it's probably not unimportant is it? But I imagine it's more advanced than what I am studying now which is why you have put it to one side...
 
  • #4
cbetanco
133
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You can write any state as [itex]\Psi[/itex]=[itex]\psi[/itex] e, where [itex]\psi[/itex] and α are real. So when you calculate the probability distribution (the thing that is actually phzsically significant) you get [itex]\Psi[/itex]*[itex]\Psi[/itex]=[itex]\psi[/itex][itex]\psi[/itex]e-iαe=[itex]\psi[/itex][itex]\psi[/itex]. That is what I mean when I say the phase e is unimportant.
 
  • #5
cbetanco
133
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actually, I was wrong, [itex]\psi[/itex] does not need to be real, and in the end zou get [itex]\psi[/itex]*[itex]\psi[/itex]
 
  • #6
jewbinson
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So photons remain undisturbed (in the same state) if they don't interact with matter? Does "disturbed" include a change of a field or a change in density of an object? Would a photon ejected from a star/galaxy in red-shift relative to Earth traveling freely in space stay in the same state if it does not experience a force? If so, is this what we rely on when measuring red-shift - that the vast majority of photons from a source go through space not interacting with matter or any other field (I imagine the gravitational field has little effect on a photon actually because they have such minute mass)...?
 

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