Physics, asked by Mandlik3370, 1 year ago

Can a measurement partially “collapse” a wavefunction?

Answers

Answered by Anonymous
0

Let's say I have a wavefunction ΨΨ which can be decomposed into a sum of it's energy eigenstates:  Ψ=a|1⟩+b|3⟩+c|8⟩+d|10⟩ Ψ=a|1⟩+b|3⟩+c|8⟩+d|10⟩ Where, of course, |a|2+|b|2+|c|2+|d|2=1|a|2+|b|2+|c|2+|d|2=1.  And let's say I have a device which can measure the energy of this wavefunction. Unfortunately, the device has an inherent uncertainty of ±3±3.  I measure ΨΨ and find it to have an energy of 7±37±3. After my measurement the wavefunction has "collapsed" (to some extent?). I can think of a few possibilities for post-measurement ΨΨ:  1) ΨΨ really is in either |8⟩|8⟩ or |10⟩|10⟩. The problem statement is wrong: any uncertainly in energy measured is a laboratory issue. It must have an exact physical answer.  2) Ψ=c′|8⟩+d′|10⟩Ψ=c′|8⟩+d′|10⟩ Where, |c′|2+|d′|2=1|c′|2+|d′|2=1.  I might even go so far as to say c′=c/(|c|2+|d|2)12c′=c/(|c|2+|d|2)12 My immediate answer: the measurement simply eliminated the possibility of the |1⟩|1⟩ and |3⟩|3⟩ eigenstates.  3) Ψ=e|4⟩+f|5⟩+g|6⟩+h|7⟩+k|8⟩+m|9⟩+n|10⟩Ψ=e|4⟩+f|5⟩+g|6⟩+h|7⟩+k|8⟩+m|9⟩+n|10⟩ The "measurement" isn't really a measurement; just a disruption to the wavefunction.

Answered by GhaintMunda45
0

Hey !

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Can a measurement partially “collapse” a wavefunction?

Collapse in wave function

♣ Many theory cannot prove it.

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Thanks !

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