Can a measurement partially “collapse” a wavefunction?
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Hey mate ^_^
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 wave function....
Unfortunately, the device has an inherent uncertainty of ±3±3....
#Be Brainly❤️
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 wave function....
Unfortunately, the device has an inherent uncertainty of ±3±3....
#Be Brainly❤️
Answered by
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Hello mate here is your answer.
the probability of collapsing the system to state.In these cases, the wave function will partially collapse to a linear combination of "close" eigenstates (necessarily involving a spread in eigenvalues) that embodies the imprecision of the measurement apparatus.
Hope it helps you.
the probability of collapsing the system to state.In these cases, the wave function will partially collapse to a linear combination of "close" eigenstates (necessarily involving a spread in eigenvalues) that embodies the imprecision of the measurement apparatus.
Hope it helps you.
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