Finite Well: Measuring Bound Particles w/ E > 0?

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In a finite potential well, a bound particle with energy less than zero can theoretically be measured with energy greater than zero due to the nonzero probability of finding it outside the well. However, measuring the particle's position alters its state to a position eigenstate, which can lead to a subsequent measurement of energy yielding a positive value. This indicates that the measurement process can effectively "knock" the particle out of the well. The discussion emphasizes the distinction between energy eigenstates and position eigenstates in quantum mechanics. Ultimately, the measurement process plays a crucial role in determining the energy state of the particle.
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If you constructed a wavepacket for the finite well representing a bound particle i.e. one with <E> < 0, would you ever measure it with E > 0?

The finite well we are discussing has a potential 0 except between -a and a where it is -V_0.

I would say yes because there is nonzero probability that the particle could be found outside the well i.e. |psi(x)|^2 is nonzero for x > |a|. But I am not sure whether that implies that the energy is greater than 0 as well?
 
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You would be wrong. If you have an energy eigenstate and you measure the energy, you get the same energy. If you measure the position then you change the eigenstate to a position eigenstate. And yes, then you can measure the energy and get E>0. In this case the measurement process has knocked the particle out of the well.
 
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