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QM - finite square well

In the problem of a finite potential well, we search for bound states, in which E<V.
Say the potential is defined to be 0 outside of the well, and -V0 inside it.
Analyzing the case when -V0 < E < 0, one finds that the probability of finding the particle outside the well is not zero. This result means that a particle might have a negative kinetic energy - a situation which is impossible from a classical point of view.
So if such a situation is acceptable, why won't QM allow a situation where E < -V0 ?
Why a particle with such energy cannot exist, if we permit negative kinetic energy?
 

Answers and Replies

George Jones
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In the problem of a finite potential well, we search for bound states, in which E<V.
Say the potential is defined to be 0 outside of the well, and -V0 inside it.
Analyzing the case when -V0 < E < 0, one finds that the probability of finding the particle outside the well is not zero. This result means that a particle might have a negative kinetic energy - a situation which is impossible from a classical point of view.
So if such a situation is acceptable, why won't QM allow a situation where E < -V0 ?
Why a particle with such energy cannot exist, if we permit negative kinetic energy?
Remember, this is not a classical situation. One way to to measure kinetic energy is to measure p and take kinetic energy as p^2/2m. Since any measurement of p yields a real number, p^2/2m is always measured as positive, even in the classically forbidden region.

Solving the Schrodinger equation for E < -V_0 results in a state the is not normalizable.
 
I see my mistake, the kinetic energy must always be positive. But then how is it possible that the equation E=T+V is still relevant? if V=0 and T>0 it's impossible for E to be <0. The only explantion I can think of is probability - maybe the equation is valid only on average? namely, if we measure E and T in X identical experiments the mean values will satisfy the equation? but then what about V? it depends on the coordinate anyway, and does not change from one experiment to another, and therefore the mean value is V itself... :bugeye:...So I guess my explanation doesn't make sense at all. How do you explain this?:frown:
 
siddharth
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I see my mistake, the kinetic energy must always be positive. But then how is it possible that the equation E=T+V is still relevant? if V=0 and T>0 it's impossible for E to be <0. The only explantion I can think of is probability - maybe the equation is valid only on average? namely, if we measure E and T in X identical experiments the mean values will satisfy the equation? but then what about V? it depends on the coordinate anyway, and does not change from one experiment to another, and therefore the mean value is V itself... :bugeye:...So I guess my explanation doesn't make sense at all. How do you explain this?:frown:
I understand it this way. Even if the eigenvalue of [tex]\hat{p}^2/2m[/tex] is negative outside the well, the corresponding wavefunction is not actually an eigenstate of the kinetic energy operator , because the equation is true only for a part of space. The measured momentum and KE depends on the *whole* wavefunction, and since the momentum operator is hermitian and has real eigenvalues, the measured KE is positive.

Also, the energy and the position operators do not commute. So, I'd guess that you can't precisely say that the particle has some KE in some region, because for example, if you measure the energy of the particle accurately enough it may be in the well region where the potential is non-zero.
 
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