Silly conceptual question

1. Feb 1, 2008

Atomos

In a finite square well, there exists a non-zero component outside the well for all bound eigenstates. What is the physical nature of this non-zero component? How can you actually observe an electron outside the well when it is in a bound energy eigenstate? It would have an imaginary wave number.

2. Feb 1, 2008

Mentz114

I'm not sure if I understand the problem.
The square of the wave function gives the probability of finding the electron at at particular place. So the non-zero component outside the well represents a probability of finding the electron outside the barrier. If it did escape, it would no longer be in an eigenstate of the bound electron.

3. Feb 2, 2008

Atomos

Out side the well, for a bound state, the wave function isnt zero, but is a decaying exponential because the wave number is imaginary.

4. Feb 2, 2008

JesseM

Why do you find Mentz114's answer inadequate? As he said, "the non-zero component outside the well represents a probability of finding the electron outside the barrier". Doesn't that give you the "physical nature" of the non-zero component?

5. Feb 3, 2008

Atomos

ok, suppose you are certain it is in some energy eigenstate, and you know that that eigenstate is a bound state. All of those states oscillate inside the well, and decay exponetially outside of it, so there is some probability of finding it outside the well. If you make a second observation and find it outside of the well, how is that possible? Its kinetic energy would have to be negative.

6. Feb 3, 2008

malawi_glenn

nope its kinetic energy would be p^2/2m (non relativistic particle)

How did you reason to get negative kin E ?

7. Feb 3, 2008

Lojzek

It makes no sense to say: if the particle is on the point x, then it's momentum is p and it's kinetic energy is p^2/2m, because x and p can't be both exactly known (Heisenberg's uncertanity principle).
Vave function (and the probability density) can extend to the place where potencial energy is larger than the energy of the state. A typical example is quantum tunneling, where a current of particles can travel through a potencial barrier, higher than particle's energy.

Last edited: Feb 3, 2008
8. Feb 3, 2008

malawi_glenn

So why use p^2/2m as kinetic energy in the hamiltonian...?

I only said that the particle outside the well has kinetic energy p^2/2m. Did not said that we have made a measurment..

9. Feb 3, 2008

Atomos

My problem is that if we KNOW the particle is in a bound state (suppose me measured its energy at some point and put into some bound eigenstate) and we try to measure its position, and find that it is outside of the well, by conservation of energy its kinetic energy must be negative.

10. Feb 3, 2008

JesseM

Energy is only conserved statistically in QM, I believe the only circumstance where it's guaranteed to be conserved exactly is if you measure the system in an energy eigenstate and then don't perform any measurements of variables that don't commute with energy between that measurement and your next measurement of the system's energy. But position doesn't commute with energy since kinetic energy is a function of momentum and momentum doesn't commute with position, so if you measure the position you can't be sure that your next measurement of energy will show the same energy as your previous measurement (but if you prepared the system in the same energy eigenstate initially on many trials and then on each trial you measured its position and then measured its energy a second time, the average energy of the second measurement should be the same as the initial measurement over all the trials). For example, see the discussion on page 93 of Gordon Baym's Lectures on Quantum Mechanics.