Understanding the Significance of 'Square' in Quantum Mechanics Potentials

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SUMMARY

The discussion centers on the concept of 'square' in quantum mechanics, particularly in the context of square potential wells. The term 'square' refers to the discontinuous nature of the potential energy graph, which resembles a square corner. In the infinite square potential well model, the wavefunction outside the well is zero due to the infinite potential energy, which violates conservation of energy if the particle were to possess finite energy. The conversation also touches on quantum tunneling and the implications of the Heisenberg uncertainty principle in relation to energy and time.

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Square potentials (finite or infinte, well or barrier) are used in intro Quantum Mechanics courses. My question is:

What does the word 'square' signify?
 
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My other question:

In the infinite square potential well model, the wavefunction outside the well is zero. Two reasons are usually put forward:

1. V is infinite. So Vu is infinite unless u = 0. (I don't understand what's still wrong with Vu being infinite.)

2. The particle can't be in a region of infinite potential if it lacks infinite energy. (Why not? Why should classical principles dictate quantum rules? Moreover, in the finite square well model, there is a finite probability of the particle being within the potential barrier? Isn't that a violation of the classical principle we are using in the above argument?) Therefore, u = 0.

Thanks for any help.
 
Square refers to the appearance of the plot of V versus x.

failexam said:
My other question:

In the infinite square potential well model, the wavefunction outside the well is zero. Two reasons are usually put forward:

1. V is infinite. So Vu is infinite unless u = 0. (I don't understand what's still wrong with Vu being infinite.)

2. The particle can't be in a region of infinite potential if it lacks infinite energy. (Why not? Why should classical principles dictate quantum rules? Moreover, in the finite square well model, there is a finite probability of the particle being within the potential barrier? Isn't that a violation of the classical principle we are using in the above argument?) Therefore, u = 0.

Thanks for any help.

Energy is conserved in both classical mechanics and quantum mechanics. Assuming that the particle inside the well has finite energy, and the potential outside the well is infinite, then when the particle is outside the well, it will have infinite energy, which will violate the conservation of energy. Vu represents the potential energy of a particle outside the well. The other term in the Schroedinger equation represents the kinetic energy. Assuming that the total energy is finite and positive, if Vu were infinite, the kinetic energy would have to be negative infinity (steady state), which is impossible.
 
Rap said:
Square refers to the appearance of the plot of V versus x.

There is nothing of the plot of V versus x that reminds me of a square, actually!

Rap said:
Energy is conserved in both classical mechanics and quantum mechanics. Assuming that the particle inside the well has finite energy, and the potential outside the well is infinite, then when the particle is outside the well, it will have infinite energy, which will violate the conservation of energy.

But isn't that exactly what happens in quantum tunneling? How does the particle get the extra energy to jump into a potential barrier. Isn't that in violation of the principle of conservation of energy?

Rap said:
Vu represents the potential energy of a particle outside the well. The other term in the Schroedinger equation represents the kinetic energy. Assuming that the total energy is finite and positive, if Vu were infinite, the kinetic energy would have to be negative infinity (steady state), which is impossible.

Howcome Vu is potential energy? I thought V was?
How can the other term represent the kinetic energy?
How can positive and negative infinity add to give a finite positive number?
What does 'steady state mean'?

Thanks for all the help!
 
failexam said:
Moreover, in the finite square well model, there is a finite probability of the particle being within the potential barrier? Isn't that a violation of the classical principle we are using in the above argument?)

You're not using any classical principles. Since you've apparently solved the finite square well, why don't you just let V \rightarrow \infty and see what happens?

How does the particle get the extra energy to jump into a potential barrier. Isn't that in violation of the principle of conservation of energy?

You're thinking about it classically. How do you know its potential energy (i.e. the location) and the kinetic energy (i.e. the momentum) at the same time?
 
failexam said:
There is nothing of the plot of V versus x that reminds me of a square, actually!

Square refers to the fact that the potential jumps discontinuously from one value to another, so that the graph of V(x) has a "square" corner. Compare to e.g. a harmonic potential well, whose graph is a parabola.
 
failexam said:
There is nothing of the plot of V versus x that reminds me of a square, actually!

When the potential is finite, the "well" forms three corners of a rectangle. I wouldn't worry too much about the fact that it doesn't form a square. Its just a name.

failexam said:
But isn't that exactly what happens in quantum tunneling? How does the particle get the extra energy to jump into a potential barrier. Isn't that in violation of the principle of conservation of energy?

I may be wrong here, but as I remember, the time that a particle spends outside the well (delta t) is such that the Heisenberg uncertainty principle is obeyed. In other words, the energy deviation times the time interval is less than Planck's constant/2.


failexam said:
Howcome Vu is potential energy? I thought V was?
How can the other term represent the kinetic energy?
A better way to say it is that they are the potential energy and kinetic energy terms in the Schroedinger equation.

failexam said:
How can positive and negative infinity add to give a finite positive number?
Its a limiting process. For example, as n grows larger, (n+1)^2- (n^2+2n) equals 1. Each term on the left grows larger without bound.

failexam said:
What does 'steady state mean'?
It means no change in time.

Thanks for all the help![/QUOTE]
 

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