Bound state of finite square well, why do we make this statement?

[Outrageous]

Reading from http://quantummechanics.ucsd.edu/ph130a/130_notes/node150.html
Again we have assumed a beam of definite momentum incident from the left and no wave incident from the right.

Why is the above statement made?
What does the reflected wave mean? There is now all why reflected?
Please guide .thanks

 

[jtbell]

This is not a bound state problem, which would have E < 0. It is a scattering problem, which has E > 0. We assume that we have a wave coming in from one side, corresponding to a beam of particles coming in from that side. We assume they are coming in from only one side, for simplicity.

What does the reflected wave mean?

It means that some of the particles "bounce back" from the well instead of passing through.

There is now all why reflected?

I'm sorry, I don't understand this question.

 

[Outrageous]

This is not a bound state problem, which would have E < 0. It is a scattering problem, which has E > 0. We assume that we have a wave coming in from one side, corresponding to a beam of particles coming in from that side. We assume they are coming in from only one side, for simplicity.



It means that some of the particles "bounce back" from the well instead of passing through.

I'm sorry, I don't understand this question.

Thanks
Yup, it is scattering state. So it doesn't mean we don't have wave from the right, we take only left for simplicity.
Reflected means bounce back, but in this scattering state, there is no potential wall or any wall, why the wave bounce back?

 

[Outrageous]

One more to ask, for bound state, do we still have wave transmitted and reflected? No enough energy to transmit? What is the negative energy mean?

 

[kith]

Reflected means bounce back, but in this scattering state, there is no potential wall or any wall, why the wave bounce back?

Just like a part of the wave gets transmitted through a potential wall, a part of the wave is reflected by a potential well. Classical intuition fails here.

One more to ask, for bound state, do we still have wave transmitted and reflected?

The wavefunction of a bound state shows an exponential decay outside the well. There are no incident waves and thus neither a transmitted nor a reflected wave.

 

[kaplan]

Just like a part of the wave gets transmitted through a potential wall, a part of the wave is reflected by a potential well. Classical intuition fails here.

Actually no, it doesn't. Classical waves can reflect off wells or dips in a very similar way. For example, the sound waves inside a tube of a wind instrument (like a flute) partially reflect off an open end of the tube.

 

[kith]

You are right. What fails is the classical intuition about particles.

 

[Outrageous]

The wavefunction of a bound state shows an exponential decay outside the well. There are no incident waves and thus neither a transmitted nor a reflected wave.

Why there is no incident wave ,yet we still have exponential function?

 

[jtbell]

Why do you think there is a problem with that?

 

[Outrageous]

Why do you think there is a problem with that?

The exponential function ( is not complex ) here doesn't represent wave that is why no wave outside the well.
But why do we have exponential function outside the well? What does it represent?

 

[kith]

But why do we have exponential function outside the well? What does it represent?

We have it because it appears in the solution of the time-independent Schrödinger equation. Like the other wavefunctions, it is related to the probability to find the particle at a certain position. So the exponential decay indicates that there's also a small probability to finde the particle outside the potential well although it is in a bound state. You can see a visualization of the solutions here