Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Nov 11, 2013 #1
    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
     
  2. jcsd
  3. Nov 11, 2013 #2

    jtbell

    User Avatar

    Staff: Mentor

    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. :confused:
     
  4. Nov 11, 2013 #3
    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?
     
  5. Nov 11, 2013 #4
    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?
     
  6. Nov 11, 2013 #5

    kith

    User Avatar
    Science Advisor

    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.

    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.
     
  7. Nov 12, 2013 #6
    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.
     
  8. Nov 12, 2013 #7

    kith

    User Avatar
    Science Advisor

    You are right. What fails is the classical intuition about particles.
     
  9. Nov 12, 2013 #8
    Why there is no incident wave ,yet we still have exponential function?
     
  10. Nov 12, 2013 #9

    jtbell

    User Avatar

    Staff: Mentor

    Why do you think there is a problem with that?
     
  11. Nov 12, 2013 #10
    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?
     
  12. Nov 13, 2013 #11

    kith

    User Avatar
    Science Advisor

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Bound state of finite square well, why do we make this statement?
  1. Finite Square Well (Replies: 1)

Loading...