Wave function in linearly increasing potential well from x=0 to E=V0

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 3K views
Brydan
Messages
2
Reaction score
0
[Note: no template because this post was moved from the QM subforum]

I was working on problem #41 and was confused about what the wave function would look like from the time x = 0 to when E=V0. (See image in attachments)https://mail.google.com/mail/u/0/?ui=2&ik=be26b6c94d&view=fimg&th=14c66f8d9fef8e09&attid=0.1&disp=inline&realattid=1497006575826501632-local0&safe=1&attbid=ANGjdJ9_sVoeG4InQCcYyJgM9GdXq5BJUBa-YH0cv1QCny82hco_1lvkH8I3AvpB_3B_zcwK5VVtj8BIZFl_yafUAFkA85LYwtvbi40WKWRnP6ch-psjeDgoFr6Vf7k&ats=1427656728466&rm=14c66f8d9fef8e09&zw&sz=w1896-h859In class we did this problem and my professor drew something on the board that looked similar to the image below, where both the amplitude and wavelength decrease from when x = 0 to when E = V0.

20150328_234815.jpg


However, I don't understand this solution. as according to the equations below, the wavelength should be increasing:
20150328_235319.jpg
Additionally, I don't understand why the amplitude would be decreasing.

I was wondering if anyone could perhaps clarify what the amplitude and wavelength should be doing and why. My test is tomorrow so any help would be greatly appreciated. Thanks!
 

Attachments

  • upload_2015-3-29_15-34-6.png
    upload_2015-3-29_15-34-6.png
    11.4 KB · Views: 767
Last edited by a moderator:
Physics news on Phys.org
In this problem, ##E## and ##V_0## are constants so thinking about what happens as the one approaches the other is distracting you from the point of the problem.
 
Nugatory said:
In this problem, ##E## and ##V_0## are constants so thinking about what happens as the one approaches the other is distracting you from the point of the problem.
But since V0 is increasing linearly wouldn't it not be considered a constant?
 
Suppose you made a crude approximation of that linearly-varying potential with (say) three sequential finite-wells...(which must be joined appropriately at their boundaries)...
 
The graph is wrong for two reasons, just to the left of x=0 the particle has its greatest kinetic energy and moves the fastest and is least likely to be found in some small interval. Near the point where the energy is nearly all potential the particle moves slowly and is more likely to be found. You should know which region has the shortest wavelength.
 
Hello Brydan, I hope you did well on the test. Your own hand-written thingy is along the lines of Spinnor's input and seems healthy to me. It's just that you used V0 instead of V there that confused Nugatory and yourself. So you should be able to qualitatively improve on teacher's rough sketch. And explain. Which is all the exercise requires.

Must say I have no idea what the composer of the exercise means with 'outside the potential well'. All I can distinguish is a region E > V and a region E < V.