Delta Function Well

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  • #1
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So I'm studying that part right now. I only get parts of it though, it seems.

The first thing the book goes over (This is intro to QM by Griffiths) is a potential that has the form -A*deltafunction. Okay, that's just something he plucked for simplicity.

But then if the potential is lower than the particle's energy at that point, why does the math work out so that there's a higher chance of finding it there? When we did problems with a modified infinite square well, where the bottom was V = 0 up to some point, then V = X afterwards, there was a higher probability of finding the particle where there is more potential energy (V = X), just like there would be classically.

I don't understand why the particle would "hover" at that point instead of leaving even faster.

The other thing is that this relates to scattering, right? I get it when you flip A and just get V = A*deltafunction, so that V is infinite, you have a barrier, and then you get scattering proper, i.e. the particle either hits the wall and bounces back or goes through.

How come the same thing applies when the well goes below the particles energy like in the first case? I would have thought it would just zoom past the potential well, since there's nothing stopping it.
 

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  • #2
blechman
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But then if the potential is lower than the particle's energy at that point, why does the math work out so that there's a higher chance of finding it there? When we did problems with a modified infinite square well, where the bottom was V = 0 up to some point, then V = X afterwards, there was a higher probability of finding the particle where there is more potential energy (V = X), just like there would be classically.

I don't understand why the particle would "hover" at that point instead of leaving even faster.

For an even, finite square well, the ground state of the system has the particle living in the center of the well; for an infinite square well, the particle wants to live in the middle of the well far away from the walls. I think you're confused about something! You always want to MINIMIZE your energy, classically and quantum mechanically. So you would rather go to lower potential while in the ground state.

The other thing is that this relates to scattering, right? I get it when you flip A and just get V = A*deltafunction, so that V is infinite, you have a barrier, and then you get scattering proper, i.e. the particle either hits the wall and bounces back or goes through.

How come the same thing applies when the well goes below the particles energy like in the first case? I would have thought it would just zoom past the potential well, since there's nothing stopping it.

Nope! That's where QM and CM differ! It's purely a consequence of QM. A better example of falsified classical intuition in the quantum world I cannot imagine. ;-)
 
  • #3
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Okay. I just have to trust the math, then. Thanks.
 
  • #4
blechman
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Okay. I just have to trust the math, then. Thanks.

well, yes, you should always trust the math, assuming you don't make a mistake! But you shouldn't give up on the physics either! For the scattering problem, it's probably easiest to see what's happening in the wave-mechanics picture. If you send a water wave in the direction of an obstruction, even if it's not directly incident on the obstruction there will be some diffraction (scattering). This just follows from the nonlocal nature of the wave.

Hope that helps with the intuition.
 
  • #5
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Well it helps if there is a barrier there, when V = inf at some point X. But when it's -inf, why would it spend it's time there? It has a momentum, so it can't get "stuck" in the well, can it?

Well, in any case, it doesn't really matter, because that kind of wave function is non-normalizable, right? It goes from -inf to inf, so it's a moot point. That's more to do with scattering probability at that point. So I guess if the energy can fluctuate like the position and momentum can (it would have to...), then there's a chance it could jump into the well, not make it all the way, and then come back? Or something...
 
  • #6
blechman
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Well, wait a second: if it's a negative delta function potential, then there is an attractive force and the particle would like to hang out near the origin. Thus there is a bound state. However, if it's a positive delta function potential, then it's a repulsive force, and there is no bound state. Go ahead and look for one: you'll find that you can never match boundary conditions unless the wavefunction vanished everywhere.

However, for both positive and negative delta functions (or potentials in general) there *is* always scattering. But a scattering state does NOT "spend all its time [at the origin]", so I'm not sure what you're saying there.
 
  • #7
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Okay, you're right, I totally understand it now.

I guess what was messing me up is this:

http://www.phys.washington.edu/users/karch/324/2007/hw3.pdf

The first problem. The TA's and the prof. both said that in the "step" after 2a the frequency of the wave function goes down (less energy so that makes sense) and the amplitude goes up because there's a higher chance of finding it there (because it has less energy).

Is that because it was already in a bound state?
 
  • #8
blechman
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The first problem. The TA's and the prof. both said that in the "step" after 2a the frequency of the wave function goes down (less energy so that makes sense) and the amplitude goes up because there's a higher chance of finding it there (because it has less energy).

that's a funny sort of argument, since the particle has the same energy on both sides of the bump ([itex]E_0[/itex]). It is true that the particle has less kinetic energy, which is probably what you mean. Then the particle will spend more of its time in the high-potential region since it can't escape that region as quickly the low-potential region (it's "moving slower"). That's a *very* sloppy way of saying it, but that's the idea.

The fundamental difference between this problem and the other problems you were talking about is that this problem has infinite wells on each side, so the particle can't escape to infinity. This is why the result is different than what you were asking above.

Hope that helps.
 

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