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Oct 7 (Wed) delta-function potential well - reflection/transmission

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Green Suit

What makes the wide/deep well and the shallow/narrow well, as Griffiths puts it the “two limiting cases of special interest”?(pg 80) What about the wide/shallow well and the narrow/deep wall??

David Hilbert's hat - 10/6 3pm

The wide/deep well and the shallow/narrow well correspond with the endpoints of the graph on page 80 - the two cases you're asking about are sort of like “mixed” cases: it seems like if a potential was wide it would also be deep (for a generally “large” potential), or if it was shallow it would also be narrow (small). This is described by what Griffiths calls z0, which has terms for both a and square root of V0. It seems like as z0 and tan(z) intersect, only the product of V0 and a determine specific points, not one or the other. So in the limit of high or low z0, the well must be both wide and deep, or shallow and narrow, or a z0 with either a or V0 dominating the other.

Put another way, if you want to look at a well that is both wide and shallow, you have to decide - is it primarily wide, or primarily shallow? Which term dominates z0, the width or the depth? The question of a wide/shallow well is kind of ambiguous in that sense: in general, depending on which term is more important, the question can have more than one answer.

Ekrpat - 12:54 - 1172

In discussion, Justin mentioned that the first excited state of the double delta potential we worked on only sometimes exists. I believe he said it was dependent on the spacing <math>a</math> (and <math>\alpha</math>?). How can you determine whether this second solution exists for a particular <math>a</math> and <math>\alpha</math>?

liux0756 17:31 10/06/2009

If you solve the wave function and apply the boundary conditions, the following relationship can be achieved:<math>\frac{\hbar^2k}{m\alpha}-1=\pm exp(-2ak)</math>. Then sketch the left-hand-side and right-hand-side respectively to solve k. Note k must be positive. When <math>\frac{\hbar^2}{m\alpha}<2a</math> (line a), there are two possible values for k, and in this condition the second solution exists. When <math>\frac{\hbar^2}{m\alpha}\ge2a</math> (line b or c), there is only one possible value for k, in this condition no excited state exists.

Blackbox 18:31 10/06/2009

In the case of the infinite square well, I understood the particle can be found out in the square well because the energy is larger than potential energy, E>V(x) (0<x<a). That's why we try to get the Shrodinger equation. Compared to this concept, Could you explain the physical concept of the Schrodinger equation for the Delta function Well? For the bound state, E<0, this means that particle can be found out within the narrow Delta function Well?

Daniel Faraday 10/6 12:30

The way I see it, there are two big conceptual differences between the two models:

First: For square wells, the bottom of the well is at V=0, so for a wavefunction to exist anywhere it has to have E>0. But for the delta well potential, the energy at the top of the delta well is zero, so a particle with E > 0 will not 'bounce off' the well and a particle with E < 0 will bounce off the well (except for quantum tunneling effects, that is).

Secondly, for the infinite square well, the particle can only exist in the well because the potential everywhere else is infinite. In other words, <math>\psi(x) = 0</math> outside the well. But for the delta well potential, the wavefunction could have a nonzero value anywhere.

In practice, you don't really try to find the value of the wavefunction “in the well”. You find the wavefunction on the left side of the well, and also on the right side of the well, and use boundary conditions at the well to find the unknown coefficients of the wavefunctions.

I hope that made sense.

Blackbox 19:31 10/06/2009

I can understand them mathematically. What is the physical meaning of 2.121?

Chap0326 20:58 10/06/2009

I don't know if it has any physical significance…I think it's just what we know to be mathematically true so that we can find enough equations to solve the unknowns.

David Hilbert's Hat 22:30 10/06/2009

The 2000 level quantum book says this must be true because ψ(x) is an eigenfunction; to be an eigenfunction ψ(x) and dψ(x)/dx must be finite, single valued, and continuous. The 2000 level book explains the need for these things because it makes measurable quantities “well-behaved” if ψ and dψ/dx are, because then all the terms containing ψ in the Schrodinger equation are also well-behaved. This seems to be given for finite-ness and single valued - if the wavefunction is not single-valued then there's no way to normalize it or take expectation values. As far as continuity goes, I am not entirely sure.

prest121 22:50 10/6/2009

As far as the physical meaning of 2.121 - I think it makes sense that a wavefunction that describes a real particle must be continuous. Since the probability amplitude is a product of the wavefunction, it seems intuitively that a non-continuous wavefunction would be unnatural. A discontinuous jump in the w.f. would imply that at a certain x position, the probability of finding the particle drastically changes, and that doesn't seem correct to me intuitively.

A little bit off subject - an interesting thing to think about here is how this relates to the bound potentials. In the infinite square wells, the wavefunctions have to go to 0 at the boundaries (to ensure continuity of the wavefunction) because there is no chance of finding the particle outside of the well (since the potential is infinite at the boundaries). But for the finite square well and the simple harmonic potential, the wavefunction doesn't go to zero at the boundaries because the binding potential is nonzero and there is some probability of finding the particle past the boundary due to tunneling.

Chap0326 21:02 10/06/2009

Is there a resource I can review to understand how one is able to normalize through building a linear combination. I'm a bit shaky on understanding Griffith's explanation of it.


Correct me, but i think, In eq. 2.121 this are the same old boundary conditions. The significants of this conditions allow use to understand special case and general problems.#2 is the restriction we apply to the infinite square well to understand the basics of the wave behavior in a simple case and leads to the idea of leakage i believe. However, for #1 of these B.C. should it say ψ is always continues except at points where the potential is infinite?

Zeno 10/7 10:30AM

On pg. 75 of our text Griffiths reminds us that “it's impossible to create a normalizable free-particle wave function without involving a range of energies” so “R and T should be interpreted as the approximate reflection and transmission probabilities for particles in the vicinity of E.” Is there a way to describe R and T in terms of the range of energies specifically? Or are they stuck just being estimates? If they are destined to only approximate the probabilities, how can we be sure of their accuracies?

Zeno 10/7 11AM

One more conceptual question: I understand how particles can penetrate potential barriers, reflecting and transmitting the incident wave, but I don't understand the reflection by a potential well. The delta function well in Fig 2.15 has me wondering how a well reflects part of a wave instead of “trapping” it in a bound state or transmitting it. I feel like there's something simple conceptually that I'm missing, so a little help with interpretation would be much appreciated.

chavez 10/8 6PM

It makes sense (to me) if you divide the space under consideration into regions associated with specific potentials. There will then be a discontinuity in the potential energy between each region of space. It is a general property of waves to scatter when a discontinuity in potential is encountered.

Pluto 4ever 10/7 7PM

Essentially, what I get from this is that this deals with the probability of a particle being able to get out of a well. The more energetic a particle is the greater the probability it has of escaping from the well. So if a particle does not have the energy to make it through the barrier then it will be reflected back into the well (trapped). Of course, that is not to say it will never be able to get out. Although, it energy is less than the barrier it can still escape by means of tunneling.

Dark Helmet 10/08

I think of it like waves on a string of different materials. Any time the wave changes medium there will be both reflection and transmission. If you think of the potential well as just a part of the string that is less dense and a wall as part that is more dense, it makes sense. Anytime the wave changes potentials, it's like the string changing mediums. At least that's how i think of it, but i could be missing something too.


Are there other resources that explain the delta function and the associated limit of a sequence.


When we have a delta-function barrier, why do we no longer have a bound state?

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