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So, I finally have a question. It's more of a comment.

I have decided to treat this class more like a math class than a physics class. The concepts are so far out there that I will have to reverse my usual thinking. Normally I try to focus on understanding, and from there the math comes more easily, and any required memorization will be easier as well. Here, I think that process would be way too slow. We have a scant few weeks. For me, anyway, the understanding will take longer than that, if it ever truly comes, and I have exams to take that are more math than anything else.

I guess my question would be does anyone see anything wrong with this idea? Is it a good idea?

I think there's wisdom in your idea however I believe it will only provide you limited improvement. Consider the first quiz. At first glance it appears to be heavy on the mathematical computations but at second glance some of the math can be bypassed. For example the first problem, Prof. Kubota stated, “guesses based on sound physics intuition to simplify your calculations are encourages” and indeed the problem can be solved with little to no math. Without the understanding all your are left with is the math and memorized equations which can leave you stranded. But, I suppose having a strong understanding of the math and memorized equations and a weaker understanding of concepts is better then a weak understanding of all of it!

Well, now I'm curious. How do you propose to do this problem with “little to no math?” I can see getting <x> and <p> - those were pretty obvious, especially in hindsight, but <x^2> and <p^2> were not trivial results I didn't think.

On second glance, too, I don't see how the math can be “bypassed,” either, without some memorization especially. #2 was the easiest, but I got that one pretty well on test day I thought. Number three required either some memorization of at least one key relationship between a+ and a- (otherwise you would have to do some extra math). And number four? Took me 23 lines, most of complicated integrals. How do you propose to do that one with little to no math?

Bottom line, I personally can't see any way to solve even just #1 and #4 with little to no math, which is clearly what would have to happen in order to finish in 50 minutes or less.

Well here's the other thing you need to notice. A lot of these problems are only solved based on 'physical properties of the system'. For example, when we solve the equations for the finite square well, we x out some of the coefficients, because physically, some of the solutions don't exist – or the Dirac equation, which I imagine most mathematicians are made sick by, the way we 'create' it and handle it simply to express a physical system. When I Was taking some of my math classes, I found math to be deliciously abstract and beautiful. When I use math in this class, I feel like I'm turning a screw driver upside down and using it as a hammer. It's a tool used to get a job done, but often not beautifully. Just kind of works to get at what we need. I think you'll be disappointed if you took this class as an advanced math class. There's too much screw-driver hammering. In terms of math, it often frustrates me. Ultimately, like Green Suit said, you'll end up not being able to solve some problems if you take only the math side.

I was going over mondays lecture and was wondering how did we get D=-A and C=-B? was it just the fact that we were looking for odd solution which is Antisymmetric? and since we know this we could eliminate some unknowns this way. and another thing is that the imaginary k still bothers me a little and it came up today as well. what does it really (physically) mean to have an imaginary p?

I also have this general question…I think it's something to do with the fact that we can have E > 0 and E < 0, and the boundary conditions for these cases, giving us the two options of C = +/- B, but I'm not very sure on that.

I think you can just determine that based on symmetry. Why would the particle behave differently on the left side of the well than on the right side? This isn't the same as the case where we are “shooting” particles at a barrier or at a well, when we have different coefficients, because E < V0 everywhere except -a < x < a. So there isn't going to be any tunneling (eventually, all particles that penetrate through the well will be reflected back. This is like one of our homework problems from last week, where R = 1.).

one more thin; we say that because the potential is symmetric we can reduce the problem and look at the continuity of psi and its derivative only at one point (a or -a). would this work if the potential was -V between 0 and 2a rather than -v between a and -a. (and 0 everywhere else) ?

In response to your first question, the reason we chose C = -B and D = -A is mostly in part to symmetry but also we were looking for the odd solutions of the finite square well. Had we stuck with even solutions then C = B and D = A.

In response to your last question, I think that you can only use symmetry to reduce the number of variables when you're looking at +a and -a. If you're looking at a function which is symmetric but it's not centered at zero, you'd want to shift your variables so you have the function centered, so you can take advantage of the symmetry.

As far as imaginary p goes, I think of it like the particle is in 'tunneling mode'. If it were behaving classically, it could not enter such a region at all, but since it has its 'imaginary p' it has a chance of passing through a classically forbidden space by tunneling through it into an allowed space.

For example, in the step potential problem we did for homework, the particle had some amount of wavefunction which penetrated into the barrier. The particle still couldn't be observed in the barrier, but if the barrier were narrow, there could have still been a nonzero value of <math>\psi</math> on the far side of the barrier, thereby allowing tunneling.

I don't think about p and just try to look at <math>\psi</math>, remembering that only <math>\psi^2</math> has any physical meaning.

Why is the scattering problem inherently asymmetric? Why does the wave only come in from one side only? Do scattering problems have even and odd functions? If so what are some examples? How can they be identified?

To answer the question of why the wave comes in from one side is because the scattering problem is based on a wave coming into contact with a barrier. This can only occur if you define an origin of the wave outside of said barrier.

Correct me if I'm wrong, but I believe the scattering problem is asymmetric because of the condition that for scattering, you have E > 0 and the boundary conditions works it out such that you have asymmetry, whereas the bound states occurs when you have E < 0.

The scattering problem is asymmetric because we choose to define it that way. If we look at the situation of a particle interacting with a potential barrier or well, we obtain equations for the wavefunction in each region. We *choose* to look at the problem in this way: we are sending particles in from the left side. Therefore, the particles hit the barrier or cross the well heading to the right. Some might be reflected back to the left by the barrier/well. Or they will pass over the well or through the barrier and continue on to the right. Once the particles are past the barrier/well, there is no reason for them to ever be traveling to the left.

I am interested in figure 2.19. We can see that at certain energies the transmission coeffiecient is 1, while in the energies between them the transmission coefficient is less than 1. Is this property useful for some practical applications?

I'm pretty sure this is why some materials are transparent, while others are opaque. More correctly, that materials are transparent at some frequencies and to some particles and opaque for others. Tons of practical applications, eh? Think transparent aluminum, a la Star Trek.

In thinking about this graph, and looking up some stuff, I read that transmission coefficients can be composed of subsequent transmission coefficients, T(a)… perhaps what the graph is showing us is that there are certain “Eigenstates” of the function, and that there are other compositions of these states – the points between the peaks and troughs. Are the coefficients themselves similar in form to the wave equations their in? <math>\Psi(x)=\psi<x_1>+\psi<x_2>…</math>

Interesting question. I'd like to add that I thought of a way to interpret the figure classically, and I wonder if someone can confirm or deny my interpretation: so the transmission coefficient is (approximately) the probability that a particle of energy E will cross the potential barrier. If you consider thousands of “bouncy balls” each with energy E (that don't lose energy as they bounce) bouncing around inside a physical well similar to thousands of waves oscillating in a potential well, a few of the balls whose phase was correct to reach a maximum at or near the boundary could cross the physical boundary. The higher the energy, the more likely the balls are to escape; however, even a highly energetic ball with the a certain phase could bounce into the barrier wall and be reflected back. As the energy of the balls increase the barrier becomes less significant, and as E approaches infinity the potential barrier becomes negligible and all of the balls will escape (T→1). I know there are a few slight differences between the concept of bouncing balls and oscillating waves, but does this sound like a reasonable classical interpretation? By the way, I like East End's explanation regarding the phases of photons and barriers of solids translating to opacity.

But why would there be those levels of energy? In general, I like our picture, that higher energy creates less significance of the well – but why would there be those discrete levels that are more or less transparent?

Consider eq. 2.171 the energies are the same old energies associated with energy quantization. Which occurs at discrete allowed energy levels represented by 2.171. So i feel like the corresponding energy levels represented in figure 2.19 come from this eq. En=n²± Vo this will only give discrete energy levels when this difference is precisely one of the allowed energy states of the system. Leading to complete transparency.

Can anyone tell me why differentiability boundary condition doesn't necessarily hold for a infinite square well?

I'm not entirely sure as to why they wouldn't since the only thing we really changed between the two systems is that V is now infinite. Therefore, the chance of a particle escaping the well is non-existent making R=1. Unless, of course, if the walls become transparent.

That's a very good question… the wave function has to be zero outside the infinite square well, and the slope of the sine function is definitely not zero at the well walls; the transmission coefficient could never be greater than zero for an infinite square well, suggested by intuition as well as eqns [2.169] and [2.171]. Griffiths, on pg 71. eqn [2.121], states that <math>\frac{d \psi}{d x}</math> must be continuous except at points where potential is infinite, such as the delta function or the wall of the infinite square well.

This wiki keeps boot'en me off when i go to save what i have edited. is there some limited time frame to editing are is this a common problem, seems to be a new one for me.It happens for creating lecture notes as well, frustrating!

===Yuichi=== You have 15 minutes of inaction. Then if someone else wants to edit the same wiki, s/he will have a precedence. but as long as you are typing, you should be able to keep your edit. If you are losing the edit even if you are not leaving the edit for more than 15 minutes, let me know. I will check with the systems person.

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