Return to Q&A main page: Q_A
Q&A for the previous lecture: Q_A_1007
Q&A for the next lecture: Q_A_1012

If you want to see lecture notes, click lec_notes

Main class wiki page: home

I am sure this is just a misunderstanding of parameters, but when one has the second solution to the delta function wells, the one with the positive peak, and the negative peak… but it was my understanding, that for instance, in the SHO, at the top of hte peak was where the highest probability of finding the particle was. However, when you have a negative value peak and a positive value peak… how is this different. I think I'm just looking at a graph thinking it's graphing something else… but …. none the less, I'm a little unsure, conceptually, why we can have a positive and negative value peak, and both have the same meaning in terms of where the particle is found…

The graphs we did today are all graphs of <math>\psi</math>. To get the expected value of the position, or any other measurable quantity, you have to look at <math>\psi^2</math> which gives you nonnegative probabilities everywhere.

of course. I knew it'd be straight forward. I think I'm just used to assuming <math>\psi^2</math> for graphs these days…

Just remember that superposition of two waves with negative and positive peaks would make them cancel and then <math>\psi^2</math> would be zero! So they are not the same thing even if the probability of finding them individually is the same.

This whole discussion has been moved to Q_A_1012

About the Delta Function. I guess I just don't feel comfortable with it because I've used it extremely little… but do we never actually define the delta function, besides just the intuitive case? I went and googled it but do we only use it in terms of limits… solving integrals where the limit is a value vs. 0? Is it more like we're borrowing the concept of the delta function?

I'm also confused about the delta function. To me, it only seems that the delta function's only real purpose is to make the potential well normalizable so we can get a practical function for the well.

The Dirac delta function is defined explicitly in Eqn. 2.111. It's not so much a function as it is a mathematical construct. The way we are using it is analogous to how we used an infinite square well to simplify the finite square well, but instead of a well we are modeling something like an impluse (maybe a point charge/mass).

The delta function isn't technically a function because functions that are zero everywhere except at one point must have a total integral of zero, not one. It seems to just be an abstract concept used to simplify calculations and approximate things that we can't explain in a more accurate way. Something to just get the job done i guess.

I know this was asked on Wednesday's Q&A, but I'm still confused about how to form linear combinations and how this makes the scattering wave functions normalizable (pg. 75).

The only place Griffiths really talks about this is on p.61 in the text and the footnote on that page. He says that the individual stationary states of a free particle are sinusoidal and extend to infinity, so they can't be summed and aren't normalizable. But if you add together many sinusoidal functions with different values of k, you can make those waves cancel out everywhere except in a very small region (which is where the particle is).

And, of course, by 'add together many…with different k' Griffiths means take an integral over all possible values of k. That's what gives you equation 2.100.

Somebody correct me if I'm wrong, but I don't think that he proves that this works. He just states it and gives a qualitative explanation as to why it works.

This same reasoning also applies to solutions to the dirac delta potential.

It seems like the proof of what is said on p.61 is just a fourier analysis trick; given some initial wavefunction, you can find the values of φ(x) by taking the fourier transform of the initial wavefunction, and you know this works from Plancherel's theorem. Which is a quite opaque argument, but put it this way: any given sine or cosine function will be indeterminate at infinity; but in general you can compose any well behaved function by superposition of sine and cosine functions, which is the basis for fourier analysis. But since some well behaved functions do converge at infinity, the proper superposition of sine and cosine functions will converge as well. To build the proper superposition, you use 2.103.

If the dirac delta function is defined as being either 0 or infinity, what is the point of multiplying it by some constant α? It seems like α wouldn't matter much, because the potential will either be 0 or infinite, regardless of α. Yet α still somehow defines the strength of the potential. Does it just come in because it is related to the derivative of ψ like in 2.125?

I think although delta function is 0 or infinity, the integral of the function somehow reflects the 'strength'. While the delta function integral is 1, the integral of delta function multiplied by α is α.

Ah, okay, that makes much more sense.

Yuichi mentioned that the scattering problems focus on when E>0. Why is that?

I'm not completely sure but I think we used E>0 today because we chose <math>V(x)=- \alpha \delta (x)</math> and had to limit E to positive to get the scattering we wanted.

Yes because V=0 when x is not 0, E>0 is for scattering state and E<0 is for bound state.

Ok, I just need some verification on the delta-function….. It is not so much a “function” but instead a distribution. And we know that a distribution represents a probability….. But since the delta-function is spiked at one point only….does this mean there is only one probable outcome? I apologize for my fragmented question full of dotted pauses….. but I think it effectively reflects my confusion & frustration with this seemingly hand-waving method. It works, but why?

Yup, you got everything down! Why does it work, well integrate over P of a delta function. You will find that you get 1, for some x=a, where delta(x-a). Since delta is only defined at one point to be 1 and zero on all the other points, you find that at that one point we get P=1.

I think Griffiths makes a reference to the delta function when it's used as a distribution of a point particle's mass/charge. Everywhere that's not located at the particle it's zero, but at the particle it's infinity; when you integrate over all space it comes out to be 1, because you have one particle. That seems to be the easiest way to see the delta function - it is very peculiar because it is zero everywhere except at one point it's infinity, but it integrates to 1, like a point particle's distribution (which seems a lot more familiar).

This question is about problem 2.34. The problem asks us in part c to show that <math>T=\sqrt{\frac{E-V_0}{E}}\frac{|F|^2}{|A|^2}</math>. However, Eq. 2.139 on page 75 says <math>T\equiv \frac{|F|^2}{|A|^2}</math>. That's with three lines, as in identically equal, as in any statement that they're not equal (such as that in the prompt in 2.34c) is incongruent. Now, I remember the professor said in class that <math>T=\frac{|F|^2}{|A|^2} \frac{k_2}{k_1}</math>, but how do we show that? We have no definition of T other than the (patently false) one in 2.139. For shame, Griffiths. For shame.

Return to Q&A main page: Q_A
Q&A for the previous lecture: Q_A_1007
Q&A for the next lecture: Q_A_1012