Which value of c would you choose?

Tuesday, September 23, 2008

Homework 1 and welcome to 139b






Hi. This is your first Homework assignement. Welcome to Physics 139b.

42 comments:

Trapezoidal said...

should be fun

Anonymous said...

test

Zack Schlesinger said...
This comment has been removed by the author.
Anonymous said...

better test

Jill said...

I think what Zack means to communicate here about the crucial role of the blog dialogue in this class are the following clarified points:

*All homeworks (in addition to miscellaneous other class materials) will be posted here for you to download and discuss through "comments"

*Your grade in the class will be positively impacted by your contributions to the blog discussion of class concepts and homeworks

*lack of engagement with the discussion board and failure to contribute comments at least once a week (and hopefully more) may negatively impact your final grade

*most importantly, this is meant to be FUN! and to get you thinking more deeply about the concepts in a collaborative, interactive forum.
Don't worry so much about the content and quality of your comments--focus more on being relaxed and engaged with the material!

Isotope said...

question, are we also doing 11 & 12 for this tuesday's assignment?

Zack Schlesinger said...

Yes, 1-5, 11 and 12

captain said...

question, for problem 1e, when finding what the potential is, do you want us to keep the seperation constant E as it is or should it be in some specific form because of the form of the potential?

Zack Schlesinger said...

I think i see what you mean. There will be an "E" in the time- independent Schrodinger equation that the potential goes into. I think that my response would be that you may expect to find relationships between: "E", "a" and the parameter that establishes the "strength" of your potential. Exploring those relationships, e.g., graphically, is probably worthwhile. Hope that helps.

I guess one way to answer would be to say, "don't keep E as it is. it will depend on something else"

Bobby said...

I'm having difficulty coming up with a satisfactory answer for 2d, why the expectation value of the kinetic energy in an ISW is always positive. My current argument involves a restriction that the total energy must be a positive non-zero value but this seems like an arbitrary restriction. Does anybody else have a better idea?

Second, this is more a basic math question, does anyone remember how to do the integral of cos^2 by hand? This appears in problem 2a. I got the actual answer by mathcad but I feel that the solution to this integral is something I should know by now.

Lastly, when making the < T > operator, T = p^2/m, why does that p^2 become < p^2 > and not < p >^2? It never sat well with me that Griffiths never explained that. Maybe he did and I missed it.

Thanks.

captain said...

well regarding the second question you had cos^2 with argument x can be rewritten as [1+cos(2x)]/2 this is so because cos(2x)=cos^2-sin^2 according to the trig formulas given in the back flap of griffiths.

Anonymous said...

Bobby, in regards to the < T > question, I would just say that we start with the assigment T = (p^2)/m, and then when we calculate the expectation value we see that < T > = <[(p^2)/m]> = <(p^2)>/m.
If m were an operator we'd have a harder problem :)

captain said...

just a slight correction to T is that it should be p^2/(2m) and not p^2/m

Trapezoidal said...

Well Bobby, I think the reason it becomes < p^2 > and not < p >^2 may have a lot to do with the same reason we choose to use the spacial variance as < x >^2 - < x^2 >. Remembering back we needed some way to determine not only the spacial average < x > but also a measure of the spatial extent of the object. For the extent we needed some sort of absolute measurement where < |x| > would've worked apart from being unweildy. So then we chose the easier to use < x^2 > instead.
In the same way < p >^2 and < p^2 > serve similar purposes. Since < p > is commonly zero for bound states you can't use it to determine anything about the kinetic energy of the object. On the contrary < p^2 > which measures the absolute distribution of the momentum will give you something useful. Additionally as kinetic energy is considered postive we need to take this absolute value of the momentum to get what we're after.

Or something like that.

Bobby said...

Thanks everybody. I got it now. Is this assignment taking anybody an unexpectedly long time? I just finished 1-5, 11, and 12 and it all took ~8 hours. Granted, I'm still rusty at this stuff after such a long break.

Zack, is homework going to be a typical length for our homework in the future?

Zack Schlesinger said...

First, let me say, "great discussion!" and thanks for getting things going with raising some very nice discussion topics.

Regarding typical length for HW, there really is no typical length, however, I would say that in the beginning of the class there tend to be a large number of problems focused on review and preparation, and as the course goes on there tend to be fewer problems, but the problems tend to be longer and perhaps more difficult. This is not by design so much as necessity--even the simplest problems one can come up with to illustrate some points, can be pretty difficult.

I am not sure really how long this "should take", but i tend to think that 8 hours is not too long and really seems quite modest to me. I mean, there are a lot of problems and some difficult aspects...

Oh, also, i don't think that total energy has to be positive. E.g., in an H-atom the gs (and other states) have negative energy...

Zack Schlesinger said...

How was the section with Paul Graham? Did a lot of people come? Was it helpful? Which problems or other things did you cover?
(i stopped by room 235 around 4:45 but it must have just ended.)

Is everyone doing ok with this HW? I hope so. Feel free to post comments or questions here. What are the hardest problems?

Derek said...

how do you go about proving that a wavefunction is a bound state of a certain potential?ie 5C does this consist of just finding the energy level that corresponds to the wavefunction?

Vince said...

If the math is correct and the kinetic energy is negative for 3c, what is that telling us? I'm having a bit of trouble reasoning why its negative. Thanks.

Trapezoidal said...

The section with Paul was pretty helpful from my perspective. About ten people showed up and her covered a bit of problems 2, 3 and 12.

He also talked about getting an intuitive sense for some aspects of the S.E. and the infinite square well in particular.

Chaos said...

well carlos, that was a very good question. I've been wondering about that for a long while.

I'm not sure if this is right, but it seems like the the double derivative in the kinetic energy operator (-h(bar)/2m) would give you a negative sign because the sharp peak is the max of the wave function. grouped with the negative sign from -h(bar)/2m and the negative sign from doing the integral, you end up with 3 negative signs, which ultimately give you a negative kinetic energy expectation value.

However, i don't understand what this negative kinetic energy is telling us qualitatively. Maybe somebody here has a greater insight to this problem?

Jon said...

Carlos: One thing we talked about in the discussion section is that if you graph that wave function you'll notice that the curvature is always positive for all x (along with the value of the wave function). When this is true, the expectation value of T should be negative. As Chao pointed out you can mathematically get the negative KE using the operator.

Another thing I noticed is that if you plug in the wave function to the s-eq and compare the powers of x, you'll see that the energy (which is constant) is a negative value. Furthermore, you'll see that V(x) is a function of |x| so it is always going to be positive. Therefore KE has to be negative for all x. This is just another way of showing mathematically why KE is negative here. However, if there's a more intuitive physical explanation (aside from that bit about curvature mentioned above) then I'm missing it. Hope this is helpful

Jon said...

I screwed up in the second part of my above post. Plugging into s-eq doesn't show you positive V(x), it shows you V(x) = 0. This makes more sense I guess, since this wave function reeks of the delta potential with is zero except for at one location. Anyway, the conclusion is still the same - KE must be negative. See everyone tomorrow

Vince said...

Thanks Chao and Jon.

Bobby said...

It was my understanding that a negative KE meant that your wavefunction (in the regions with KE < 0) is one of a tunneling particle. If you look very closely at problem 3 you'll realize that the wavefunction is the piecewise part we use in the finite square well to describe the behavior right after the well ends. This might not be the *only* scenario in which this happens but something just feels right about this conclusion.

I'm not sure if this is right. Does it sound right to anybody else? Zack?

Derek said...

for 11 does finding the expression for psi(x,t) just finding the Cn's in integral form?

Isotope said...

since prob. 11 only asks to find the expressions for certain quantities, i took that to mean writing them in integral form, so i would say yes.

I'm not sure if this is useful, but could the negative KE be related to positrons/hole theory? In a vacuum the negative E states are filled while positive states are open. but if a neg. E electron undergoes a transition to a pos. E state, this opens up a "hole" in the vacuum which we observe as a positron. I'm not sure if this relates to the problem however.

Zack Schlesinger said...

reminder: 6 and 7 are due tomorrow (Oct 2)

Zack Schlesinger said...

How is it going with those?

Tim said...

5 and 6 seem pretty ok. I'm just on 5 and I'm not shure what Zack means by 5b "...looking suitably normalized". I'm not shure how to figure this out from a sketch

Anonymous said...

Tim, what Sam and I interpreted that portion of the question to mean was that we should make sure, while drawing our sketch, to keep the total area (ie. probability when squared)under the curve equal to the same amount. So this meant that the 'bumps' for the each state got progressively lower.
The orthogonality would follow if it appeared that if when you multiplied (inner product) any two different states together, you would get zero.

Zack Schlesinger said...

word

Anonymous said...

zack, how much detail are you looking for in the solutions to problem 6a?

captain said...

question, for problem 8b it asks for you to make your wavefunction in terms of "a" without m,w, h-bar, and a bunch of other constants but the only thing is that the book already does that when it lists the general form of psi for hydrogen in terms of the bohr radius. is there anything else you want us to do with that problem?

Zack Schlesinger said...

I don't think so. that sounds fine.

captain said...

zack, I am having trouble reading problem 9b. could you re-type what it says?

Zack Schlesinger said...

It says: describe each state. Include sketches, e.g., of amplitude contours or other useful forms of representation. Discuss phase, amplitude, planes of nodes (i.e., where the w-f is zero). Whatever else is useful or needed to give the reader a sense of the nature of the w-f (wave-function).

This tends to be difficult for many people. This sort of visualization and description approach is perhaps not a skill that you have developed yet. It will be of great value in this class.

captain said...

zack, for the amplitude contour would you like us to hold theta and phi constant and graph with respect to r or do have another preference?

Megan said...

For problem 8 (a) what luke was commenting about above, Is the bohr radius the characteristic length scale you are looking for? should we be evaluating the bohr radius in the limit that h->0 ?

Zack Schlesinger said...

just the regular old Bohr radius. Sorry, i think i made that seem more complicated than necessary.

Zack Schlesinger said...

regarding the amplitude contours, I can't really answer too clearly, but I am pleased that you are thinking about that and would encourage others to think about this too and to comment here regarding this question.

Tim said...

I made my plots with Gnuplot. This worked quite well. A good starting point for this program is always http://t16web.lanl.gov/Kawano/gnuplot/index-e.html

There are some useful hints for 3D plotting.