Added problems:
[ note: both 4's are optional. Mike or Kelsey is supposed to post something explaining the problem with 4.o]
4.o (original) Calculate the expectation value of r (the vector), in the mixed state that is an equal mix of the state 00 and one of the (correctly normalized) +-2hbar eigenstates. You know, something like: (20-02 + i 11) or its friend.
4.1 (alt.revised version) It was cool how r went around in circles in problem 3, right? Create a problem in which r goes around in circles that involves a 2hbar eigenstate (of the 2D HO).
(Problems 5 to X are probably more important to understanding H.)
5. Write Del^2 in:
a) 2D
b) 3D
(oh, i mean in both cartesian and in cylindrical (a) and spherical (b) coordinates.)
6. Can part of Del^2 in 5 be expressed in terms of L^2, where L is the appropriate angular momentum "operator"? (you can answer this with a yes or no, and by writing down the part. You don't have to prove it or derive it.)
7. What is the commutator of Lx and Ly? [hint: use the definition of L and the commutators for px and x, etc.
8. Show that if phi is an eigenstate of Lz, then (Lx+iLy)phi is also an eigenstate of Lz.
(most of the time.., i.e., unless it is zero)
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old:
This assignment involves polar plots of 2D H.O. eigenstates (so learning about polar plots is a good idea). If someone would outline a description here, that would be much appreciated.
If you find polar plots confusing, please feel free to do "contour plots" (plots of surfaces of constant value) instead. The main thing is to develop tools to enable you to understand and visualize the nature of a wave-function in 2 and 3 dimensions. (In 1D, we just plotted Psi(x) vs x. In 2 and 3 D we need to work a little harder on visualization techniques...) This will be essential to understanding the H-atom wave-functions.
Do those 1D H.O. states look like correct, normalized states? I think they are all right, but it is always nice to get a 2nd opinion. Problem 3 is time dependent, right?
So I initially had no idea on how I should begin attacking this...I had some thoughts on conversions to polar coordinates, which is probably the proper way to do it, but abandoned it in favor of a quick visualization from mathematica.
ReplyDeleteSo I got this:
http://people.ucsc.edu/~egarzali/139a/psi0xpsi0y.png
The starter notebook, if anyone is interested, is here:
http://people.ucsc.edu/~egarzali/139a/hw7pt1.nb
Which has crazy hashes going on, probably for the image.
This picture seems to jive with what chemistry I remember - the pictures I had in mind are similar to this:
http://en.wikipedia.org/wiki/File:Hydrogen_Density_Plots.png
So yes, I think the ground state looks good. The 1d state is symmetric about the horizontal axis, so it would only make sense that the 2d state be symmetric about both the horizontal and the vertical axis. Or, in polar coordinate terms, symmetric for any fixed r and a variable theta.
If the usual "nodal analysis" applies, then I suspect for the excited state in x, ground in y, there will be the usual circular symmetry in y, and one node in x - sort of a bow-tie shape. I suppose now the question is one of normalization. Do we impose the normalization condition by saying the double integral in x and y is one, or do we set up two separate integrals for x and y, and set each one equal to one?
To generalize this to three dimensions, ground in x,y,z would suggest a sphere, with the indeterminancy principle making it sort of a fuzzy ball.
Any thoughts? Too long?
So for part 2, I tried putting in a time-dependence term and animating that to get a feel for what was going on. This was probably a bad idea, as I have no idea what it is I am seeing.
ReplyDeleteI did time-dependence for each spatial term as e^((-itE_n)/hbar), where E_n = (n + 1/2)omega*hbar. Four spatial terms yielded four associated time terms.
If anyone wants a look-see:
http://people.ucsc.edu/~egarzali/139a/hw7pt1anim.nb
Talking to myself here...
ReplyDeleteI ended up recalculating the wavefunctions to get them into polar coordinates, and got cleaner-looking pictures. The downside is I get crazy results for part 2.
a) http://people.ucsc.edu/~egarzali/139a/20-02.png (seems reasonable enough)
b) http://people.ucsc.edu/~egarzali/139a/20+02.png - this looks similar to what happens with my 3-body simulation when the code starts getting close to dividing by zero (physically, a collision).
Adding in a time-dependence and animating that yields a static animation: http://people.ucsc.edu/~egarzali/139a/hw7pt1polar.nb
Any thoughts from the hive?
Your plots for #1 and the first part of #2 seem to match what I got. I'm not sure why the second part went so crazy on yours; I did it as a 3-D plot in Mathematica and got a surface that would I think translate to a four-lobed parametric plot, as in the first part. I didn't use polar coordinates, though.
ReplyDeleteSo you did something like z = f(x,y) = [1/sqrt(2)](02 + 20)? That is, the z-axis corresponds to the vertical displacement for both psix and psiy? I guess that makes sense, if we think of the wavefunctions as disturbances on a membrane of some sort...without any boundary conditions. Like an infinite drum!
ReplyDeleteYeah. I'm not sure how valid it is, but it gave me something to visualize while drawing the parametric plots.
ReplyDeleteFor #2, I agree with kcollier, I found plotting functions in 3d to be a good idea.
ReplyDeleteHere are the images that I got from Maple
(mathemtica folks, let me know if yours look the same):
for part a:
http://pics.livejournal.com/amexist/pic/0001bwf6
for part b:
http://pics.livejournal.com/amexist/pic/0001c23s
The second one reminds me of the Bessel functions and vibrations of circular drum (and that makes sense, since we are dealing with the wave equation in 3d). See: http://upload.wikimedia.org/wikipedia/commons/2/23/Drum_vibration_mode03.gif
Max
Max - I like your graphs, they look like what I was expecting to see. Not sure where mine went wrong...I'll keep trying.
ReplyDeleteMax - I just realized why my plot of 20 + 02 looks so screwed up. I think it's exactly the behavior I thought it was - division by something getting close to zero. You're showing a huge well in the center - I regraphed in 3d and got the same thing. I think my 2d plot is screwing up around the singularity at the bottom of the well.
ReplyDeleteSo these look similar to your graphs:
ReplyDeletehttp://people.ucsc.edu/~egarzali/139a/20+02.3d.png
http://people.ucsc.edu/~egarzali/139a/20-02.3d.png
Edolfo, yeah nice graphs.
ReplyDeleteAnd I know what you are talking about - I got the same spiky thing in 2D. I believe, 3D is as way to go.
Max
There's no way I'm getting this without some background. I've been working on the first few by hand and it's just not happening.
ReplyDeleteWTF?
ReplyDeleteNo background?
No notice?
Due tomorrow?
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ReplyDeleteI am wondering if you are thinking that you need to calculate something and maybe you are stuck on how to do that? That is not really the way I am thinking of this assignement. What you are being asked for is a type of sketch that illustrates the nature of a stste in a simple, very minimalistic manner. It is an approach that helps you realize if you understand a state or not. Even if you are good at plotting, like Edolfo, you want to also be able to do and see and understand these quick sketches.
ReplyDeleteThey are the language that physicists use to communicate about states. I don't think that there is any "background" needed. I think this is a matter of general skills related to plotting, visualization and representation.
If you can't do the assignement, that is okay. You will see what the graphs look like soon.
Who will be the Hipp-o-Plot-amus??
I'm using ( ) to represent the arrow brackets, since the comment box seems to think I'm trying to input HTML.
ReplyDeleteThis may be a dumb question... But I'm not sure how to proceed with [(x),(y)]. Commutator? I'm kind of lost, because thinking about it, I would just get (x)(y)-(y)(x). Not sure how to cast our state in the form of (x) either. Is it just the expectation value?
What am I supposed to do with that?
I am thinking it is not the commutator because it is a vector r. The (x,y) are the coordinates of r vector, so the exspect(r) = (expect(x), expect(y))
ReplyDeleteFor #4, I'm getting 0 for both x and y. Is anyone else getting this?
ReplyDeleteHey Mike, I'm getting zero for four as well.
ReplyDeleteHere is a cool website that Ethan showed us in office hours. http://falstad.com/qm2dosc/directions.html
ReplyDeletehttp://falstad.com/qm2dosc/
have fun!
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ReplyDeleteDel^2 is Del dot Del, i think. An inner product. So it turns the vector, Del, into a scalar operator. (not a a vector) (Maybe someone can explain this better...)
ReplyDeleteL is a vector so it is just (Lx, Ly, Lz). A 3 vector with an operator in each spot. L^2 is L dot L (scalar operator).
For these problems you do not have to calculate much. Just look it up somewhere and write it down and stare at it a bit to become familiar..
If some of the mathematical people could comment and clarify the questions Jody raises, that would be great.
Sorry for the delay, I forgot I was supposed to post about the original 4. Yeah, we've decided it's going to be zero, which is a bit boring, and so not really worth doing, unless you just want the practice. If you think about it, it makes some sense--the "00" state is even in x and y, and so are "20" and "02"; so multiplying any of these by "x" or "y" and integrating would yield zero (odd function, symmetric integration region). As for the "11" term, it's odd in both x and y; so, multiplying by x to get the x expectation value would still leave it odd in y (so integral = 0) and likewise multiplying by y to get the y expectation value leaves the function odd in x, and so it integrates to zero.
ReplyDeleteFor 8, isn't "Lx + iLy" an operator, not a state? So how could it be an eigenstate of Lz?
ReplyDeleteOops, forgot about the dot product between the two "del"s in my last post. Actually the calculation was no problem...it just took a second. Sorry, I was just rambling last night on coffee and trying to conceptualize if there is a way to have a non-zero case.
ReplyDeleteMike: true, and that's not the only thing that doesn't make sense in that question. What should the question say?
ReplyDeleteZack could you post some class notes from last Thursday. I couldn't make it because of the 'suspicious device' scare. I understand you went over IP #3. I'm unable to do it.
ReplyDeleteRegarding 8:
ReplyDeleteAre you asking if eigenstates of Lz are necessarily eigenstates of Lx + iLy?
Any resolution on #8? If we're saying the Potential energy is an eigenstate of the L_z operator, are we basically finding a relationship between [L_x+iL_y] and V=const./r? I don't know how to prove something is an eigenstate given another thing is an eigenstate.
ReplyDeleteOriginal version: "Show that if V is an eigenstate of Lz, then Lx+iLy is also an eigenstate of Lz, or else...?"
ReplyDeleteThis is an If-Then formulation. The "Then" part has to be related to the "If" part, right?, otherwise it is meaningless nonsense. The "If" part specifies something about the eigenstate, V. Therefore V has to be mentioned in the "then" part. So anyway, it should have said:
"Show that if V is an eigenstate of Lz, then (Lx+iLy)V is also an eigenstate of Lz.
(btw. V is a vector. It has nothing to do with any potential.)
for number 8...
ReplyDeletedoes that iLy only change the phase of the state, as we have shown in the midterm practice problems? changing the phase doesn't change the energy, and if the energy of vector V is an eigenstate, then vector V + phase factor is at the same energy as vector V, thus vector V + phase change is also an eigenstate.
something like that?
Not really. the combination Lx+iLy is a raising operator relative to Lz, meaning if "V" is an eigenstate of Lz with eigenvalue hbar m, then (Lx+iLy)V is an eigenstate of Lz with eigenvalue hbar (m+1). Actually we will not get to this until about Thursday so it is best to work on HW 8 to prepare for Tuesday.
ReplyDelete