Which value of c would you choose?

Monday, November 10, 2008

Homework problems to work on ASAP: revised Sunday, more comments please

Many, if not all, of these problems are vague and not clearly articulated. I think they need clarification. Go to it. Clarifying comments and questions are needed, e.g., I think what Zack is asking for here is..., or I don't understand. Are we supposed to ...? That doesn't seem possible, doesn't make sense...

Also through discussion perhaps we can discuss HW priorities and relevance?

I am hoping for and expecting comments from each person actively involved in this class.

41. Suppose a hydrogen atom, for which the electron is initially in the ground state, is perturbed by an oscillating electric field polarized in the x direction, at a frequency which is resonant with the energy difference between the gs and 1st excited states. What are the probabilities, as a function of time, of transitions to any of the 1st-excited states (in the n,l.m basis, unless you prefer another one) and what are the branching ratios (relative probabilities)?
How would the branching ratios change for polarization along y or z?
How could you most effectively stimulate a transition to the 2,1,1 state?
(This is a question, i think, related to the electric field (of the "photon").

42. For variational problems involving two electrons, suppose you write the variational state as the product: Psi_a(r1) Psi_b(r2), where Psi_a and Psi_b are two states that differ in some way, and r1 and r2 are the vector positions of the two electrons, respectively. Discuss both the conceptual and practical significance of symmetrizing the wave function, which can be done via a 2x2 "determinant", i.e.,
Psi_sym = Psia*Psib+Psib*Psia .

You can explore this question with specific examples, e.g., wave-functions for H2 or He. Does symmetrization make a difference in calculating energy expectations values?
Please discuss here...

43. Regarding what we worked on in class today, perhaps you can think of the homework associated with that in terms of the question: Starting from the Schrodinger eqn, and the one-electron orbitals that follow as its solutions (for the -1/r potential): how can we account for and explain the patterns of bonding associated with carbon. Please explain with appropriate math and a few paragraphs of cogent discussion.

44. Consider equal strength delta functions symmetrically arranged around a hydrogen atom center (and of course equidistant from "r=0"). Determine their effect on the degeneracy and order of the 2nd excited states.

45. extra credit review problem: For the 2D harmonic oscillator, analyze the accuracy of the perturbation theory for either: 1) the 4 delta function perturbation from the midterm or 2) the problem with the perturbation V proportional xy problem that we did a long time ago. (hint: i only have some idea of how to actually do something for one of these choices. the other is sort of a decoy. Wouldn't it be funny if that one turned out to be more interesting, haha?

30 comments:

Stokes said...

Zach,

There is no class today.

Derek said...

so because of the orthogonality of the spherical harmonics doesnt this mean that the c's will now become c(n,l,m)? i took an inner product of the total psi with a specific n,l,m state when plugged into TDSE. this allows us to find the c to a specific nlm and thus the probability. this seems pretty crazy. does anyone else seem to find it like this or any other way?

any idea on the difference of branching ratios and probability of a specific n,l,m state?

Stokes said...

Derek,

I have reached the same conclusion. This means that each constant associated with the given nlm state is expressed with the other four constants associated with the remaining states. Not to mention, each has a time dependent derivative also. this a lot of algebra after an integral is taken to each side to get rid of the derivative.

Stokes said...

I am curious, is there a breakdown in degeneracy?

Bobby said...

Adam, what do you mean "the other four constants?"

Stokes said...

Bobby,
The other four constants are associated with the other three first excited states and the one ground state.

Zack Schlesinger said...

Can anyone suggest how this problem (41) might be made (much) less difficult than it seems based on the above comments.

Trapezoidal said...

Well, I think the way to make problem 41 much easier is to first of all not use the nlm basis. A better basis to use would probably be the s, x, y, z basis, where s is the symmetric first excited state psi_200.

Thinking about it in terms of these states it's easy to see that many of the coefficients will go to zero once the TD Perturbation is considered.

Bobby said...

I disagree. If we go back to Cartesian coordinates this problem will become really messy and much more difficult. We have radial symmetry in this problem and to go to Cartesian would forsake that.

So far: we know the frequency is resonant with the energy gap between n=1 and n=2 we can restrict out basis to Psi{1,0,0}, Psi{2,0,0}, Psi{2,1,0}, Psi{2,1,+1}, and Psi{2,1,-1}. But this is still far too many states to be considered simple.

We have asserted in the past that Psi{2,L,M} have the same energy so we know that the electron will be able to go to any one of these. But what determines L and M? Should there be some sort of conservation of angular momentum that we haven't considered so far? What about our rules for electric dipole emission? From these: we know L must change to have an electron collapse from n=2 to n=1 and emit a photon (right?); if true, this should be the reverse process. So this rules out Psi{2,0,0} and then we can treat Psi{2,1,+1} to be a mirror image of Psi{2,1,-1} and just say that they have equal probability(?). If that is correct then we would be in a basis of Psi{1,0,0}, Psi{2,1,0}, and Psi{2,1,+-1} which is probably easier.

I really have no idea if any of this is right but it might get things rolling.

Zack Schlesinger said...

Adam, Chaos, Captain, Tim, Meagan, Sam, Nina, Isotope, Jon: where are your comments?

Zack Schlesinger said...

PS. the latest revision includes some editing of 45 for clarity and a remark about comments in the intro.

Tim said...

To 41: I'm still thinking about the comments. Obviously we can't (or only with a lot of effort) start with a linear combination of five states. The crucial question in this problems seems to me "which state/states is/are favored by a transition from n=1 to 2 and why". The first excited states differ only in angular momentum among themselves, as Bobby noticed. I guess this will be important.

I wonder if one could only consider two states but this would not yield reasonnable ratios of the the probabilities.

Zack, which of the problems are due to tuesday?
Tim

Zack Schlesinger said...

Which problems do you think should be due Tuesday?

Trapezoidal said...

I see no problems with using the sxyz basis, but several with using the nlm. One of the main problems with using the original nlm basis is that you have two states that are imaginary; for the sxyz all the states are real.
The sxyz states are no less radially symmetric than your nlm; additionally since we're dealing with an electric field aligned along a cartesian coordinate using a cartesian basis is natural.
When calculating the < V' >_state1,state2 you have many zeroes as V' ~ x gives you many combinations odd in x over a symmetric interval.
Although you will eventually get the same from starting with the nlm basis, it will be much more work when you could have used a much more natural basis to begin with.

Jon said...

I agree that there's no reason to use the original nlm basis. The polarization in the x direction only makes it easier to do these integrals if you write wave functions in cartesian. However, I think there's some confusion on what Jesse's talking about with the sxyz basis.

The original nlm basis has a couple of complex states. If you remember we made linear combinations of these complex states in order to make everything entirely real. These new basis vectors had terms looking something like r*cos(theta)*cos(phi). This term = x and calls for using cartesian cords. This is what Jesse means by the sxyz basis. First use linear combinations to make everything real, then write everything in cartesian and label appropriately. It makes a lot of the V' integrals very easy because you get odd powers of x integrated over symmetric regions giving 0. Don't be scared of the root(x^2+y^2+z^2) etc. Using this basis is considerably less work than the original nlm basis.

What should be due Tuesday? I dunno, the first 3 maybe? Maybe just the whole thing due Thursday? I dunno, whatever people want is fine with me.

Isotope said...

Regarding 41, i'm still not convinced that changing to sxyz would make the problem easier. And are we assuming that the diagonal matrix elements for this perturbation vanish? i.e. H'_aa = 0 and H'_bb =0 ?

And about the branching ratios...i think this means the ratio of the modulus of c_b and c_a (absolute value squared). or, is it referring to another quantity?

For 42, zack did you mean Psi_sym = Psia*Psib - Psib*Psia? because this is the determinant...but actually your psi_sym is similar to Eqn. 7.37 in the book. and also, you say Psi_a(r) and Psi_b(r) are different in "some way", but it's unclear what they should then be since both represent radial state functions for Hydrogen.

hw should be due on thursday, or just whenever we can each individually get them done.

Megan said...

So we have deduced that by symmetry the only states that contribute to this problem are the 1s and 2x (using Jesse's notation) which makes since because the polarization is in the x direction). Now that we know this we avoid a lot of math and focus in on the time dependent coefficients of the 1s and 2x states. (remember 2x is a linear combination of 211 and 21-1 as jon was saying) Is there a way to qualitatively reason out what these branching ratios are without math and just by thinking about what they should be physically, or is that too hard and complicated, or even not possible?? Zack, what do you think about this? --Megan

captain said...

I agree with Jesse and Jon. Using the x,y,z basis definitely has its advantages. I feel in the sense its easier to see the orthogonality in the x,y,z basis rather than the n,l,m basis when evaluating the expectation value of V.

I think that the homework due on Tuesday should be the first two or three problems.

Zack Schlesinger said...

That sounds good. 1st 3 problems. 41, 43 and 42. 41 everyone has been working on and thinking about; and 43 is an important extension of what we have been doing in class, so we want to keep our momentum on that; and 42 is kinda related to 43 in a way, I guess.

44 and 45 are newer, so they could be due Thursday or something?

Zack Schlesinger said...

How's everybody doing on 43? Is it easy or difficult?

captain said...

Isotope, I think that using the cartesian basis just helps you know which matrix elements are 0.

Nina said...

I understand how to go about finding the linear combination of rotated orbitals, and their relative probabilities. I still feel like single vs. double bonds and valence electrons need to be addressed and i don't know how to go about doing that in the language/math we have been using. Any thoughts?

Also, 44. implies that it does not matter how many delta functions are present as long as they are equidistant from the center...can we extend this to thinking about a continuous delta function at some radius. Does that make sense in terms of the structure of atoms/molecules.

I'm sorry if this is nonsense, i have a head cold...

Chaos said...

Dear Isotope,

I agree with Capitan. The whole point of working in x,y,z basis is to make it easier to identify the non-zero matrix elements. Once these elements are identified, the integrals are easier to evaluate in polar coordinates.

Trapezoidal said...

My current progress on problem 41 leads me to the following conclusions. My branching ratios from the ground state into the first excited state are 0 into psi_2S, 0 into psi_2Y, 0 into psi_2Z, and 1 into psi_2X.

If the polarization were to change to be in the Y, or Z direction, the branching ratio in psi_X would goto 0 and the ratio into psi_Y, or psi_Z would goto 1 respectively.

I got to these conclusions as every coefficient c_blank(t) is a combination of the matrix elements of the other 4 states combined with state "blank". However, since we have a perturbation aligned in a cartesian direction using a cartesian set of basis would clear up which matrix elements are zero and which are not. From this point we then use our cartesian basis vectors in polar to more easily evaluate the values of the matrix elements. This shows that the matrix elements for transitions from the ground state into each of the first excited states except psi_X can't occur. This gives us our branching ratios.

If we want to effectively stimulate the ground state into the nlm basis state psi_211 we should look for some TD perturbation that exercises the orthogonality of the nlm system in the same way as our original x perturbation did. It seems to me that a linearly polarized photon will just not do...

Jon said...

I agree with Nina, I understand the portion we did in class involving finding the new basis vectors. It's pretty straight forward to combine the original basis vectors with the appropriate coefficients. These were the possible wave functions for a Hydrogen bonded to carbon. However, I'm not sure how to go further with this without thinking about other aspects of this problem.

This problem seems to involve a lot of different interaction terms. Hydrogen's electron with the carbon valence electrons, the protons interacting with protons of the other nuclei, etc. It seems like you should think about shielding so that these interaction terms become fewer and simpler. Does that make sense, or am I thinking in the completely wrong direction? Also, there's this issue with double bonds that I don't really understand. Do our basis vectors change in that sort of a situation? I'm not sure.

Anyway, forming the basis vectors with linear combos seems simple enough - but I'm really not sure what to do beyond that. So far I'm just jotting down on my paper whatever thoughts pop into my head.

Derek said...

havent even started 43, but regarding 41, all that it entails you to do is find the matrix elements? so what is the meaning of the off diagonal terms? just kind of curious. it seems like to me that the diagonal terms will give the probability of being in the ground,s,x,y,z, states, but i cant really understand what the off diagonals mean. are they kind of like transitions between similar states?

as well it seems to me that the integrals are the same in cartesian and spherical, just the zero terms are just given by the phi dependence.

any ideas? thanks

Trapezoidal said...

Yeah the integrals are the same, they just seem a little simpler to do in polar. Dealing with Exp[-Sqrt[x^2 +y^2 +z^2] and the like would be a pain in the butt I'd imagine...

I think the matrix elements correpsond more to like row or column elements as they all share the state they're originating from but are different otherwise.

Nathan said...

Can anyone clarify the question being posed in problem 42? It seems to be that both the H2 and He examples have wavefunctions that use both r1 AND r2 since there are 2 electrons in these situations, making some Psi_a(r1,r2) and some Psi_b(r1,r2), not just functions of one of the position vectors alone as the problem states. Am I missing the boat here? Maybe that's the point... any ideas?

Bobby said...

Problem 42 asks you to take what we usually use Psi_1(r_1) + Psi_2(r_2) and mix this around a bit so that we get Psi_1(r_1)*Psi_2(r_2) + Psi_1(r_2)*Psi_2(r_1). Griffiths talks about this sort of method when he does two particle systems in chapter 5. We didn't quite get there in 139A but we should have.

This form is sort of like squaring (and interchanging r_1 and r_2) the normal variational wavefunction but without the (Psi_1(r_1)*Psi_2(r_2))^2 terms. These wouldn't help us describe a system of identical particles. To do that we want to be able to completely swap the two particles' locations and the answer will remain the same. This is why Zack drew the analogy to the determinant, but I found that to just confuse me since there isn't really a matrix here (at least I don't think).

The matrix (if it exists) would be:
|psi_1(r_1) -psi_1(r_2)|
|psi_2(r_1) psi_2(r_2)|

I've never encountered a matrix like this but if anyone has it might give us a clue about two electron systems.

Nina said...

Bobby, thanks for directing us to chapter 5. I think that because electrons are fermions, the combination should actually be the difference between the two terms i.e. Psi_a(r_1)*Psi_b(r_2) - Psi_a(r_2)*Psi_b(r_1), making the matrix a little more standard...Also for fermions, the Pauli exclusion principle applies. (this is also a very late and probably useless response to isotopes comment.