Sunday, May 12, 2019

HW 7 with solutions.

Please turn in HW 7 to Michael Saccone's mailbox in the physics department mailroom by 6 PM on Monday,  May 20
The solutions for problem 7 are in a separate post, posted May 22, entitled Dirac notation and sp2 states.
Solutions link: https://drive.google.com/file/d/1WZdGvZbKe-ilqPXrynWSYkz0DuESglmZ/view?usp=sharing


1. The interaction potential for the hydrogen atom system is V(r)=e2/(4πϵor). This is an attractive potential with a strength of e2/(4πϵo).  Note that e2/(4πϵo)=1.44eVnm.  One can calculate a length scale for an electron in this potential as a=2m(e2/(4πϵo)). (Notice that this has a form very similar to the attractive delta-function length scale, with the strength of the interaction in the denominator and 2 in the numerator.)
a) Graph V(r) as a function of x from -6a to 6a for y=z=0 (that is, along the x axis). What are its values in eV at x= a, 2a and 4a, respectively.  Show on your graph.
b) Illustrate the nature of this function with an x-y plane contour plot which includes specific contours (lines of constant value) at the energies you found in part a.

2. The ground state wave-function of the hydrogen atom electron is: ψ1s=1πa3er/a.
a) Graph this wavefunction as a function of x from like -4a to +4a for y=z=0 . What are the its units? 
b) Illustrate the nature of this wave-function using an x-y plane contour plot.
c) Calculate the expectation value of the potential energy of an electron in this ground state. In the context of that calculation, what value did you find for the expectation value of 1/r?
d) Given that a=.053 nm, what is the PE Eval in eV?
e) Show that the KE expectation value is for this localized electron is 2/(2ma2). Add the PE eval and KE eval to get the total energy in eV. What does it mean that this energy is negative?  [It means that the electron is _____.  (I am thinking of a 5 letter word that starts with b.)]  Note that the ratio of the PE to KE, -2, is associated with the virial theorem for a 1/r potential.
f) extra credit: What is the actual value of the WF at x=0 in nm3/2 and why is it so big?

3. 1st ES. Three mutually orthogonal first-excited-state wavefunctions are:
 ψ2x=132πa3(x/a)er/2a,   ψ2y=132πa3(y/a)er/2a,   ψ2z=132πa3(z/a)er/2a.
This choice of 1st excited states is not unique. I think it is a really good choice, but that is just an opinion. There are many ways to choose and organize first excited states. For example,
ψ2,1,1=12(ψ2x+iψ2y) is an eigenstate of Lz as we will see in problem 8. Anyway, let's look at these three.
a) Graph ψ2x as a function of x from like -6a to +6a for y=z=0.
b) Illustrate the nature of ψ2x using an x-y plane contour plot.
c) Illustrate the nature of ψ2y using an x-y plane contour plot.

4. Just like in 1D, we can make a state that models an electron moving back and forth using a superposition of the ground state and a first excited state. For example, consider an electron in the state, Ψ=12(ψ1seiE1t/+ψ2xeiE2t/)=eiE1t/2(ψ1s+ψ2xei(E2E1)t/).
a) Calculate the expectation value of x at t=0.
b) Illustrate the nature of this wave-function at t=0 using an x-y plane contour plot.
c) Graph Ψ at t=0 as a function of x from -6a to 6a for y=z=0 (that is, along the x axis). (Don't forget that r represents the positive branch of the square root...).
d) extra credit: Calculate the expectation value of x as a function of time.

5. Three orthogonal 1st-excited states of the form x, y, and z, as above, is what you expect in a spherically symmetric potential. The states ψ2x, ψ2y, and ψ2z solve the TISE and are mutually orthogonal. It makes sense that they have the same energy since x, y and z are all equivalent directions. This is essentially what you see in a harmonic oscillator in 3D. However, the weird thing is, for the 1/r potential, there is a fourth linearly-independent 1st-excited state,
ψ2s=132πa3(r/a2)er/2a.
The importance of this 4th 1st-excited state cannot be overstated.
a) Graph ψ2s as a function of x from like -6a to +6a for y=z=0. Where is it zero?
b) Illustrate the nature of ψ2s using an x-y plane contour plot.

6. ψ2s can be used, in conjunction with one of the other 1st excited states, to construct an off-center energy eigenstate. This is something we have not encountered before for a potential that is centered at 0 and symmetric. Specifically we can make a energy eigenstate for which x0 despite that fact that the potential is centered at 0 and symmetric. Consider, for example,  the state: ψoc1=12(ψ2s+ψ2x)
a) Calculate x for an electron in this state.  Is x time dependent?
b) Is this an energy eigenstate?
c) Graph ψoc1 as a function of x from like -6a to +6a. Does it extend out further on one side than the other?
d) Illustrate the nature of this state using an x-y plane contour plot of ψψ.

7. Bonding involving carbon in graphene and in biological systems involves a basis for the first excited states in which there are three off-center states in the x-y plane, each with the same amount of ψ2s mixed in.
a) What are the expectation values of x and y for the state ψsp21=13ψ2s+23ψ2x.
b) Consider the state ψsp22=13ψ2s16ψ2x+12ψ2y.  Are this state and the state from part a mutually orthogonal?  Calculate the expectation values of x and y for an electron in the state  ψsp22.
c) On an x-y plot, put a dot where your expectation values (x,y) lie for each of these two states.
d) Find a third state, ψsp23. Show that is orthogonal to ψsp22. Show that it is normalized. (Any integral that comes up is either zero or 1 since your original basis was orthonormal, so this should take very little time once you understand the concept.)
e) extra credit: The 4 states, ψsp21 ψsp22 ψsp23 ψ2z form a orthonormal basis of the sub-space of 1st-excited states of hydrogen. Discuss your feelings about that. Does this basis help you understand bonding in graphene or in carbon rings? (Note added: if you feel like this is maybe a start, but you don't really understand how bonding works or exactly how these states fit in, I feel that that is wise. I think there is more to it that just these states. For example, what does the phrase "singlet-coupled electron pair bond" mean? That sounds like something that involves spin and suggests that we need to learn about spin in order to understand the quantum origins of covalent bonding.)

8. Show that ψ2,1,1=12(ψ2x+iψ2y) is an eigenstate of Lz.
Note that the states: ψ2,1,1=12(ψ2x+iψ2y), ψ2,1,1=12(ψ2xiψ2y), ψ2z and ψ2s form a orthonormal basis of the sub-space of 1st-excited states of hydrogen. (note added: See guide for help with this problem.)

Problem 9, below, is an important problem and once we do it, it can be something that we refer to and discuss, but I am not sure if it really belongs with this HW set. If problems 1-8 take a lot of time, this might not fit. Suggestions and feedback about what to do with problem 9 and when it should be due are welcome.
9. The double well is of great interest because it has two states which are very close in energy to each other and very far in energy from any other states. Long story short, this makes the double well popular in quantum computing research for use as a qubit (the fundamental logic element of quantum computers). There are several other reasons it is an important system to study and understand. For example, when we introduce the "spin" of an electron and electron-electron interactions, the double well may be an interesting system in which to begin to learn about the meaning of spin and how it influences electron-electron correlation. Also, it explains the nature of tunneling as derived directly from the Schrodinger wave equation. This is the fundamental theory of tunneling.
Previously we studied a double well system with 2 identical wells, each 300 meV deep and 2nm wide, separated by a 1 nm wide barrier between them.   You will not need all of the parameters we found for that double-well system, but just for reference they are*:
k1=1.128nm1, A1=.593nm1/2, B1=.238nm1/2, C1=.139nm1/2 and b1=2.57nm1;
k2=1.169nm1, A2=.607nm1/2, B2=.253nm1/2, C2=.134nm1/2 and b2=2.55nm1
 For a couple reasons let's shift the energy at the bottom of the wells to -48.5 meV and have the top at +251.5, so the wells are still -300 meV deep, but with the zero of potential shifted.
a) In this case, are any of the parameters different?  What are the energies of the gs and 1st ES in this case? (This is a short calculation that should only take a minute or two.)
b) Consider an electron in the mixed state Ψ=12(ψ1eiE1t/+ψ2eiE2t/).  Graph Ψ at t=0.
c) At t=0, calculate the fraction of the probability density that lies inside the right hand well.
d) Consider the later time t=π/(E2E1). What is Ψ at that time? Graph Ψ at this time. Graph ΨΨ. Calculate the fraction of the probability density that lies inside the left-hand well at this time.
e) extra credit: Create graphs of Ψ and or ΨΨ at the intermediate time,  t=π/2(E2E1).

* Corrections and comments are welcome.
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 Problems below this line are pure extra credit problems in case someone is really into quantum and has a lot of time.
10. extra credit: a) Are  ψ2x, ψ2y, and ψ2z eigenstates of L2=L2x+L2y+L2z  ?  If so, what is their L2 eigenvalue?
b) Calculate (Lx+iLy)ψ2z.

25 comments:

  1. For 7b, if the states are to be orthogonal, the last term in ψsp22 should be ψ2y, not ψ2x.

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    Replies
    1. We are told the (normalized) states ψ2s,ψ2x,ψ2y are orthogonal, so the inner product between them is 0 or 1.

      For example, consider the wavefunctions:

      ψ1=aψ2s+bψ2x
      ψ2=cψ2s+dψ2x

      <ψ1|ψ2>=ac<ψ2s|ψ2s>+bd<ψ2x|ψ2x>+ad<ψ2s|ψ2x>+bc<ψ2s|ψ2x>

      =ac+bd, since the self inner product terms are 1, and the cross terms are 0

      Delete
  2. Good call. Thank you! Excellent point!

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  3. For anyone interested, the formula I posted for HW5,

    ψ(2x2)ψdx=|ψx|2dx

    generalizes nicely to three dimensions:

    ψ(2)ψdτ=|ψ|2dτ

    Where dτ is a volume element.

    The proof for 3 dimensions is a little more involved than the 1d case. You need to integrate by parts using the formula:

    fA=(fA)Af

    Then apply the divergence theorem, and use the fact that ψ goes to zero at infinity. From there you can show

    (ψ)(ψ)=|ψ|2

    This again has the advantage of only needing to take the first derivatives, rather than the second derivatives. In particular, if ψ is a function of only r, compare the laplacian and gradient operators:

    2r=1r2r(r2r)

    r=ˆrr

    ReplyDelete
  4. To help with 2.b), I have compiled a list of 5 letter words that begin with b (not comprehensive), listed in no particular order:

    bunny
    baked
    burnt
    bagel

    bento
    boxes
    begin
    beefy
    belly
    below
    belts

    billy
    bared
    bling,
    bambi
    bites
    booty

    bored
    baric
    booms
    bongo,
    biped
    bebop

    beach
    bound
    boing
    bacci
    balls
    bogus
    boogy
    board

    bachs
    buddy
    built
    basic
    baton
    badge

    bulky
    bison
    beast
    badly
    begat
    bilbo

    ReplyDelete
    Replies
    1. This comment has been removed by the author.

      Delete
  5. One of my favorite, and yet versatile "b" words that didn't make the list - be-ehs, useful as both a noun ("that's just be-ehs!") and an adverb ("stop be-ehsing me").

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    Replies
    1. I just realized the question gives the answer: The electron is blank

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  6. This comment has been removed by the author.

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  7. 7 may look time consuming with 4 expectation values to calculate, but if you have already done 6, no more integration is necessary for 7. Doing as much work as you can symbollically before using the actual expressions for the wavefunctions can save time as well.

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  8. Is there a 2c? I see the problem jumps from b to d

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  9. Use 3D del^2 in spherical coords. That is in the video posted today.

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  10. Just the 2nd part. Calculate Lzψ. technically Lz can't be a hamiltonian because it does not have units of enrgy.

    Also, maybe check the guide. It has notes on this problem

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  11. I didn't understand what you meant at first, but now i see. This a really important and excellent question. It is a 3D integral. What keeps in from diverging is the 4πr2.

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  12. That's pretty much the idea. Any function of only r will be even with respect to x,y, and z. If you multiply by x it becomes odd, while if you multiply by x2 it stays even. It only needs one "odd factor" for the integral to be zero, e.g. yx2er would integrate to zero.

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  13. I think there are only 2 types of non-zero integrals. They both involve cross terms. 2s*2x and 2s*2y.
    The way to check that is in spherical coordinates. It will be the theta or phi integral that is most likely to make it zero. Once you have those integrals you can evaluate them in seconds on WA. They are definite integrals from 0 to pi or from zero to 2pi, right?

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  14. Any confusion at all regarding theta and phi will totally sink you.

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  15. Correct. Also, in Mathematica, you can use Plot3D to help with the visualization. Not required, but I found it helpful.

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  16. How do other people feel about that? Also, I am adding a note on this problem to the HW7 guide. It could make it a bit easier (that is, less tedious and more intuitive).

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  17. I think that this postponing problem 9 would be a good idea

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Midterm 2 solutions

Here are solutions to midterm 2. I see that I did not write it in the solutions, but it is helpful for sp2 type integrals to recall that \(\...