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 \(V(r) = -e^2/(4\pi \epsilon_o r)\). This is an attractive potential with a strength of \(e^2/(4\pi
\epsilon_o)\). Note that \(e^2/(4\pi \epsilon_o)= 1.44 \: eV*nm\). One
can calculate a length scale for an electron in this potential as
\(a=\frac{\hbar^2} {m (e^2/(4\pi \epsilon_o))}\). (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 \(\hbar^2\) 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: \(\psi_{1s} = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}\).
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 \(\hbar^2/(2ma^2)\). 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 \(nm^{-3/2}\) and why is it so big?
3. 1st ES. Three mutually orthogonal first-excited-state wavefunctions are:
\(\psi_{2x} = \frac{1}{\sqrt{32\pi a^3}} (x/a) e^{-r/2a}\), \(\psi_{2y} = \frac{1}{\sqrt{32\pi a^3}} (y/a) e^{-r/2a}\), \(\psi_{2z} = \frac{1}{\sqrt{32\pi a^3}} (z/a) e^{-r/2a}\).
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,
\(\psi_{2,1,1} = \frac{1}{\sqrt{2}} (\psi_{2x} + i\psi_{2y})\) is an eigenstate of \(L_z\) as we will see in problem 8. Anyway, let's look at these three.
a) Graph \(\psi_{2x}\) as a function of x from like -6a to +6a for y=z=0.
b) Illustrate the nature of \(\psi_{2x}\) using an x-y plane contour plot.
c) Illustrate the nature of \(\psi_{2y}\) 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, \(\Psi = \frac{1}{\sqrt{2}} (\psi_{1s} e^{-iE_1 t/\hbar} + \psi_{2x} e^{-iE_2 t/\hbar}) = \frac{e^{-iE_1 t/\hbar}}{\sqrt{2}} (\psi_{1s} + \psi_{2x} e^{-i(E_2-E_1) t/\hbar})\).
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 \(\Psi\) 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 \(\psi_{2x}\), \(\psi_{2y}\), and \(\psi_{2z}\) 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,
\(\psi_{2s} =
\frac{1}{\sqrt{32\pi a^3}} (r/a-2) e^{-r/2a}\).
The importance of this 4th 1st-excited state cannot be overstated.
a) Graph \(\psi_{2s}\) as a function of x from like -6a to +6a for y=z=0. Where is it zero?
b) Illustrate the nature of \(\psi_{2s}\) using an x-y plane contour plot.
6. \(\psi_{2s}\) 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 \(\langle x \rangle \neq 0\) despite that fact that the potential is centered at 0 and symmetric. Consider, for example, the state: \(\psi_{oc1} = \frac{1}{\sqrt{2}} (\psi_{2s} + \psi_{2x})\)
a) Calculate \(\langle x \rangle\) for an electron in this state. Is \(\langle x \rangle\) time dependent?
b) Is this an energy eigenstate?
c) Graph \(\psi_{oc1}\) 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 \(\psi^*\psi\).
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 \(\psi_{2s}\) mixed in.
a) What are the expectation values of x and y for the state \(\psi_{sp2_1} = \frac{1}{\sqrt{3}} \psi_{2s} + {\frac{\sqrt{2}}{\sqrt{3}}}\psi_{2x}\).
b) Consider the state \(\psi_{sp2_2} = \frac{1}{\sqrt{3}} \psi_{2s} - {\frac{\sqrt{1}}{\sqrt{6}}}\psi_{2x} + {\frac{\sqrt{1}}{\sqrt{2}}}\psi_{2y}\). 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 \(\psi_{sp2_2}\).
c) On an x-y plot, put a dot where your expectation values \( (\langle x \rangle, \langle y \rangle)\) lie for each of these two states.
d) Find a third state, \(\psi_{sp2_3}\). Show that is orthogonal to \(\psi_{sp2_2}\). 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, \(\psi_{sp2_1}\) \(\psi_{sp2_2}\) \(\psi_{sp2_3}\) \(\psi_{2z}\) 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 \(\psi_{2,1,1} = \frac{1}{\sqrt{2}} (\psi_{2x} + i\psi_{2y})\) is an eigenstate of \(L_z\).
Note that the states: \(\psi_{2,1,1} = \frac{1}{\sqrt{2}} (\psi_{2x} + i\psi_{2y})\), \(\psi_{2,1,-1} = \frac{1}{\sqrt{2}} (\psi_{2x} - i\psi_{2y})\), \(\psi_{2z}\) and \(\psi_{2s}\) 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*:
\(k_1=1.128 \:
nm^{-1}\), \(A_1=.593\:nm^{-1/2}\), \(B_1= .238\:nm^{-1/2}\),
\(C_1=.139 \:nm^{-1/2}\) and \(b_1= 2.57\:nm^{-1}\);
\(k_2=1.169 \:
nm^{-1}\), \(A_2=.607 \:nm^{-1/2}\), \(B_2= .253\:nm^{-1/2}\),
\(C_2=.134 \:nm^{-1/2}\) and \(b_2= 2.55\:nm^{-1}\)
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 \(\Psi = \frac{1}{\sqrt{2}} (\psi_{1} e^{-iE_1 t/\hbar} + \psi_{2}
e^{-iE_2 t/\hbar})\). Graph \(\Psi\) 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 \(t=\hbar \pi/(E_2-E_1)\). What is \(\Psi\) at that time? Graph \(\Psi\) at this time. Graph \(\Psi^*\Psi\). Calculate the fraction of the probability density that lies inside the left-hand well at this time.
e) extra credit: Create graphs of \(\Psi\) and or \(\Psi^*\Psi\) at the intermediate time, \(t=\hbar \pi/2(E_2-E_1)\).
* Corrections and comments are welcome.
-------------
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 \(\psi_{2x}\), \(\psi_{2y}\), and \(\psi_{2z}\) eigenstates of \(L^2 = L_x^2+L_y^2+L_z^2\) ? If so, what is their \(L^2\) eigenvalue?
b) Calculate \((L_x + i L_y) \psi_{2z}\).
Subscribe to:
Post Comments (Atom)
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 \(\...
In #2g it asks for the value of the wavefunction in units nm^-1/2, but I would've expected the units to be nm^-3/2 due to the square root of a^3 in the denominator out front - which is correct?
ReplyDeleteGood call. Thank you! Excellent point!
DeleteFor 7b, if the states are to be orthogonal, the last term in \(\psi_{sp2_2} \) should be \(\psi_{2y} \), not \(\psi_{2x} \).
ReplyDeletegood point. I fixed that.
DeleteIs there an easier way of showing orthogonality besides multiplying both states together and integrating?
DeleteWe are told the (normalized) states \(\psi_{2s},\psi_{2x},\psi_{2y} \) are orthogonal, so the inner product between them is 0 or 1.
DeleteFor example, consider the wavefunctions:
\(\psi_1=a\psi_{2s}+b\psi_{2x}\)
\(\psi_2=c\psi_{2s}+d\psi_{2x}\)
<\(\psi_1\)|\(\psi_2\)>=ac<\(\psi_{2s}\)|\(\psi_{2s}\)>+bd<\(\psi_{2x}\)|\(\psi_{2x}\)>+ad<\(\psi_{2s}\)|\(\psi_{2x}\)>+bc<\(\psi_{2s}\)|\(\psi_{2x}\)>
=ac+bd, since the self inner product terms are 1, and the cross terms are 0
thank you lance!
DeleteFor anyone interested, the formula I posted for HW5,
ReplyDelete\(\int_{-\infty}^{\infty}\psi^*(-\frac{\partial^2}{\partial x^2})\psi dx=\int_{-\infty}^{\infty}|\frac{\partial \psi}{\partial x}|^2dx\)
generalizes nicely to three dimensions:
\(\int\psi^*(-\nabla^2) \psi d\tau=\int|\nabla \psi|^2d\tau\)
Where \(d\tau\) 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:
\(f\nabla \cdot\vec{A}= \nabla \cdot(f\vec{A})- \vec{A}\cdot\nabla f\)
Then apply the divergence theorem, and use the fact that \(\psi\) goes to zero at infinity. From there you can show
\((\nabla \psi^*)\cdot(\nabla\psi)= |\nabla\psi|^2 \)
This again has the advantage of only needing to take the first derivatives, rather than the second derivatives. In particular, if \(\psi\) is a function of only \(r\), compare the laplacian and gradient operators:
\(\nabla^2_r=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}) \)
\(\nabla_r= \hat{r} \frac{\partial}{\partial r} \)
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:
ReplyDeletebunny
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
Sorry I meant 2f
DeleteThere are at least 393 words that start with lower case b... The ones that you did not have, including but not limited to,
Delete"baaed", "babel", "babes", "baccy", "backs", "baggy",
"bahts", "bails", "bairn", "baits"
BTW, how did you forget BACON ??!!
This comment has been removed by the author.
DeleteBlaze baked blunt, block brain blood, begin bluff. Boson? Bangs. Bound betas!
DeleteOne 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").
ReplyDeleteI just realized the question gives the answer: The electron is blank
DeleteHa. You are right!
DeleteThis comment has been removed by the author.
ReplyDelete7 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.
ReplyDeleteIs there a 2c? I see the problem jumps from b to d
ReplyDeleteOops. That is a typo. I'll fix that.
DeleteThis comment has been removed by the author.
DeleteHi, I have a question for deriving the expected kinetic energy for 1e) using given the wave function as exp(-r/a), should the first and second order derivative be with respect to (d/dr) and (d^2/dr^2)?
ReplyDeleteUse 3D del^2 in spherical coords. That is in the video posted today.
DeleteFor #8a, do we show this by calculating H*Psi = Lz*Psi, where H is the hamiltonian?
ReplyDeleteJust the 2nd part. Calculate \(L_z \psi\). technically Lz can't be a hamiltonian because it does not have units of enrgy.
DeleteAlso, maybe check the guide. It has notes on this problem
Hi there, I know I'm doing something obviously wrong, but I was wondering if anyone could help me figure out why my integrals for 2 c) and 2 e) do not converge? For example, I am solving \(const* \int \frac{1}{x}e^\frac{-|x|}{a} \) from -infiniti to infiniti and it doesn't seem to converge to a PE Eval. Again, I'm sure I'm missing something very obvious! Thanks.
ReplyDeletesorry there should be a 2 in the nominator of the exponent!
DeleteI am now putting together that the answer to my question might lie in today's video lecture!
DeleteI 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 \(4\pi r^2\).
DeleteFor #7b we are having some trouble thinking through the integrals and figuring out which ones are zero or nonzero. We are trying to think in terms of even/odd functions over symmetric intervals but quickly confused ourselves, is there a better way of thinking about this?
ReplyDeleteThat'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 \(x^2\) it stays even. It only needs one "odd factor" for the integral to be zero, e.g. \(yx^2 e^{r}\) would integrate to zero.
DeleteI think there are only 2 types of non-zero integrals. They both involve cross terms. 2s*2x and 2s*2y.
DeleteThe 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?
Any confusion at all regarding theta and phi will totally sink you.
DeleteSo for the x-y plane contours, we are plotting the wf where z=0 right?
ReplyDeleteCorrect. Also, in Mathematica, you can use Plot3D to help with the visualization. Not required, but I found it helpful.
DeleteI think it would be better to have problem 9 be part of the next homework perhaps. Some of the other problems are pretty time consuming and it seems that the theme of problem 9 does not fit perfectly with the remaining problems. (:
ReplyDeleteHow 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).
DeleteI agree. :)
DeleteAgreed
DeleteI think that this postponing problem 9 would be a good idea
ReplyDelete