Please point out any errors, and ask questions and share thoughts here freely. Problem 7 is now finished (updated) following a discussion of that in the comments section (and it includes a special extra-credit part!).
The assignment had two distinct parts. Problems 1-4 relate to finding the bands (of electron states) for our microscopic 1D model. If you get stuck on 4, which is pretty difficult, I would recommend skipping it and going to 5. Also, you can ask questions (about 4) here, and that is encouraged.
Problem 5 is a transitional problem. It introduces the Fermi function. It is important at this stage to be able to visualize the Fermi function as a function of energy! The energy dependence is the important part.
Problems 6-8 use a somewhat simples DOS and involve calculations of what states are occupied, using the Fermi function. While this DOS is simpler that the one you would get in problem 4, it is still not trivial and understanding the juxtaposition of bands (regions of non-zero DOS) and gaps (regions of zero DOS) is very important. In fact, that, along with visualizing the Fermi function vs energy, is probably the most important thing.
PS. I think 8 is a pretty difficult problem, really 2 hard problems molded into one. Leave yourself enough time for that. Get actual numbers; accurate numbers. The part where you calculate the number of empty states in the valence band involves an approximation for the Fermi function that is somewhat more difficult than the approximation we use for the Fermi function in the conduction band.
1. Our key result from Tuesday's class was:
\( E(k) = E_1 - 2t \, cos(a k) \)
where
\( t = - \int^\infty_{-\infty} \phi_1(x) v(x-a) \phi_1(x-a) dx \)
a) What does \(E(k) \) represent?
b) What does \(\phi_1(x) \) represent?
c) What does \(v(x) \) represent?
d) What does \(E_1\) represent? Thinking about its origins, would you think that \(E_1\) would be positive or negative? Why?
e) What does \(\psi(k) \) represent?
2. Assume that t is positive and that the magnitude of \(E_1\) is about 10x larger that the magnitude of 2t.
a) With just your understanding of the cos function and the relative magnitude of \(E_1\) and t, sketch a thoughtful plot of \(E(k)\). Graph only from \(k= - \pi/a\) to \(+\pi/a\). Take your time on this. Do it thoughtfully and look at it.
b) Where is \(E(k)\) largest? Where is it smallest?
c) What is the difference, in energy, between those highest and lowest points?
d) Looking at this graph, what might you call the bandwidth? (That is, how would you define bandwidth?)
e) extra credit. How would you define the effective mass associated with this band? To what part of the band does effective mass refer? How are effective mass and bandwidth related?
3. Extra credit: Suppose that \(\phi_1(x) = \frac{1}{\sqrt{b}} e^{-|x|/b} \) (This is a reasonable atomic wave-function in the sense that it decays exponentially away from a cusped center (like an H-atom ground state). Additionally, suppose that \(v(x) = -\alpha \delta (x) \) (a delta function).
a) Do you know how to do an integral involving a delta function? If no, ask someone about it or pst a question here. (A lot of people don't seem to know how to do this.) Computationally, it is really easy to do the integrals once you know what to do. Conceptually, think of the delta function potential as a very, very narrow square well. So narrow that \(\phi_1(x)\) is pretty much the same everywhere inside the well, which is all you need to know to integrate since \(v(x)\) is zero outside the well.
b) Calculate t. (The overlap integral)
c) Use t to obtain an expression for \( E(k)\) that is based on the approximation of keeping just the n=j=1 and n=j=-1 terms from "the sum".
d) Plot \(E(k)\) from \(k= - \pi/a\) to \(+\pi/a\). What is the bandwidth? Assuming \(E_1\) is -16 eV, what is the lowest energy state in the band and what is the highest?
e) What are the largest terms form the sum involving \(v(x\) and also from the other sum (over just n). Explore whether they are actually smaller than the terms we kept (e.g., n=j=1) and how including them might alter the band, \(E(k)\).
4. Starting with the 1D band-structure:
\( E(k) = E_1 - 2t \, cos(a k) \)
where \(E_1\) is the atom state energy of the atom state from which the band is derived,
t is not time but is an overlap integral, as we discussed in class on Tuesday (except that we changed the sign of t so that it would be positive for a band made from a symmetric state. a is the lattice parameter, and k is the key variable.
a) What are the units of density of states as a function of energy for our 1-dimensional crystal?
b) What is the density of states as a function of energy? To calculate this you can start by assuming that states are uniformly distributed along the k axis. Also, you can assume that the total number of the states in the band, per cm, is equal to the number of atoms per cm.
[thoughts on problem 4: One can find the energy dependence by differentiating followed by
substitution to eliminate k (and get things just in terms of E). I think
you will get a density of states (as a function of energy) with
"integrable singularities" (at the band edges). The integral of D(E)
over the band should be related to the number of atoms in the crystal,
since there are two states per atom if you include spin (one state per
atom if not). Please post questions, thoughts or comments here.] There are some solutions here and in the link.
https://drive.google.com/file/d/0B_GIlXrjJVn4eXg0NlNzV21JMk0/edit?usp=sharing
There are more problems below this break.
5. Have you heard of the Fermi function? I think it is \( f(E) = \frac {1} {e^{(E-E_F )/KT} +1}\) or, better, \( f(E) = \frac {1} {e^{(E-\mu )/KT} +1}\), where \(\mu\) is called the chemical potential.
a) Plot the fermi function as a function of E.
b) Consider and describe how the function is influenced by \( \mu\).
c) Consider and describe how the function is influenced by kT.
---- part II
6. The density of states that you would get from problem 4 is relevant to some physical phenomena, and emerges naturally from our simple microscopic 1D-crystal model. However, it is not the simplest thing to use in conjunction with the Fermi function.
Using the Fermi function along with a density of states is essential to a lot of physics, including p-n junctions which we will cover soon. So to facilitate understanding the role and use of the Fermi Function, let us now use a density of states (DOS) that is mathematically simple. Let's use a DOS that is constant within each band (and zero between bands). That is, for the ground state derived band let us suppose that \(D(E)=D_1\) for \((E_1-2t_1)\leq E \leq (E_1+2t_1)\).
Additionally, we'll suppose that the density of states contribution for the band that comes from the atomic 1st-exited state is: \(D(E)=D_2\) for \((E_2-2t)\leq E \leq (E_2+2t_2)\),
and \(D(E)=D_3\) for \((E_2-2t_3)\leq E \leq (E_3+2t_3)\) is the DOS contribution from the 2nd-excited state derived band. These 3 non-overlapping contributions constitute the total DOS of a crystal. Does that make sense? Can you picture that?
a) Sketch a plot of the total D(E). (Assume the bands don't overlap.)
b) Suppose each constituent atom, when isolated, has only one electron. Where would \(\mu\) tend to be? (at low T)
c) Suppose each atom had two electrons. Where would \(\mu\) tend to be? (at low T)
d) Suppose each constituent atom had 4 electrons. Where would \(\mu\) tend to be? (at low T) What bands would be full?
7. Continuing with the case from d), which is an important case for semi-conductor physics, suppose each constituent atom has exactly 4 electrons to start and these atoms come together to form a crystal. Use the density of states form from the previous problem. (Constant within in band; zero outside/between bands.)
a) Suppose kT = 25 meV. What else would you need to know in order to calculate the density of electrons in the 3rd band? This is a pretty open ended, vague question. Let's discuss this here so that we can collectively create a precise question.
b) (precise version: added April 7) Suppose kT = 25 meV and, suppose further that \(E_3-2t_3\), the bottom of the 3rd band, is 0.5 eV higher energy than \(E_2+2t_2\), the top of the 2nd band. Can you picture that? We can call that the gap between the 2nd and 3rd bands. Dos that make sense? (see, for reference, your plot from 6a). Further suppose that \(D_2 = D_3 = 4 \times 10^{21} \frac {states}{eV-cm^3}\). I think with all that given, it should be possible to calculate the number of electrons in the 3rd band. Let me know if you need something else, if I have left anything out. [Hint: Where is \(\mu\) on the energy axis? That is a good place to start. Feel free to post thoughts, guesses, results, etc regarding where \(\mu\) is.
c) Extra credit challenging question. If you get this, and I think you can, I will be impressed. Based in what you are given in part b), what can your infer about \(t_2\) and \(t_3\)? (email me at zacksc@gmail.com if you get this.)
8. (This is a hard problem.) Doping in 3 dimensional semiconductor. There are ways to influence the position of \(\mu\), to some degree, that play a very big role in semiconductor physics (mostly doping by chemical substitution). Suppose that \(\mu\) is 0.7 eV above the top of the 2nd band and 0.3 eV below the bottom of the 3rd band. Further suppose that with each band the DOS is \(10^{22}\) states per \(cm^3\) and that both bands have a band-width of 4 eV.
a) At room temperature how many electrons would there be in the 3rd band?
b) At room temperature how many unoccupied states would there be in the 2nd band?
c) How would those numbers be different (and what would they be) if \(\mu\) were instead 0.2 eV above the top of the 2nd band and 0.8 eV below the bottom of the 3rd band?
d) how would you describe the two cases above? What names could you give them?
e) What is the band gap for this semiconductor?

The density of states ends up being positive so it makes sense that there is a minus sign there. I'm not sure why though.
ReplyDeleteThe density of states ends up being positive so it makes sense that there is a minus sign there. I'm not sure why though.
ReplyDeleteAlso, the minus sign reflects the definition of t. I'll add some notes in on that above.
ReplyDeleteYes, the Fermi function is used to describe Fermi-Dirac statistics. In Phys 102, Hee-Sun called it the occupation number for the fermi-dirac distribution.
ReplyDeleteWow, thank you very much for that! It was definitely helpful to see this derived a second time step by step. I noticed you tried using TeX... is that how you embedded math into your comment? How did you do that?
ReplyDeleteNevermind, I found a useful site that explains it: http://mathjaxtest.blogspot.com/
ReplyDeleteJohn, you should try using double dollar signs for inline math. Apparently, it should work.... test: $$ \pm $$ .
Awesome post!
ReplyDeleteFor fellow students who took Physics 102 with me or are simply unfamiliar with Dirac notation, the
ReplyDeleteWikipedia (http://en.wikipedia.org/wiki/Bra%E2%80%93ket_notation)
and Wolfram MathWorld (http://mathworld.wolfram.com/DiracNotation.html)
pages do a good job at explaining it.
So I'm still confused about the "projection on phi" and as to what that means.
ReplyDeleteSo E_1 is equal to the kinetic energy + the potential energy in the center well?
ReplyDelete"So E_1 is equal to the kinetic energy + the potential energy in the center well?"
ReplyDeletePretty close. Not exactly, but that is a good way to envision it and is very close to the exact answer in many cases.
"So I'm still confused about the "projection on phi" and as to what that means."
ReplyDeleteIt is natural to find that confusing. It is a hard thing. Mostly, as I see it, it is a sophisticated mathematical "trick". It is a way to get what we want. E(k)
Anyone know anything about the integrals with the delta functions (cause I don't)?
ReplyDeleteFor anyone else who is confused, I kind of answered my own question through the magic of Google. This page was fairly straightforward:
Deletehttp://mathworld.wolfram.com/DeltaFunction.html
It appears that, for even indices, the integral of the overlap should just equal the value of the function at 0.
regarding math stuff, i think you can also use slash( to start it,
ReplyDeleteand then slash) to end it. (slash means the slash key under the vertical bar.
I'm working on the amendments to problem 4. I think the assumptions we make do make sense, allowing us to integrate over energy and finding an answer that's related to atoms in the crystal. I computed the indefinite integral and got an answer in terms of an inverse trig function, and right now I'm looking at the taylor expansion of this inverse trig function, trying to convince myself that it makes physical sense.
ReplyDelete"I computed the indefinite integral and got an answer in terms of an inverse trig function, and right now I'm looking at the taylor expansion of this inverse trig function, trying to convince myself that it makes physical sense."
DeleteYou have a lot of mathematical skill, but I don't think you need to do that here. Just do the definite integral over the entire band and the put in a multiplicative factor to make that equal to 2n, where n is the density of atoms.
I would not do an indefinite integral or an expansion. It is less difficult than that. (Although it is pretty difficult.)
For 4, I think you may end up with a density of states proportional to \(\frac{1}{2ta (1-(E-E_1)^2/(2ta)^2)^{-1/2}}\). Does that look like what you got?
DeleteAnd you want to integrate that from \(E_1 -2t\) to \(E_1 + 2t\) to get the total number of states in the band, right?
DeleteTry making a change to a unit-less variable \(u =(E-E_1)/2t\) and then I think the integral is from -1 to 1 and the integral has a very simple value (like pi or something like that).
Deleteright. good point!
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DeleteYeah, I'm confused about the mathematical derivation as well. Its says on the problem to differentiate and we're gonna have to integrate but what exactly are we differentiating?
Deletesince DOS is constant k, then number of states in a small range of energy is delta k/ delta E (sketch that). So the density of states vs E is proportional to 1/(dE/dk), where E is E_1-2t cos(ak).
DeleteSo take that derivative and you get a sin(ak) and you can use sin^2= cos^2-1 to get that in terms of E. (Critical step!!)
Then you'll have D(E) proportional to \(\frac{1}{\sqrt{1-(E-E_1)^2/4t^2}}\) more or less. (not guaranteed, but that may be correct). That would give you the dependence of D(E) on energy
The definite integral of that from E1-2t to E1+2t , which is finite, should be equal to 2 n_atom (the number of states in each band). Include a multiplicative factor to make it so.
Does that make sense?
Since DOS is constant as a function of k i meant.
Deletesin^2= 1-cos^2 is even more crucial!
DeleteYeah that makes sense. You said in the problem to differentiate and I was pretty sure it was E(k) that I had to differentiate but I didn't know what variable to differentiate E(k) to or why I had to differentiate E(k) in the first place. I think this was because I didn't exactly know what k was until now. Thank you professor!
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ReplyDeleteCould I ask when this homework is due? I would assume it would be tomorrow, but the title of the assignment still says Thursday.
ReplyDeleteThursday April 10.
DeleteGreat, thanks!
ReplyDeleteThese are really good ideas. I incorporated them into an update of this problem in the post.
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ReplyDeleteI don't think it's possible to solve for t_3. John Shin and I haven't been able to solve for it t_3. Are we overlooking something?
ReplyDeleteIs there any relationship between band gap and band width?
ReplyDeletemaybe indirectly. not really in any direct straighforward way that one can use. short answer: no.
DeleteSo I'm working through number 4 and through some reading my understanding of the density of states is that of the density per volume and energy of the number of solutions to the Schrödinger's. To calculate this value with respect to energy we need to find the total number of solutions to the Schrodinger's equations that have a wavenumber less than k. I'm pretty lost as to how I might go about doing this... Any advice?
ReplyDeleteThe number of states in each band is twice the number of atoms in the crystal. e.g., if there are 10^21 atoms/cm^3 then the integrated (over energy for just one band) number of states in a band will be \( 2 \times 10^{21}\) and the units will be states/cm^3.
DeleteThe units of density of states are state/(eV-cm^3).
Also, by the way, DOS, D(E) and density of states all refer to the same thing, in case anyone is confused about that.
I seen what you are asking now. We want density of states vs energy. Not density of states vs k.
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ReplyDeleteWhat do we do with the different number of electrons in each "constituent" atom? Isn't mu always equal to the Egap divided by 2?
ReplyDeleteI'm not completely sure, but I think mu changes location depending on which bands are filled. I read that the Fermi energy (also called the chemical potential which is mu) is used to describe the top of the electron energy levels. So if the valance band is filled, it seems to me that the Fermi level (mu) should be exactly at the very top of the valance band but its actually above the valance band halfway between the valance and the conducting band which I don't know why.
DeleteHere is the link to the article that I read:
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html
agree: my 2 cents:
DeleteWhen you have 2 electrons per atom, then only the lowest band should be filled, right? So then, I think mu should be above the 1st band, but below the 2nd band. Because f(E) is close to 1 (corresponding to filled states) for E less than mu. And
f(E) is close to zero (signifying empty states) for E greater than mu. Does that make sense?
When you have 4 electrons per atom, then the lowest two bands should be filled, right? So then mu should be between the 2nd band and the 3rd band.
Both of those cases correspond to mu in a band gap, a place between bands where D(E), the density of states, is zero
When you have just one electron per atom, then I think the lowest band would be half-filled and half-empty. Where would mu be in that case? (anyone?)
What kinds of elements would the 1 electron, 2 electron, and 4 electron atom crystal consists of?
DeleteSo if each atom had only 1 electron, than I think mu would be somewhere in the other half of the empty band. Is that right?
Delete"somewhere in the other half of the empty band."
DeleteI am not sure I understand what you mean.
For the one electron per atom situation, you said that the core band would be half filled and half empty. For this case, I think mu would be above the middle of the core band but still within the core band itself. Does that make more sense? I wish I could draw a picture, it would be so much easier to show you.
DeleteThat makes sense. I can totally picture it. Where exactly would it be? (mu, that is)
DeleteI think mu is just in the middle of the region that I mentioned, in the middle of the empty second half of the band so like 3/4 of the way into the core band
Deletemu has units of energy and represents where the crossover from filled to empty states occurs. it is the crossover point. At mu f(E) = 1/2 .
DeleteAlso, I saw that someone posted that in #7, you integrate from E3-2t3 to E3+2t3 which would make sense, but you never directly addressed this... is this correct?
ReplyDeleteI think this is an open question for everyone. what do you all think about this?
DeleteYes, that's correct. In class, Professor George Brown did the same thing, except he shifted the scale to the right by setting \ E_3 - 2t_3 = 0 \ .
DeleteQuick question about that. I thought, when we graph E vs D(E), should E_1, E_2, and E_3 be negative, and since t is much smaller than E, the graph should start from E_1 to the left, E_2 in the middle and E_3 closest to the origin and largest?
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ReplyDeletemaybe, but I think you can make those two approximations and it will be easier and you will get the same answers.
ReplyDeleteFor number 7b, I get an answer that has electrons/(cm)^3 in the denominator. How do you eliminate the (cm)^3 just so that its just the number of electrons?
ReplyDeleteget a finite size crystal?
Deleteyou don't eliminate it. That's fine. That is perfect. The question was unclear (misleading) and you triumphed over that. Good work!
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ReplyDeleteFor 8A, I get a number that is also in electrons/(cm)^3. Is it ok to leave (cm)^3 here as well or do I have to get rid of it?
ReplyDeletethat's fine. as it should be.
DeleteDon't we need the number of atoms for number 7b?
ReplyDeleteNevermind, I figured out how to do it.
DeleteFor 8(e), isn't the band gap just 1eV? The "hard" label is really screwing with my head. Isn't it just as simple as adding 0.8eV or 0.2eV or I am I missing something here, like something to do with the fact that it is 3d, not 1d?
ReplyDeleteI think that 8a, b and c are the hard part. Have you done those? PS. make up a nickname please.
Delete