There seems to be a lot of interest in both the hydrogen atom and in electron states in crystals. I think we can do both,and it is true that we can cover angular momentum in the context of studying the hydrogen atom.
Hydrogen will be pretty elaborate: separation of spatial variables in spherical coordinates (we use spherical coordinates because of the symmetry of the potential), solving the theta-phi part (thereby getting angular momentum eig.-states), and then solving the radial part, and finally putting it all together and noticing the unusual degeneracies that occur (s and p states with the same energy...) which are related to less obvious symmetries of the 1/r potential.
With regard to crystal states, I think that some of you will appreciate the opportunity to work on them yourselves, in addition to covering that in class, so I am re-posting both the earlier notes on that and IP#3 here, with the idea class can start in on that as an interactive on-line problem right away. If you work on it i think you would get a much deeper and long-laster appreciation of this important topic, and its beautiful mathematical symmetries.
Here is a suggestion for how to obtain a concrete results. On the last page of the 3-page, handwritten notes, above, there is an expression for Eq. To evaluate that: start by assuming that the denominator is 1 (we can correct that a bit later). Eq is then equal to e0 plus the inner product of phi(x), delta-V(x) and Psi_q (x),
where phi(x) is the ground state of a single, isolated delta-function potential (it should have been called phi_0; and e0 is the energy of that isolated state. (Both of these are well-known to you.) delta-V and psi are both infinite series. That might seem scary, however, try keeping only the 3 terms n=0 and n= +-1 for Psi_q,
and only j=+-1 for delta-V (there is no j=0 term). You will see that this is justified when you do it. Combine complementary terms to get something real, and it will simplify greatly. The integrals are all effortless when you use attractive delta-function potentials!
Please post comments and questions here.
Ok so assuming that the denominator goes to one I get the second term for Eq to be some stuff with some cosines. It doesn't seem right though, the q = pi/a state seems like it'll give me a larger energy than the q = 0 state. Anybody else get something like that? Anyway, just to make sure is this Psi_q supposed to be one electron? If so... how can it be normalized? I don't see how if we make it an infinite set of attractive delta function potentials...
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ReplyDeleteOh nevermind. I forgot to multiply by -alpha. So yeah it makes sense. I guess it makes sense that the ground state of the particle is of less energy than epsilon naught (energy of particle in single attractive delta potential). Since if we are assuming it is a single electron then its wave function is a lot more spread out when all the attractive delta potentials are there.
ReplyDeleteNevermind?? No.
ReplyDeleteI think your first comment raises some important issues, like the relationship between the energies of different states:
"It doesn't seem right though, the q = pi/a state seems like it'll give me a larger energy than the q = 0 state. Anybody else get something like that?"
(and also the technical normalization issue). The questions you raised must be addressed and resolved carefully and the sense of "Eq to be some stuff with some cosines." definitely needs to be elucidating further.
Imagine it is 1928 and you are presenting the first calculation of quantum states of a spatially periodic system to a scientific meeting. What is the essential nature of Eq and of the corresponding states? What is the role of phase? What does this tell us about the energies of crystal states and about the nature and importance of phase in quantum states??
PS. If anyone every told you that in physics the important thing is to understand how to start a problem and that finishing it correctly is less important?:...They were wrong! and they were probably not physicists (though they might have been posing as physicists).
ReplyDeleteGetting a solution is the beginning of the stage where you analyze what it means, what its limitations are...
Graphs and careful consideration are essential!
im not not how to solve the integrals. i used dirac notation initially, and figured the only non-zero terms (due to orthogonality) are the ground state terms :
ReplyDeletePsi_x|-alpha*delta(x-a)|Psi_x and the same thing with x+a. not sure if i did that right? and if i did, im not sure how to solve the integrals.
So just to be clear Psi_q is a many electron state? right? I guess the normalization would change depending on how many terms you're actually taking i.e. how many electrons you have in the system. The expression we have for Psi_q seems only to make sense for the lowest energy state and the highest energy state hence the bandwidth of the system. I got that this bound state of this many electron system seems to be MORE bound than the state with just a single electron and single delta potential. "More bound" meaning larger negative energy and it makes sense that the excited state has less negative energy. I got the bandwidth to be something like
ReplyDelete4(alpha)Phi(a)[Phi(0)+Phi(2a)] ... only taking terms up to -1 through 1
To Jason...
ReplyDeleteI don't think you can say the other terms go to zero due to orthogonality. The orthogonality relates eigenvectors to other eigenvectors of a certain potential system. It seems that the "shifted" bound states of the attractive delta potential aren't related in that way.
one electron. Many wells.
ReplyDeleteyeah, one electron that we can't normalize!
ReplyDeleteZack, you say to keep the n=0 state for the integral, but when I do this i get an exponent that has the same power as terms I would get if if kept the j=+/-2 and let n=+/-2 (that is, a term like exp[-2ja]). Maybe I made a mistake, but it seems like if we are going to keep one term to this power in our sum we should be obliged to keep them all. Did anybody else get a term like this or did I make a mistake?
ReplyDeleteDo 'j' and 'n' both refer to something like the well number?
ReplyDeleteJustin, I got the same expression for the bandwidth. Jason, they're delta-funciton integrals, so you just plug in x=a (or 0 or 2a, etc--the location of the delta potential) into the other part of the integral (something like phi-conjugate(x) and psi_q(x)*e^iqa). Sorry if that wasn't clear. The book for 116 discusses this. Eepter, I think that's right. Does anyone have some insight into part b? Is the redundancy just that e^iqa repeats itself for q=2*pi/a and higher multiples of pi/a? Oh, and Zack, by "atomic state" do you mean the single delta potential?
ReplyDeleteTo clarify, my previous post refers to the calculation of Phi|deltaV(x)|Psi
ReplyDeleteand not Phi|Psi
I can see why you should keep the n=0 term in the latter.
Heather: yes, atomic state just means the single delta-function state.
ReplyDeleteJohn: well as you imply, the important thing is the size of the expontial: exp{-ka} is small, exp{-2ka} is smaller. I would have though that the n=0,l=+-1 terms would give you a relatively large contribution. But maybe i was wrong about that. Hmm, yeah, i think i see now what you mean... I guess the lesson is to trust your own math over my advice.
One thing i would suggest though, not that you should necessarily do what i say, is to keep it as simple as possible and focus on identifying the thing that could be called the "bandwidth" and carefully examine the "overlap integral" that establishes the size of the "bandwidth".
And Heather, identifying a minimal basis set of eigenvectors is critical. There is a repeating, as you mention. When you have that minimal set, then you can see that: Eq is not really an oscillating function at all, and that there is a degeneracy and symmetry.
Hey Zack, I have a few questions in general about what we are doing with this state. The general form you give for the state function seems to work really well for creating a state that is all positive or a state of alternating signs with period equal to the spacing of the delta function potentials. However, shouldn't there be a state with just one node, and shouldn't this state be the first excited state. I am guessing this just based on our rules from previous multiple delta function potentials. Does the state with one node exist?
ReplyDeleteIt seems that we have not done anything with states of q values other than 0 and pi/a. What is the relevance of a general q state?
Also, I am trying to apply my previous knowledge of bandwidth to this problem, but I don't think I see the relevance in the energy difference between these two states. Shouldn't the bandwidth be the energy required to put an electron into the conduction band, which I think means a free electron. It seems that the alternating state is still bound. Shouldn't we look for the energy of the lowest unbound state?
John, I assume that "bandwidth" is a more genereal term. You are talking about the "band gap" which is a one specific example of the "bandwidth".
ReplyDeleteI think what we are doing in this problems refers to the Bloch's theorem. The difference in energies associatd with different q give us energy bands.
Max
Any ideas on what "q" represents?
ReplyDeleteI think that "q" relates to the Bloch wavevector.
ReplyDeleteMax
what q represents? Would a less formal answer be helpful. Sometimes we can get a grasp on the nature of something new, like q, by thinking about how it is similar to (and perhaps also different from) something more familiar. How would that apply to q?
ReplyDeleteJohn. Your questions and thoughts seem almost ready to answer themselves. keep stirring.
ReplyDelete"However, shouldn't there be a state with just one node, and shouldn't this state be the first excited state?..."
"What is the relevance of a general q state?"
What if you did the thought experiment of considering a finite crystal extending from -L/2 to +L/2 (with N=L/a potentials), and then letting L go to infinity?...
In response to John's comment: I think that since we are assuming an infinite lattice, it is impossible denote states by how many nodes they have because there will always be an infinite amount of nodes regardless of our choice of q and a (the complex exponential will change sign an infinite amount of times). In a real crystal, the sum is not from -infinity to infinity, so you can choose a q that is small enough such that it will only change sign once over the entire sum. Just a guess.
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ReplyDeletePerhaps this q value is analogous to the k values in regular systems in that it determines the value of the energies of the excited states.
ReplyDeleteI found this interesting article on wiki:
ReplyDeletehttp://en.wikipedia.org/wiki/Particle_in_a_one-dimensional_lattice_(periodic_potential)
It might be an interesting article, but I am concerned that it could be confusing to many people since it is different from our model and it is confusing to learn learn to different things at once. That (the KP model) is complementary to our model. An advantage it has is the ability to incorporate excited states, however, it involves a more difficult calculation and leads to less insight.
ReplyDeleteOur model is in the class known as tight-binding models. It is simpler and captures the essential role of the overlap of atomic wave-functions in physics. I would suggest focusing on solving and understanding our model and not on the Kronig-Penney model.
John & everybody: well if the ground state is the q=0 state, what is the 1st-excited state(s)?
ReplyDeletewell if we're saying q is continuous which seems reasonable since we have an infinite lattice, it seems it would be hard to have a first excited state in the usual sense but rather in terms of continuity... "first excited state" might be dq if q=0 is ground state for our infinite lattice. Looking at the "s" in paranthesis in Zack's question reminds me of the degeneracy so our first excited state(s) might be ...dq + n*2pi
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