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Question about Bloch's theorem 
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#1
Nov912, 03:26 PM

P: 146

In my textbook, it's doing an example of Bloch's theorem. They're solving it pretty generally. They first solve the problem of the wave function for a single one of the potential bumps (the potential structure that's being repeated), where the potential everywhere except the bump is 0. So they have a wavefunction something like this in the 0 potential regions:
[tex]\psi(x) = Ae^{iKx} + Be^{iKx}[/tex] And the potential repeats every [itex]a[/itex]. Then they say, due to Bloch's theorem, we know that [itex]\psi(x + a) = e^{ika}\psi(x)[/itex] "for appropriate k" (note, lowercase k). I get what the uppercase K is (the wavevector of the wavefunction, and that, but what's the lowercase k? And how is it determined? Thanks! 


#2
Nov1112, 01:30 AM

P: 24

The (lowercase) ##k## is the socalled crystal momentum. Say your system contains ##N## such copies of the potential profile that is being repeated after every ##a## distance. Then using the Bornvon Karman boundary conditions you will get
##k = m\frac{2\pi}{Na}## where ##m## goes from ##0## to ##N1##. And as you may know ##m=0## and ##m=N## are equivalent. As for the relation between ##K## and ##k## I can only think of a mathematical one. You can look at the mathematical relation in this document: http://faraday.ee.emu.edu.tr/eeng245/KronigPenney.pdf The ##\alpha## and ##\beta## (defined in eq. (4) and (5)) are the different values of ##K## in the region without and with bumps respectively. Now, these ##\alpha##'s and ##\beta##'s are given as a function of ##\varepsilon## in eq. (19) and (20) (##\varepsilon## is defined in just the next line). The relation between ##\alpha## and ##\beta## and ##k## is shown in eq. (25) and (26) (this is compactly written in eq. (27)). Now, eq. (27) represents a transcendental equation which has solutions only for specific values of ##E## and ##k##. The fact that only certain values of ##E## and ##k## are permitted can be graphically visualized in Figure 2, where the LHS and RHS of eq. (27) are plotted. In other words, only the portion of the oscillating function inside the two horizontal lines is a valid. If you recall that ##\varepsilon## (##=E_0/V##) is nothing but scaled energy. Therefore in range of ##E##'s where the oscillating function goes outside the window there is a gap in energy. Now if we resurrect our ##k## from the jungle of formulas and substitutions, you can make a plot of ##E## vs. ##k## and get Figure 3. The gap that you saw in the previous figure can also be observed here. 


#3
Nov2412, 11:46 PM

P: 146

Thanks! 


#4
Nov2512, 11:12 AM

P: 24

Question about Bloch's theorem
Yes, you can associate ##k## as the wave vectors of phonons. Lattice vibrations give an intuitive feel for what ##k## represents in that specific context. However, ##k## can be defined much more generally.
Similar to how you have quantum numbers ##(n,l,m,s)## (principal, azimuthal, magnetic, spin) in an atomic system, you can define a new set of quantum number ##(n,k,s)## (band index, crystal momentum, spin) in a crystalline system. The logic behind introducing new quantum number ##k## can be seen by looking at the three different representations of the plot of ##E## vs. ##k##. You can refer to Figure 6.1 of: http://www.springer.com/cda/content/...539p174100757 You can notice that part (a) is similar to Figure 3 of the previous document (except here they have included the negative ##k## regions as well). The gap at the points ##n\pi/a## (where ##n## is any integer) can be seen as the forbidden states which resulted from the periodic potential. You can notice that, besides the gaps, the plot is a parabola. This parabola corresponds to the dispersion of a free electron ##E_k = \hbar^2k^2/2m##. In other words, the physical momentum of the electron is given by ##p = \hbar k##. But now, if we move the branches of the plot that are green, red, purple along the ##k## axis by an amount ##2\pi/a## we would get part (b). Let us now introduce another quantum number known as the band index ##n##. The blue, green, red, purple headband indices 1, 2, 3, 4 respectively. Then note that I can write the physical momentum of the electron as ##p = \hbar k + \hbar (n1)G## where ##G = 2\pi/a## is the reciprocal lattice vector (here ##k## is a quantity defined to be limited to ##\pi/a \le k < \pi/a##. It doesn't matter where you put the ##\le## since ##\pi/a## and ##\pi/a## are equivalent). Note: we are still working under the assumption that there are no gaps. After introducing the gaps the above inequality will not hold exactly. Alternatively, I could just define the regular wave vector as ##p = \hbar k^\prime## such that ##k^\prime = k + (n1)G##. Then we have ##E_{k^\prime} = \hbar^2 k^{\prime 2}/2m##. Once we know what the band index is, there is no need to keep track of ##k^\prime##. In other words, ##k^\prime## carries a lot of redundancy. For a given band index, only the quantity ##k## can uniquely determine ##p## and as a result ##E##. This ##k## is the socalled crystal momentum. It is a property of the lattice since it is defined to lie within ##G/2 < k \le G/2##. This region in kspace is also known as the first Brillouin zone (also referred to as just the Brillouin zone). Aside: Another way to state the redundancy with using ##k^\prime## is to show that the Bloch eigenstates are equivalent if you translate along the ##k## axis by ##G##. You can replace ##k \rightarrow k+G## and plug it into the expression for the Bloch eigenstate. You will see that you will simply pick up a phase factor of ##exp(iGx)##. And as you may know, in quantum mechanics, two states differing by a phase factor are equivalent. Unfortunately, all I did was make some more mathematical arguments to define what ##k## is. But hopefully some of the physical arguments that motivated the definition of ##k## is much more insightful than noticing ##k## as just some intermediate variable in your calculation. Evidently, in the latter case one may wonder if ##k## is some arbitrary mathematical artifact. Well, in the end you could say that ##k## is an artificial construction. But as you pointed out before, you can gain some physical intuition of ##k## when you think of phonon modes. If you're interested, you can refer to the book: http://www.amazon.com/WavePropagati.../dp/0486600343 to get some more insight. The good thing about this book is that you approach this whole idea of crystal momentum and Brillouin zones from a completely different perspective. 


#5
Dec912, 03:27 AM

P: 146

Hey, thank you so much for the help. I have a related, more practical question now. I wanted to do a problem with a repeating square well potential. Just a simple ##V(x) = U_0## for ##0<x<a##, ##V = 0## for ##a<x<2a## or something. I thought to use a way in my book to solve it, but the solution I found used something totally different (perturbation theory).
In my book they show a general way of solving these  you look at a single period of the periodic potential. You then view the general solution as a linear combo of the equation of a particle scattering from the left, and a particle scattering from the right: ##\psi = A\psi_l + B\psi_r##, ##\psi_l = e^{iKx} + re^{iKx}## for x in the free space to the left of the the potential and ##\psi_l = te^{iKx}## for x in the free space to the right of the potential (and vice versa for ##\psi_r##). ##K## is just related to the energy of the electron through ##E = \hbar^2 K^2/2m##. Then they use Bloch's theorem and say ##\psi(x+a) = e^{ika}\psi(x)## where ##a## is the periodicity and k is "some k" (clearly, the one we've been talking about). From here it's assumed that you can solve your single period scattering problem for ##t##, and including boundary conditions they get a relation between ##k##, ##K##, and ##t## of the form ##cos(Ka + \delta)/t = ka## (where ##\delta## is just the phase of complex ##t##). This gives that famous picture that shows the origin of the allowed energy bands (that I of course can't find now). The question asked me, for a given ##k## (small k), find the energy gap between the first and second bands. To me, this just meant, solve for t (not hard, classic 1D scattering problem), plug it into that equation, and find the two lowest values of ##K## that satisfy it, and find the difference between their corresponding energies. Does that make sense? Thank you, sorry for the long winded reply. 


#6
Dec912, 03:07 PM

P: 24

This approach more or less seems like the one adopted by the link I pasted in my first post:
http://faraday.ee.emu.edu.tr/eeng245/KronigPenney.pdf I don't think that's making use of perturbation theory; this seems like an exact calculation. If you look at the nearlyfree electron model then you will find the use of perturbation theory. You can, however, determine the energy gap between the first and second bands even using the above method (the nearlyfree electron perturbative method is much simpler though). Once you have found ##t## you can solve for the values ##K## and ##k## that satisfy the cosine equation and make a plot of ##E## vs. ##k## similar to Fig. 3 of the above link. From now on let me start making references to that figure since the plot that you'll get will be similar to this one. In Fig. 3 you can observe that the first branch (or band) exists in the interval 0 to ##\pi##, the second one exists in the interval ##\pi## to ##2\pi## and so on. Therefore, the energy gap between the first and second bands is the (vertical spacing) energy spacing between the points at ##\pi## in those two branches (bands). 


#7
Dec912, 04:05 PM

P: 146

Actually, I made a mistake before; the equation relating ##k##, ##K##, and ##t## should be: ##\frac{cos(Ka + \delta)}{t} = cos(ka)## (where ##t## is going to be a function of ##K## usually). The graph I was talking about in my book is like this: The top and bottom lines are just 1 and 1 to show that the values of K that make the curve go above and below them (respectively) can't possibly make that equation true, so those ranges are the band gaps. So, just to make it explicit so I'm sure: If we want to find the band gap for one distinct value of ##k##, I would draw a line across this graph at ##\pm cos(ka)## and find the intersection with the curve. This would give me the values of ##K## corresponding to this ##k##, and then these are easily turned into ##E##'s, like so: Is that correct? Thank you for all the help! 


#8
Dec912, 07:54 PM

P: 24

No, that is not the correct ##\Delta K## (##= K_1  K_2##). That ##\Delta K## corresponds to range of energies ##\Delta E = \hbar^2 K_1^2/2m  \hbar^2 K_2^2/2m## that are allowed in the dispersion relation. You want to determine the ##\Delta E## for which the states are forbidden. That region will correspond to consecutive intersections of the red curve with the cyan line only or the purple line only; NOT the region between the intersection of the red curve with the purple line followed by the intersection of the red curve with the cyan line.
But then what value of ##k## should you use? For each ##k## you will get a different ##\Delta K## (or ##\Delta E##). The band gap is defined as the ##k## for which ##\Delta E## in minimum. In this example, this point will occur at the edge of the Brillouin zone. In this example, ##k = 1/2 \times 2\pi/b##, where ##b## is the size of your unit cell. Since you mentioned that the (barrier region) ##V(x) = V_0## region has a width ##a## and the ##V(x)=0## region also has width ##a##, your unit cell size will be ##b=2a##. Therefore you should find the gap at ##k = \pm \pi/2a##. You should recheck your expression (more specifically your RHS where I think you might have ##\cos(2ka)## instead of ##\cos(ka)##). For ##k = \pm \pi/2a## the RHS will be zero. In other words, your two horizontal lines will collapse onto each other. 


#9
Dec912, 09:14 PM

P: 146

So, you're saying that the ##k## at which ##\Delta E## is a minimum will be at the ##k## that makes the argument of the RHS biggest, i.e., 1. I realized I actually made a mistake in copying the problem to simplify it  in the book a whole period is ##a##, while the way I wrote the potential here it's ##2a## like you pointed out. So wait, this should give you an exact (numerical) answer if I do it right? In that case, is that what I should see the approximations and lower order corrections to the energy from the 'nearly free electron' perturbation method approach? I'd like to do this out and confirm it if that's the case. Thank you so much! 


#10
Dec1012, 04:46 PM

P: 24

Yes, the ##\Delta K## you've shown in the updated figure is correct.
Yes, if you do it the numerical calculations correctly you should get the exact solution. Now, comparing the exact solution to the perturbative one is a little tricky. In the KronigPenney model, which is discussed in the document I sent earlier (more like pasted the link to it), you treat the periodic potential as strong. I must admit that I did not read the expression ##\cos(Ka+\delta)/t=\cos(ka)## (from your second last post) very carefully. I thought that it was just the simplified form of the expression given in that document. Could you please give me the reference to the book you got it from? I'm beginning to think that your expression has some more assumptions embedded in it (in order to bring it to that simplified form). Anyways, for now let us simply assume that we're using the expression in the that document. If that is the case, then the pertubative treatment will fail! The KronigPenney model solved in that document is applicable to a strong potential. In other words the total energy (##E##) of the electron will be comparable to amplitude (##V_0##)of the periodic potential. In order for the nearlyfree electron picture to work, we need the total energy (or kinetic energy) ##E\,(\text{or}\,E_\text{kin}) >> V_0## 


#11
Dec1012, 11:55 PM

P: 146




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