Register to reply 
Periodic potential 
Share this thread: 
#1
Jul1207, 05:47 AM

P: 2,066

Is there well known solutions to a differential equation
[tex] \Big(\frac{\hbar^2}{2m}\partial^2+U\cos\big(\frac{2\pi x}{L}\big)\Big)\Psi(x)=E\Psi(x)? [/tex] 


#2
Jul1207, 09:46 AM

P: 529

You could do worse than visit this page:
http://www.falstad.com/qm1dcrystal/ 


#3
Jul1307, 08:10 AM

P: 2,066

I didn't find energy to start finding out how that java thing works. It looks like some kind of simulation. Perhaps interesting, but I would prefer analytic solutions.
But I have more problems related to this same matter: Firstly, I must say that the Bloch's theorem is a very suprising result. Before encountering it, I was quite sure that there should be localized eigenstates of the energy in a periodic potential. I mean a kind of wave functions, that have a global maximum (or an expectation value of position) in some local potential minimum, and the wave function would then vanish when you go farther from this local potential minimum. I found a following proof for the Bloch' s theorem, http://www.tcm.phy.cam.ac.uk/~pdh100...is/node18.html , and I have some problems with it. They explain, that since the translation operator and the Hamiltonian commute, there must be states that are simultaneous eigenstates of the both operators. But not so fast! If [itex][A,B]=0[/itex], and [itex]\psi\rangle[/itex] is an eigenstate of the A operator, then [itex]AB\psi\rangle = BA\psi\rangle = aB\psi\rangle[/itex] (where [itex]a[/itex] is the eigenvalue), and I can see that [itex]B\psi\rangle[/itex] is also an eigenstate with the same eigenvalue. If we also know that the eigenstates of A are not degenerate, then is follows that [itex]\psi\rangle[/itex] must be also an eigenstate of the B operator. However, is it possible to make the same conclusion if we do not know that eigenstates of A are not degenerate? Or is the truth simply, that the eigenstates do not need to be common for both operators, because B could as well take eigenstates of A to some other eigenstates of A with the same eigenvalue? If the answer to that question is, that the eigenstates must not be degenerate for the conclusion to work, then the story continues like this: Is there any reason to assume, that in a periodic potential, the Hamiltonian would have degenerate eigenstates? Well absolutely yes! If [itex]\psi(x)[/itex] is a localized wavefunction that represents an eigenstate of energy, then obviously also [itex] T_{R}\psi(x)=\psi(xR)[/itex] represents another eigenstate with the same energy. So if localized eigenstates exist, they are quite manifestly degenerate. Hups. I didn't notice that, by strange coincidence, olgranpappy had just answered this question here http://www.physicsforums.com/showthread.php?t=176037 . Well the next post still contains a question that I'm interested in. 


#4
Jul1307, 08:47 AM

P: 2,066

Periodic potential
I've tried to solve a differential equation
[tex] \big(\partial^2 + \cos(2\pi x/L)\big)\psi(x)=A\psi(x) [/tex] as the Bloch's theorem suggests. If the solutions to this are found, the solutions to the original DE can be obtained with a change of variables. Substituting an attempt [tex]\psi(x)=e^{ikx}u(x)[/tex] gives an DE [tex] u'' + 2iku' + \big(Ak^2\cos(2\pi x/L)\big)u = 0 [/tex] The Bloch's theorem says, that u(x) should be a periodic, so a Fourier series attempt [tex] u(x) = \sum_{n=0}^\infty a_n\cos\frac{2\pi nx}{L} + \sum_{n=1}^\infty b_n\sin\frac{2\pi nx}{L} [/tex] seems natural. Substituting this, and using equations [tex] \cos(x)\cos(y)=\frac{1}{2}\big(\cos(x+y)+\cos(xy)\big) [/tex] [tex] \sin(x)\cos(y)=\frac{1}{2}\big(\cos(xy)\cos(x+y)\big) [/tex] gives a following result. [itex]a_0[/itex] and [itex]b_1[/itex] may be chosen freely, and rest of the constants are defined uniquely with the following recursion relations. [tex] a_1 = 2(Ak^2)a_0 [/tex] [tex] a_2 = 2a_0 + \Big(2(Ak^2)\frac{8\pi^2}{L^2}\Big)a_1 + \frac{8ik\pi}{L}b_1 [/tex] [tex] b_2 = \Big(2(Ak^2)\frac{8\pi^2}{L^2}\Big)b_1\frac{8ik\pi}{L}a_1 [/tex] [tex] a_{n+1} = \Big(2(Ak^2)\frac{8\pi^2n^2}{L^2}\Big)a_n + \frac{8ik\pi n}{L}b_n  a_{n1} [/tex] [tex] b_{n+1} = \Big(2(Ak^2)\frac{8\pi^2n^2}{L^2}\Big)b_n \frac{8ik\pi n}{L}a_n  b_{n1} [/tex] The asymptotic behaviour [tex] a_{n+1} = \frac{8\pi^2 n^2}{L^2}a_n [/tex] [tex] b_{n+1} = \frac{8\pi^2 n^2}{L^2}b_n [/tex] for large n seems unacceptable, but how do you terminate these sequences? 


#5
Aug207, 07:13 PM

P: 2,066

I'm having some problems with Bloch's theorem now.
Having struggled with differential equations for some time, I've developed some intuition concerning the series attempts. If the recursion relations for the series coefficients are nice, then the attempt had a right form, and if the recursion relations are complicated, then the attempt was of bad kind. For example you could try to solve a harmonic oscillator with an attempt [tex] \psi(x)=\sum_{k=0}^{\infty} a_k x^k [/tex] but the recursion relations would be too complicated to be useful. The attempt [tex] \psi(x)=e^{x^2/2} \sum_{k=0}^{\infty} a_k x^k [/tex] instead leads to the correct answer. Now when I try to solve a SE for the periodic potential, using the series attempt suggested by the Bloch's theorem, I get hopelessly complicated recursion relations. That is confusing And besides this, I cannot see anywhere the SE being solved really. All texts just mention, that the solution is of the form [tex] \psi(x)=e^{ikx}u(x) [/tex] but the u(x) is left unsolved. Besides my complaints about Bloch's theorem, I have one very precise question. How is Bloch's theorem not in contradiction with this: http://www.physicsforums.com/showthread.php?p=1307616 The periodic potential is symmetric when the origo is chosen appropriately, and the solutions should be symmetric or antisymmetric, but the Bloch waves are neither. Now 30sec after posting this, I already got an idea what could be wrong in the apparent contradiction. For example V(x)=0 is symmetric also, but the plane waves are not symmetric or antisymmetric. Does the proof concerning symmetric and antisymmetric solutions somewhere rely on integration by parts, and on assumption that the wave function vanishes in infinities? I've never gone through the proof actually. 


#6
Aug207, 09:18 PM

P: 701

it's not much different in solid state. aside from the very simplest of cases (Drude oscillator, etc.) where you have a nice analytical solution to gain physical insight from, you will be stuck with solving an approximate SE numerically with a fast computer. for anything but the simple cases the math will (generally) get horrendous and intractable. 


#7
Nov2807, 01:17 AM

P: 2,066

Some comments on my previous posts:
My post #3 dealt only with my misunderstanding of the Bloch's theorem. The ending of the post #5 However, the calculations and remarks in my post #4 remain open. If the Bloch's theorem is true, then there exists a basis of solutions, from which other eigensolutions can be formed with linear combinations, that have the form as I assume in the calculation. That means, they are a product of a plane wave and some periodic term. Correct? Am I not correct, when I think that the divergence of the fourier coefficients is a big problem? Do the series terminate? Since the termination of these series is difficult problem, one might think there should be some theory to deal with it. Could the band structure enter the picture already here? Is there going to be some bands for the allowed values of [itex]a_0[/itex] and [itex]b_1[/itex] for which the fourier series terminate? 


Register to reply 
Related Discussions  
Are two signals that make up a periodic signal necessarily periodic?  Precalculus Mathematics Homework  2  
Differential equation periodic and non periodic solutions.  Calculus & Beyond Homework  0  
Periodic potential: Bloch's theorem  Quantum Physics  25  
Electric potential, potential difference, and potential energy  Introductory Physics Homework  2  
Periodic function  Precalculus Mathematics Homework  7 