3
$\begingroup$

I am studying the construction of the Wannier functions for 1D system with periodic boundary conditions and my reasoning seems to be a little bit different from what I can find in textbooks.

First of all, if my lattice has $M$ nodes then wave vector belonging to the 1st Brillouin zone is discrete and of the form: $$k_{n} = -\frac{\pi}{d}\left(\frac{2}{M}n - 1 \right)$$ where $n = 0, 1, \ldots, M$ and $d$ - lattice spacing. So, $-\pi/d \le k_n \le \pi/d$. In total we have $M+1$ different values. Once we have normalized Bloch functions $\psi_{n}(x)$ we shall define Wannier function as: $$W(x) = \frac{1}{\sqrt{M+1}}\sum\limits_{n=0}^{M}\psi_{n}(x)$$ with $1/\sqrt{M+1}$ coefficient ensuring normalization: $$\int\limits_{0}^{L}dx\ W^{*}(x)W(x) = 1$$ where $L = d\cdot M$. We use the fact that: $$\int\limits_{0}^{L}dx\ \psi^{*}_{n}(x)\psi_{m}(x) = \delta_{n,m}$$ Most textbooks define Wannier function with $1/\sqrt{M}$ which for my taste is incorrect because Wannier function would not be properly normalized. What do you think?

Second thing. Let's consider the simplest case of free particle. Then Bloch functions are the following: $$\psi_{n}(x) = \frac{1}{\sqrt{L}}e^{i k_n x}$$ According to my definition of Wannier function we get: $$W(x) = \frac{1}{\sqrt{L(M+1)}}\sum\limits_{n=0}^{M}e^{i k_n x} = \frac{1}{\sqrt{L(M+1)}}\frac{\sin\left[\frac{\pi}{d}\left(\frac{1}{M} + 1 \right)x\right]}{\sin\left[\frac{\pi}{d}\frac{1}{M}x\right]}$$ In textbooks apart from the already mentioned difference between $M$ and $M+1$ they write: $$W(x) = \sqrt{\frac{M}{L}}\frac{\sin(x \pi/d)}{x \pi/d}$$ I checked numerically that for large $M$ $$\frac{\sin\left[\frac{\pi}{d}\left(\frac{1}{N} + 1 \right)x\right]}{\sin\left[\frac{\pi}{d}\frac{1}{N}x\right]} \approx M\frac{\sin(x \pi/d)}{x \pi/d}$$ Do you know how to show that?

UPDATE

According to definition 1st Brillouin zone in 1D is a set of wave vectors confined to a region: $$-\frac{\pi}{d} \le k_n < \frac{\pi}{d}$$ where $$k_n = \frac{\pi}{d}\left(\frac{2}{M}n - 1 \right)$$ with $n = 0,1,\ldots, M-1$. With this definition Wannier function for free particles looks like this $$W(x) = \frac{1}{\sqrt{M L}}\sum\limits_{n=0}^{M-1}e^{i k_n x} = \frac{1}{\sqrt{M L}}\left\{ \cot\left( \frac{\pi}{d M}x \right) \sin\left( \frac{\pi}{d}x \right) - i\sin\left( \frac{\pi}{d}x \right) \right\}$$ For large $M$ the above function behaves as: $W(x) \approx \sqrt{\frac{M}{L}}\left\{ \frac{\sin(\pi x/d)}{\pi x/d} - i\sin(\pi x/d)\right\}$ which is not the textbook result due uncompensated imaginary part. However, in the original paper by Kohn he replaced discrete sum by integral. In my opinion it is not well defined (additional term is missing). Consider the Euler-Maclaurin formula $$\sum\limits_{n=0}^{M-1}f(n) = \sum\limits_{n=0}^{M} f(n) - f(M) \approx \int\limits_{0}^{M}dn\ f(n) + \frac{1}{2}\left(f(0) - f(M) \right) = \frac{d M}{2 \pi}\int\limits_{-\pi/d}^{\pi/d}dk \tilde{f}(k) + \frac{1}{2}\left(f(0) - f(M) \right)$$ Kohn and other authors forgot about the second term which isn't small and actually doesn't depend on $M$ in the case of free particle. Now discrete formula gives the same answer as integral representation - Wannier function is not purely real as long as we do not include two extreme points in the summation.

$\endgroup$
  • $\begingroup$ $k_0$ and $k_M$ are equivalent and you Should not count both. This gives $M$ instead of $M+1$ different values. And I think you have missed a phase coefficient in the definition of Wannier functions. $\endgroup$ – seyed Jul 24 '16 at 18:37
  • $\begingroup$ @seyed If you do not include $k_{M} $ in the summation you will not get rid of imaginary part. You wannier function won't be real $\endgroup$ – WoofDoggy Jul 24 '16 at 19:51
  • $\begingroup$ Wannier functions are localized wave functions, what's the problem with them being complex functions as other wave functions in QM? $\endgroup$ – seyed Jul 24 '16 at 20:16
  • $\begingroup$ @seyed Indeed they can be complex, but the example for the free particle at the end is real. I will not reproduce textbook result. I also checked that 1st Brillouin zone is defined without $k_0$ or $k_M$, but this lefts me with a puzzle of complex wannier function for free particle. $\endgroup$ – WoofDoggy Jul 24 '16 at 21:06
  • $\begingroup$ Actually if you use integral definition where instead of $\sum\limits_{k}$ one takes $\int\limits_{-\pi/d}^{\pi/d}dk$ the resulting function is the textbook result. Imaginary part disappears because you "sum" symmetrically over positive and negative $k$ and this cancels sine function. In discrete case it makes a difference if you count both extreme points or just one. $\endgroup$ – WoofDoggy Jul 24 '16 at 21:37
1
$\begingroup$

About the limit:

$\frac{\sin[\frac{\pi}{d}(1/N+1)x]}{\sin[\frac{\pi x}{dN}]}= \sin[\frac{\pi}{d}(1/N+1)x]\times \frac{\frac{\pi x}{dN}}{\sin[\frac{\pi x}{dN}]}\times \frac{dN}{\pi x}\approx \sin[\frac{\pi}{d}x] \times 1 \times \frac{Nd}{\pi x} $

in the last step I used $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $1/N+1\approx 1$. after rearranging the terms it would become the one you found.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.