# Proof that 1d lattice displacement by phonons is given $u_{n\pm 1}(t) = A_ke^{i\omega_k t} e^{i knd}e^{\pm i k d}$

I looked in «Kittel - Introduction to solid state physics», Wikipedia and Google for the derivation that: A phonon of wavenumber $k$ displaces the $s$-th atom in a monoatomic 1d crystal lattice by a distance $u(s,k)$ given by: $$u_{n\pm 1}(t) = A_ke^{i\omega_k t} e^{i knd}e^{\pm i k d}$$ The first two of the above sources write down the equation of motion $$m \frac{\partial ^2}{\partial t^2}u_s = C(u_{s+1} + u_{s-1} - 2u_{s}),$$ where $m$ is the mass of the $s$-atom and $C$ is the spring constant.

Kittel then goes on by just writing down the solution without going through the math, saying that it is the solution of a difference equation. Wikipedia on the other hand excuses itself by «This requires significant manipulation using the orthonormality and completeness relations of the discrete Fourier transform», and then chickens out by writing down the solution. But the internet needs to see the proof.

I hoped that someone would kindly prove the above formula, hopefully thoroughly, using first principles from Calculus, Fourier analysis, real analysis, classical mechanics, Newtonian mechanics, linear algebra, theory of ode and difference equations. I haven't used Fourier analysis for about a year, so when any theorems are used, it'd be great to mark it in the derivation. To me it seems that one way is skip Fourier analysis and instead solve the ode and difference equation, but I don't know how to do it.

This is the FULL description of the problem.

• Isn't the exponential found via trial functions? Or am I misunderstanding what it is you're confused about? Feb 21 '16 at 18:47
• This is a standard wave mechanics problem I'm 99% sure is covered in e.g. Waves by Crawford, if you want a reference.
– user10851
Feb 23 '16 at 6:03
• Ashcroft and Mermin was pretty straightforward with this. Feb 24 '16 at 15:23

I guess, as Peter Diehr put it, it would be best if you would take out your pencil. So to help you on this, I have formulated the following questions to walk you through the problem :

Preliminary work :

1) What is the definition of the discrete Fourier transform for $u_n$?

Actually wikipedia gives it : $$u_n(t) = \sum_{k=1}^{N} U_k(t)~ e^{iknd}$$

2)What is the definition of the inverse transform (i.e. $U_k = f(u_n)$?

wikipedia gives it too (just need a small tweak to adapt it to our problem) : https://en.wikipedia.org/wiki/Discrete_Fourier_transform $$U_k(t) = \frac{1}{N} \sum_{n=1}^{N} u_n(t)~ e^{-iknd}$$

3) What is the periodic condition on the position (crucial)? We suppose that the atoms are uniformly distributed (the average positions of two consecutive atoms are separated by a distance $d$, hence the $d$ in the wikipedia definition of the Fourier transform).

Without periodicity, the Fourier transform would make no sense.

$x(1)=0=x(d(N+1)) \Rightarrow \exp(-ix(1))=\exp(-ix(d(N+1)))=1$. Actually it is better to write the Fourier transform with $\exp\left(\frac{-2i\pi nd}{N}\right)$ than with $e^{iknd}$ to clearly express the periodicity (same wise for the inverse transform). Also, it is more standard to express the sum from $0$ to $N-1$ instead of $1$ to $N$.

4)Orthogonality condition : What is the result of the following sum ?

$$\frac{1}{N} \sum_{n=0}^{N-1} \exp(-iknd) \exp(-ik' nd)$$

Tip : Two cases : a) $k=k'$ and b) $k\neq k'$ (use geometric series)

Back to the problem :

Using the Fourier transform of $u_n$ and the equation of motion, you should get (simple complexe trigonometry -- please excuse the pun -- and substitution) :

$$2 C \sum_{k=1}^{N} U_k ~ \left(\cos(kd) - 1\right) ~ \exp(iknd) = m~\sum_{k=1}^{N} \frac{d^2U_k}{{dt}^2} \exp(iknd)$$

Do you also feel the urge to take out the summation symbol? This where the orthogonality comes in! Actually, in order to take out the summation symbol, you need to... sum.

step 1 : multiply the equation by $\exp(-ik'nd)$

step 2 : what summation are we supposed to do? It is better if you figure it without looking at the spoiler --- think of our preliminary work, especially question 4).

$\sum_{n=1}^{N}$ on both side => orthogonality leaves one result, the equation for $k=k'$

Final solution :

Solve :

$$2C~U_k ~ \left(\cos(kd) - 1\right) = m~ \frac{d^2U_k}{{dt}^2}$$

wikipedia gives the answer : https://en.wikipedia.org/wiki/Phonon. Don't forget that $kd$ is a constant for a given $k$ (therefore for each equation).

Edit (for solving this question) :

So the solution if course of the form :

$$U_k = A_k e^{i\omega_kt}$$

And if the system is excited at one of its normal modes (frequencies $\nu_k = \frac{\omega_k}{2\pi}$), the displacement reduces to just one component :

$$u_n = A_k e^{i \left(\omega_k t + knd \right)}$$

and there is a phase shift of $e^{\pm ikd}$ between two consecutive knots of the chain.

$$u_{n+1}=u_n e^{ikd}$$

• Thanks, I want to accept this as the answer. It was fun to go through it all, and hope others do the same. Just to make sure that, I've understood correctly, we agree that $U_k(t) = A_k e^{i\omega t}$ for each wavenumber $k$. However, if we assume oscillations are sinusoidal composed of exactly one wavenumber $k$, then $u_{n\pm 1} = \sum_{j=k}^k A_j e^{i\omega_j t} e^{i (n\pm 1)j d} = A_ke^{i\omega_k t} e^{i knd}e^{\pm i k d}$? If this is correct, perhaps add it to the reply? That way we arrive at the equation in the title, and the one Kittel states. Do you agree? Thank you for your time. Feb 21 '16 at 15:30
• Ok, I have added the reply to this comment to the main answer. So, I agree whit what you have stated. This answer is meant to have you exercice with DFT and ( just Fourier transform) to understand how to fish on your own next time --- plus some reflexion about the periodic conditions of the problem. And, it has been a pleasure.
– A.G.
Feb 22 '16 at 14:01

You start by solving the recurrence relation; this can be pretty messy, and Kittel skips it as it doesn't contribute to the physical discussion. The physics has already been captured in the development of the equation.

All about solving recurrence relationships: < https://en.m.wikipedia.org/wiki/Recurrence_relation>

• This answer is not helpful but I cannot vote you down because my rating is too low. The problem is first of all that I don't believe you, let me explain to you why that is. The solution of the occurrence relation is $u_s = r_\pm^s$ where $r_\pm = \frac{m \omega^2}{2C} \pm \sqrt{(\frac{m \omega^2}{2C})^2 +\frac{m \omega^2}{C}}$. If you take out a pencil you will discover that somehow have to prove $r_\pm = \exp(\pm ika)$. Feel free to use this to prove yourself right. Feb 20 '16 at 12:58
• Actually, there is a typo, it should be $r_{\pm} = 1- \frac{m \omega^2}{2C} \pm \sqrt{(\frac{m \omega^2}{2C})^2 + \frac{m \omega^2}{C}}$. Sorry about that, please feel free to use this information to show us that you are correct. Feb 20 '16 at 13:13
• Ahhh ... You are welcome to take out your own pencil. In my experience, one learns by doing their own work. I commend you for wanting to see the complete derivation. Post again, showing where you are stuck. Feb 20 '16 at 13:30