Subsections

• Idea
• One-Dimensional Crystal
• Modes
• Dispersion Equation. Meaning of Phonons
• Heat Capacity for 1D Crystal
• 3D Crystals

Debye Model of Crystals

Idea

We represented each atom as a harmonic oscillator. It did not work.

Let us represent our system as a set of 3N oscillators, but choose them differently, including collective motion.

One-Dimensional Crystal

We have N particles with masses m on springs k:

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Equation of motion:  

$$m\ddot{u_i} = k(u_{i+1}-u_i) - k(u_i-u_{i-1})$$ (3)

or  

$$\ddot{u}(x) = \omega_0^2\bigl(u(x+a)+u(x-a)-2u(x)\bigr)$$ (4)

Periodic boundary conditions: u₁=u_N+1. Sound velocity:  

$$c=a\omega_0,\qquad \omega_0^2=k/m$$ (5)

Modes

All u_i are entangled! To disentangle them--use a trick:  

$$u(x) = \sum_q \exp(iqx)u(q),\qquad u(q) = u(-q)$$ (6)

What can we say about q? We want u(Na)=u(0), or $\exp(iqNa)+\exp(-iqNa)=1$ for all q. This means that $\cos(qNa)=1$,or

$$q = 0,\,\pm\frac{\pi}{Na},\,\pm\frac{2\pi}{Na},\ldots,\pm\frac{\pi}{a}$$

Physics: we represented deviation as sum of waves:

$\mbox{ \InputIfFileExists{waves.pstex_t}{}{} }$

These waves are called modes

Dispersion Equation. Meaning of Phonons

Substitute (6) into (4):

$$\ddot{u}(q)=-\omega^2(q)u(q)$$

with

$$\omega^2(q) = \omega_0^2\bigl(2-\exp(iqa)-\exp(-iqa)\bigr) = 2\omega_0^2(1-\cos qa)$$

For small q

$$\omega^2\approx q^2a^2\omega_0^2=c^2q^2$$

Solution:

$$u(q,t)= \mathit{const}\, e^{i(\omega t)}$$

In real space we obtain sum of terms proportional to

$$u(x,t) = e^{i(\omega t - qx)} = e^{iq(ct-x)}$$

This describes sound waves propagating with velocity c (that is why we claimed that c is sound velocity)

Equation  

$$\omega^2(q)=2(1-\cos qa)$$ (7)

is called dispersion equation. Each q corresponds to one normal mode or one kind of phonons--in other words, to a sound wave.

Why Debye model is better than Einstein model: since $\omega(q)\to0$at $q\to0$, some phonons are excited even at low temperatures!

Heat Capacity for 1D Crystal

We obtained N independent oscillators (modes!) with frequencies

$$\omega^2(q)=c^2 q^2, \quad q = \frac{\pi}{Na}n, \quad n =0,\,1,\dots, N$$

Each gives a contribution $C_{\text{Einstein}}$ to the heat capacity. Total contribution:

$$C = \sum_q k_B\left(\frac{\hbar\omega(q)}{k_BT}\right)^2\fr... ... \omega(q)/k_BT)}{\bigl(1 -\exp(-\hbar\omega(q)/k_BT)\bigr)^2}$$

If $N\to\infty$--integration instead of summation. We have N values for q in the interval $[0,\pi/a]$--so

$$\sum_q \mapsto 2N\int_0^{\pi/a}\frac{dk}{2\pi}$$

3D Crystals

What's new for 3D?

1.

$\mathbf{q}$ is now vector--instead of dq we must integrate over $d\mathbf{q}=4\pi q^2\,dq$, q²=q_x²+q_y²+q_z².

2.

There are three waves for every $\mathbf{q}$: two transversal and one longitudinal

Instead of integrating by $d\mathbf{q}$ we integrate by $d\omega$. Number of modes:

$$g(\omega)\,d\omega= 3V 4\pi q^2\frac{dq}{(2\pi)^3}$$

Limits: lower limit $\omega=0$, upper limit should be $\Theta_E$, but we used $\omega^2=c^2 q^2$ instead of exact equation (7), so we correct this by taking $\Theta_D$instead:

$$\int_0^{\Theta_D}g(\omega)\,d\omega=3N$$

Result (for simple lattice!):

$$C=3Nk_BD(T/\Theta_D)$$

with  

$$\begin{aligned} \Theta_D \propto\frac{\hbar\omega_0}{k_B}&... ...e^x}{(e^x-1)^2}dx& \quad \text{Debye function} \end{aligned}$$ (8)

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High temperatures:

for $u\gg1$ upper limit in the integral in (8) is small (1/u),

$$\int_0^{1/u}\frac{x^4e^x}{(e^x-1)^2}dx\approx\int_0^{1/u}\frac{x^4}{x^2}\,dx = \frac{1}{3u}$$

and C=3Nk_B

Low temperatures:

for $u\ll1$ integral is constant, and $C\propto T^3$

. This is an example of a scaling, made by a classic of physics as early as 1912!