@@ -90,7 +90,7 @@ This underestimation is not obvious at first, but as we will see, this subtle di
## The Debye model
The key simplification of the Einstein model is to consider the atoms as independent quantum harmonic oscillators.
Instead of considering each atom as an independent harmonic oscillator, Peter Debye considered sound waves - the collective motion of atoms - as harmonic oscillators.
Instead of considering each atom as an independent harmonic oscillator, Peter Debye considered the sound waves in a material - the collective motion of atoms - as independent harmonic oscillators.
> ### Sound waves
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@@ -110,14 +110,16 @@ Instead of considering each atom as an independent harmonic oscillator, Peter De
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> The space containing all possible values of $\mathbf{k}$ is called the _$k$-space_ (also named the _reciprocal space_).
The sound waves are characterized by their frequency, wavevector, and polarization. The frequency of the sound wave is determined by its wavevector $\mathbf{k}$ through the _dispersion relation_
The sound waves are characterized by their frequency, wavevector, and polarization. The frequency of the sound wave is related to its wavevector $\mathbf{k}$ through the _dispersion relation_
$$
\omega = v_s|\mathbf{k}|,
$$
where $v_s$ is the _sound velocity_ of a material. In the Debye model, each wave is considered as an independent quantum harmonic oscillators.
As in the previous lecture, the quantum mechanical excitations of a harmonic oscillator are called *phonons*, and the expected number of phonons in the oscillator at temperature $T$ is given by the Bose-Einstein distribution $n_B(\beta \hbar \omega(\mathbf{k}))$. So, instead of having $3N$ oscillators with the same frequency $\omega_0$ as in the Einstein model, we now have $3N$ oscillators (the vibrational modes) with frequencies depending on $\textbf{k}$ through the dispersion relation $\omega(\mathbf{k}) = v_s|\mathbf{k}|$. Apart from this crucial difference with the Einstein model, the calculation of the expectation value of the energy stored in the oscillators proceeds in the same way:
The expected value of the total energy stored in the oscillators (which, from now on, we will simply denote as the total energy) is given by the sum of the energy stored in all the individual oscillators. These oscillators are characterized by their wavevector $\mathbf{k}$:
where $v_s$ is the _sound velocity_ of a material.
As discussed in the previous lecture, the quantum mechanical excitations of a harmonic oscillator are called *phonons*, and the expected number of phonons in the oscillator at temperature $T$ is given by the Bose-Einstein distribution $n_B(\beta \hbar \omega(\mathbf{k}))$. So, instead of having $3N$ oscillators with the same frequency $\omega_0$ as in the Einstein model, we now have $3N$ oscillators (the vibrational modes) with frequencies depending on $\textbf{k}$ through the dispersion relation $\omega(\mathbf{k}) = v_s|\mathbf{k}|$. Apart from this crucial difference with the Einstein model, the calculation of the expectation value of the energy stored in the vibrational modes proceeds in the same way:
The expected value of the total energy stored in the oscillators (which, from now on, we will simply denote as the total energy) is given by the sum of the energy stored in the individual oscillators. These oscillators are characterized by their wavevector $\mathbf{k}$:
\begin{align}
\langle E \rangle &= 3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\hbar \omega(\mathbf{k}) n_{BE}(\hbar \omega(\mathbf{k}))\right)\\
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@@ -140,8 +142,8 @@ We still have several open questions:
We can answer all the above questions by realizing the following:
$C$ is a *macroscopic property*: it should not depend on the material's shape and should only be proportional to its volume.
Therefore, we consider a material with a simple shape to make the calculation for $C$ easier.
The heat capacity $C$ is a *macroscopic property*: it should not depend on the material's shape and only be proportional to its volume.
Therefore, we consider a material with a simple shape to make the calculation of $C$ easier.
The easiest option people have invented so far is a box of size $V = L^3$ with **periodic boundary conditions**[^2].
