- Explain why each harmonic normal mode of a crystal becomes a quantum harmonic oscillator.
- Define phonons as bosonic quanta of lattice vibration and interpret the meaning of zero-point motion.
- Derive the Bose–Einstein occupation factor and the mean energy of a phonon mode.
- Construct the phonon density of states from mode counting in \(\mathbf q\)-space and interpret the Debye approximation.
- Write the lattice internal energy as an integral over the phonon DOS and prepare the route to heat capacity.
7.1 Roadmap
- Start from the harmonic normal modes obtained in Lecture 6.
- Quantize each mode and introduce phonon creation and annihilation operators.
- Derive thermal occupation and mean mode energy from equilibrium statistical mechanics.
- Define and interpret the phonon density of states.
- Use the Debye approximation as a controlled low-frequency DOS model.
- Bridge to the next lecture: internal energy and heat capacity from the DOS.
| Symbol | Meaning |
|---|---|
| \(\mathbf q\) | wavevector in reciprocal space |
| \(\lambda\) | phonon branch / polarization index |
| \(r\) | number of atoms in the primitive cell |
| \(N\) | number of primitive cells |
| \(V\) | crystal volume |
| \(\omega_{\lambda}(\mathbf q)\) | phonon angular frequency |
| \(Q_{\lambda}(\mathbf q)\) | normal coordinate of a mode |
| \(P_{\lambda}(\mathbf q)\) | conjugated momentum of a mode |
| \(a_{\lambda}(\mathbf q), a_{\lambda}^{\dagger}(\mathbf q)\) | phonon annihilation and creation operators |
| \(n_B(\omega,T)\) | Bose–Einstein occupation factor |
| \(g(\omega)\) | phonon density of states per unit volume |
| \(\omega_D,\Theta_D\) | Debye frequency and Debye temperature, with \(\hbar\omega_D=k_B\Theta_D\) |
7.2 From Harmonic Normal Modes to Phonons
In the previous lecture we diagonalized the harmonic lattice problem and expressed the crystal motion in terms of independent normal modes. The essential next step is conceptual rather than structural: each of these decoupled harmonic modes is quantized in exactly the same way as a single quantum harmonic oscillator.
7.2.1 Harmonic Hamiltonian in Normal Coordinates
In normal coordinates, the harmonic lattice Hamiltonian is a sum of independent oscillators, \[ H_{\mathrm{harm}} = \frac{1}{2} \sum_{\mathbf q,\lambda} \left[ P_{\lambda}^{\dagger}(\mathbf q)P_{\lambda}(\mathbf q) + \omega_{\lambda}^{2}(\mathbf q) Q_{\lambda}^{\dagger}(\mathbf q)Q_{\lambda}(\mathbf q) \right], \] where \(Q_{\lambda}(\mathbf q)\) is the normal coordinate of a vibrational mode \(\lambda\) and \(P_{\lambda}(\mathbf q)\) its conjugate momentum.
In this picture, a crystal is no longer described as \(3rN\) coupled atomic displacements, but as \(3rN\) independent harmonic modes labeled by \((\mathbf q,\lambda)\).
A useful sanity check is immediate:
- one allowed \(\mathbf q\) value for each primitive cell,
- \(3r\) branches for each \(\mathbf q\),
- hence \(3rN\) modes in total.
7.2.2 Canonical Quantization of a Single Mode
For one harmonic mode of frequency \(\omega\), write the Hamiltonian \[ H=\frac{1}{2}\left(P^2+\omega^2Q^2\right) \] and recall the corresponding commutation relation \([Q,P]=i\hbar\).
We now introduce the ladder operators \[ a=\sqrt{\frac{\omega}{2\hbar}} \left( Q+\frac{i}{\omega}P \right), \qquad a^{\dagger} = \sqrt{\frac{\omega}{2\hbar}} \left( Q-\frac{i}{\omega}P \right). \]
Then \[ [a,a^{\dagger}]=1, \] and the Hamiltonian becomes
Therefore the energy eigenvalues are \[ E_n=\hbar\omega\left(n+\frac{1}{2}\right), \qquad n=0,1,2,\dots \]
The operator \(a^{\dagger}\) raises the excitation number by one quantum, while \(a\) lowers it by one.
7.2.3 Quantized Lattice Hamiltonian
Applying the quantized harmonic oscillator mode by mode gives
The number operator \[ \hat n_{\lambda}(\mathbf q) = a_{\lambda}^{\dagger}(\mathbf q)a_{\lambda}(\mathbf q) \] has eigenvalues \(0,1,2,\dots\), so the crystal energy is built from discrete quanta \(\hbar\omega_{\lambda}(\mathbf q)\).
This is the origin of the phonon picture.
7.2.4 Phonons as Elementary Excitations
A phonon is the quantum of a lattice normal mode. For a mode \((\mathbf q,\lambda)\):
- creating one phonon raises the energy by \(\hbar\omega_{\lambda}(\mathbf q)\),
- annihilating one phonon lowers the energy by the same amount.
Because the operators satisfy bosonic commutation relations, phonons are bosons. Multiple phonons may occupy the same mode.
A further important point is that even the ground state is not completely motionless. Each mode contributes the zero-point energy \[ E_{0,\lambda}(\mathbf q)=\frac{1}{2}\hbar\omega_{\lambda}(\mathbf q). \]
Thus the harmonic crystal retains quantum zero-point motion even at \(T=0\).
7.3 Thermal Population of Phonon Modes
Once the normal modes are quantized, thermal physics reduces to the thermal population of independent bosonic modes.
7.3.1 Partition Function of One Harmonic Mode
For one mode of frequency \(\omega\), the energy levels are \[ E_n=\hbar\omega\left(n+\frac{1}{2}\right). \]
The canonical partition function is therefore \[ Z_{\omega} = \sum_{n=0}^{\infty} e^{-\beta E_n} = \sum_{n=0}^{\infty} e^{-\beta \hbar\omega(n+1/2)} = \frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}, \] where \(\beta=\frac{1}{k_B T}\).
From this, we obtain the mean occupation number as \[ \langle n\rangle = \sum_{n=0}^{\infty} n p_n =\frac{1}{Z_{\omega}} \sum_{n=0}^{\infty} n e^{-\beta \hbar\omega(n+1/2)} =\frac{1}{e^{\beta\hbar\omega}-1}. \] Notice that phonons follow the Bose-Einstein statistics. Therefore it is convenient to name this quantity
Also note that for phonons, the chemical potential is zero because phonon number is not conserved.
7.3.2 Mean Energy of One Mode
The mean energy follows immediately: \[ \langle E_{\omega}\rangle = \hbar\omega\left(\langle n\rangle+\frac{1}{2}\right) = \hbar\omega \left[ \frac{1}{e^{\beta\hbar\omega}-1} + \frac{1}{2} \right]. \]
It is useful to split this into two pieces: \[ \langle E_{\omega}\rangle = \underbrace{ \frac{\hbar\omega}{e^{\beta\hbar\omega}-1} }_{\text{thermal part}} + \underbrace{ \frac{1}{2}\hbar\omega }_{\text{zero-point part}}. \]
The thermal part vanishes as \(T\to 0\), while the zero-point part remains.
7.3.3 Two Limits Worth Remembering
High temperature
At high temperature, \(\beta\hbar\omega\ll 1\), so \[ n_B(\omega,T)\approx \frac{k_B T}{\hbar\omega}, \] and hence \[ \langle E_{\omega}\rangle \approx k_B T+\frac{1}{2}\hbar\omega. \]
The temperature-dependent part becomes classical.
Low temperature
At low temperature, \(\beta\hbar\omega\gg 1\), so \[ n_B(\omega,T)\approx e^{-\beta\hbar\omega}, \] and thermal excitation is strongly suppressed.
This already hints at the key role of low-frequency modes: they are the easiest to populate thermally and dominate low-temperature thermodynamics.
7.4 Constructing the Phonon Density of States
The phonon spectrum is usually too complicated to handle mode by mode. The density of states reorganizes the same information in the context of thermodynamics.
7.4.1 From Discrete \(\mathbf q\) Values to the Continuum
For a macroscopic crystal, allowed \(\mathbf q\) values are extremely dense. The sum over wavevectors may be replaced by an integral, \[ \sum_{\mathbf q} \longrightarrow \frac{V}{(2\pi)^3}\int d^3q. \]
This replacement is the reciprocal-space version of passing from a finite crystal to the thermodynamic limit.
A good sanity check is the dimensionality: the factor \(V/(2\pi)^3\) has units of reciprocal-space density, exactly what is needed to count modes in a \(d^3q\) volume element.
7.4.2 Definition of the DOS
Define the partial DOS of branch \(\lambda\) and the total DOS by \[ g_{\lambda}(\omega) = \int \frac{d^3q}{(2\pi)^3} \delta\left(\omega-\omega_{\lambda}(\mathbf q)\right), \] \[ g(\omega)=\sum_{\lambda}g_{\lambda}(\omega). \]
With this definition, for any function \(f(\omega)\), \[ \frac{1}{V}\sum_{\mathbf q,\lambda} f\left(\omega_{\lambda}(\mathbf q)\right) = \int_{0}^{\infty} d\omega g(\omega) f(\omega). \]
So \(g(\omega) d\omega\) is the number of phonon states per unit volume with frequencies between \(\omega\) and \(\omega+d\omega\).
7.4.3 Geometric Interpretation
Let \(S_{\lambda}(\omega)\) be the constant-frequency surface in \(\mathbf q\)-space defined by \[ \omega_{\lambda}(\mathbf q)=\omega. \]
Then \[ g_{\lambda}(\omega) = \frac{1}{(2\pi)^3} \int_{S_{\lambda}(\omega)} \frac{dS}{\left|\nabla_{\mathbf q}\omega_{\lambda}(\mathbf q)\right|}. \]
This formula is physically very revealing:
- where the dispersion is steep, \(\left|\nabla_{\mathbf q}\omega\right|\) is large and the DOS is small;
- where the dispersion is flat, the DOS is enhanced;
- if the group velocity vanishes on special parts of the spectrum, sharp DOS features can appear.
Thus the DOS is high where many nearby \(\mathbf q\) values correspond to nearly the same frequency.
7.4.4 Debye Approximation
For low-frequency acoustic phonons, we can approximate the crystal as an isotropic elastic continuum. Then \[ \omega_{\lambda}(\mathbf q)=c_{\lambda}|\mathbf q|, \] with one longitudinal branch and two transverse branches.
For one branch, \[ g_{\lambda}(\omega) = \frac{1}{2\pi^2}\frac{\omega^2}{c_{\lambda}^3}. \]
Summing the branches gives \[ g(\omega) = \frac{1}{2\pi^2}\omega^2 \left( \frac{1}{c_L^3}+\frac{2}{c_T^3} \right). \]
If one introduces an effective Debye sound velocity \(c_D\) through \[ \frac{3}{c_D^3} = \frac{1}{c_L^3}+\frac{2}{c_T^3}, \] then \[ g(\omega)=\frac{3}{2\pi^2}\frac{\omega^2}{c_D^3}. \]
This cannot hold to arbitrarily large frequency, so the Debye model imposes a cutoff \(\omega_D=c_D q_D\) chosen such that the total number of modes is still correct: \[ \frac{V}{(2\pi)^3}\frac{4\pi q_D^3}{3}=rN. \]
Therefore the Debye DOS is
A crucial sanity check is normalization: \[ V\int_0^{\omega_D} g_D(\omega) d\omega = 3rN. \]
So the Debye model preserves the correct total number of vibrational modes even though it smooths out the detailed branch structure.
7.4.5 What the Debye Approximation Captures and What It Misses
The Debye DOS gets the low-frequency part right because in three dimensions:
- acoustic branches are linear near \(\mathbf q=\mathbf 0\),
- the area of a sphere in \(\mathbf q\)-space scales as \(q^2\),
- and therefore \(g(\omega)\propto \omega^2\) for small \(\omega\).
However, the Debye DOS misses:
- the detailed separation of acoustic and optical branches,
- narrow peaks from weakly dispersive optical modes,
- and sharper DOS features associated with Van Hove singularities.
Real phonon DOS curves therefore agree with the Debye form at low frequency, but deviate strongly at intermediate and high frequency.
7.4.6 Reading a DOS Plot Physically
When you look at a phonon DOS curve, several quick interpretations are useful:
- the total area counts all vibrational modes;
- the low-frequency onset is controlled by acoustic branches;
- broad smooth parts reflect rapidly varying dispersions;
- sharp peaks indicate flat dispersions and slow group velocity;
- nearly dispersionless optical modes often show up as narrow high-frequency structures.
This is why the DOS is such a useful quantity for representing the vibrational spectrum.
7.5 Bridge to the Next Lecture: Internal Energy as a DOS Integral
In the next lecture, we will combine the quantized-mode picture and the DOS into a compact formula for the thermal energy of the lattice.
For the full harmonic crystal, \[ U(T) = \sum_{\mathbf q,\lambda} \hbar\omega_{\lambda}(\mathbf q) \left[ n_B\left(\omega_{\lambda}(\mathbf q),T\right) + \frac{1}{2} \right]. \]
Using the DOS definition, this becomes
Separating thermal and zero-point parts, \[ U(T)=U_{\mathrm{th}}(T)+U_0, \] with \[ U_{\mathrm{th}}(T) = V\int_{0}^{\infty} d\omega\ g(\omega) \frac{\hbar\omega}{e^{\beta\hbar\omega}-1}, \] \[ U_0 = \frac{V}{2}\int_{0}^{\infty} d\omega\ g(\omega) \hbar\omega. \]
This is how heat capacity will come into play:
- \(U_0\) is temperature independent in the harmonic approximation, so it does not contribute to \(C_V\);
- the low-temperature behavior is controlled by the low-frequency DOS;
- the high-temperature limit is controlled by DOS normalization.
In the next lecture, we will evaluate these integrals, especially with the Debye DOS, and extract the temperature dependence of the heat capacity.
7.6 Optional / Appendix: Einstein Model as a One-Frequency DOS
Open optional appendix
Another useful model for the DOS is the Einstein model, where all oscillators have the same frequency \(\omega_E\). This corresponds to \[ g_E(\omega)=\frac{3rN}{V} \delta(\omega-\omega_E). \]
Then \[ U_{\mathrm{th}}(T) = 3rN \frac{\hbar\omega_E}{e^{\beta\hbar\omega_E}-1}. \]
This captures quantization correctly but misses the acoustic low-frequency modes. That is why it gives a qualitatively wrong low-temperature heat capacity, whereas the Debye approximation succeeds there.
- Quantizing lattice normal modes turns the harmonic crystal into a gas of independent bosonic excitations called phonons.
- Each mode behaves like a quantum harmonic oscillator and retains zero-point motion even at zero temperature.
- Thermal phonon populations follow Bose–Einstein statistics because many phonons can occupy the same mode.
- Low-frequency modes are thermally the most important because they are easiest to excite.
- The phonon DOS repackages the dispersion information into a form directly suited to thermodynamics.
- In three dimensions, the low-frequency DOS grows quadratically with frequency for acoustic phonons.
- The Debye model is a low-frequency approximation, not a detailed description of the full phonon spectrum.
- Heat capacity will follow from differentiating the DOS integral for the internal energy.
7.7 Problem Set
One mode, one oscillator. Starting from \[ H=\frac{1}{2}\left(P^2+\omega^2Q^2\right), \qquad [Q,P]=i\hbar, \] define ladder operators and show that the Hamiltonian can be written as \[ H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right). \] What are the energy eigenvalues?
Bose occupation of a phonon mode. For one phonon mode of frequency \(\omega\), compute the partition function \[ Z_{\omega}=\sum_{n=0}^{\infty} e^{-\beta \hbar\omega(n+1/2)} \] and derive the mean occupation number \[ n_B(\omega,T)=\frac{1}{e^{\beta\hbar\omega}-1}. \] Then obtain the mean energy of the mode.
Debye DOS and normalization. Assume an isotropic elastic continuum with one longitudinal sound velocity \(c_L\) and two transverse sound velocities \(c_T\). Derive \[ g(\omega) = \frac{1}{2\pi^2}\omega^2 \left( \frac{1}{c_L^3}+\frac{2}{c_T^3} \right). \] Then show that the Debye cutoff can be chosen so that the total number of modes is \(3rN\).
High-temperature limit from the DOS. Starting from \[ U_{\mathrm{th}}(T) = V\int_{0}^{\infty} d\omega g(\omega) \frac{\hbar\omega}{e^{\beta\hbar\omega}-1}, \] show that if \(k_B T\gg \hbar\omega\) over the relevant spectrum and the DOS is normalized to \(3rN\) modes, then \[ U_{\mathrm{th}}(T)\approx 3rN k_B T. \] What follows for the heat capacity?