- Explain why a finite crystal allows only a discrete set of phonon wavevectors.
- Use Born-von Karman boundary conditions to count allowed states in reciprocal space.
- Define the phonon density of states in reciprocal space and in frequency space.
- Interpret the frequency-space density of states as a histogram of phonon frequencies.
- Derive the general density-of-states formula using constant-frequency surfaces.
- Explain why flat phonon dispersions produce large density-of-states features.
7.1 Why We Need a Density of States
So far, phonon dispersions have been described as functions \(\omega_r(\mathbf q)\): for each wavevector \(\mathbf q\) and branch \(r\), there is a normal-mode frequency. This is the most detailed description of the harmonic lattice spectrum.
For thermodynamics, however, we often do not need to know which wavevector produced a given frequency. We need to know how many modes lie in a given frequency interval. This motivates the phonon density of states.
A useful analogy is a histogram. The dispersion relation is like a detailed list of every student’s exam score together with their name. The density of states is the histogram that tells us how many scores fall into each interval. It loses some detailed information, but it is exactly the right object when we want to compute averages over the whole class.
For phonons, the corresponding question is:
This question is central because the thermal energy, heat capacity, and many scattering intensities depend on sums over all phonon modes.
7.2 Boundary Conditions and Allowed Wavevectors
A real crystal is finite. Therefore not every wavevector is allowed. Boundary conditions select a discrete set of wavevectors, and this discreteness is the starting point for mode counting.
For a macroscopic crystal, the detailed physical boundary condition at the surface should not affect bulk thermodynamic quantities. We therefore choose mathematically convenient boundary conditions, provided they give the correct density of allowed modes in the large-system limit.
7.2.1 Dirichlet Boundary Conditions: Fixed Ends in One Dimension
Consider a one-dimensional chain with lattice spacing \(a\). The equilibrium positions are
\[ x_n=na, \qquad n=0,1,2,\dots,N, \]
and the length of the chain is
\[ L=Na. \]
The end atoms at \(n=0\) and \(n=N\) are fixed, so only the \(N-1\) interior atoms can oscillate.
For a standing wave, we use a superposition of waves traveling to the right and to the left,
\[ u_n = A_1 e^{i(qna-\omega t)} + A_2 e^{-i(qna+\omega t)}. \]
The fixed-end boundary conditions are
\[ u_0=0, \qquad u_N=0. \]
The first boundary condition gives
\[ A_1=-A_2. \]
Therefore
\[ u_n = 2iA_1\sin(qna)e^{-i\omega t}. \]
The second boundary condition requires
\[ \sin(qNa)=0. \]
Thus
\[ qNa=p\pi, \qquad p\in\mathbb Z. \]
Since \(L=Na\), the allowed wavevectors are
The endpoints of this list do not correspond to physical vibrations:
\[ p=0 \quad\Rightarrow\quad u_n=0, \]
and
\[ p=N \quad\Rightarrow\quad u_n\propto \sin(\pi n)=0. \]
Hence the number of nontrivial modes is
\[ N-1, \]
which is exactly the number of atoms that are free to oscillate.
7.2.2 Born–von Karman Boundary Conditions: Periodic Boundary Conditions in One Dimension
Fixed boundaries are physically intuitive, but they make the algebra slightly inconvenient because they produce standing waves. For bulk crystals we usually use periodic, or cyclic, Born-von Karman boundary conditions.
The condition is
\[ u_n=u_{n+N}. \]
Physically, this is like bending a very long chain into a ring. For large \(N\), the local physics is unchanged, but the ends disappear.
Use the traveling-wave ansatz
\[ u_n = A e^{i(qna-\omega t)}. \]
The boundary condition gives
\[ A e^{i(qna-\omega t)} = A e^{i(q(n+N)a-\omega t)}. \]
After cancellation,
\[ e^{iqNa}=1. \]
Therefore
\[ qNa=2\pi p, \qquad p\in\mathbb Z, \]
or
Restricting the wavevector to the first Brillouin zone gives
\[ -\frac{\pi}{a}<q\le \frac{\pi}{a}. \]
Equivalently,
\[ -\frac{N}{2}<p\le \frac{N}{2}. \]
This gives exactly \(N\) allowed values of \(q\), as expected for a one-dimensional chain with \(N\) moving atoms.
The spacing between neighboring allowed wavevectors is
\[ \Delta q=\frac{2\pi}{L}. \]
Thus each allowed state occupies a length \(2\pi/L\) in one-dimensional reciprocal space. The reciprocal-space density of states is therefore
Equivalently, over the first Brillouin zone,
\[ Z(q) = \frac{\text{number of allowed states}}{\text{length of the first Brillouin zone}} = \frac{N}{2\pi/a} = \frac{L}{2\pi}. \]
7.3 Mode Counting in Three-Dimensional Reciprocal Space
The one-dimensional result generalizes directly to a crystal with primitive lattice vectors \(\mathbf a_1,\mathbf a_2,\mathbf a_3\). Let the crystal contain
\[ N=N_1N_2N_3 \]
primitive cells. A Bravais-lattice vector is
\[ \mathbf R=n_1\mathbf a_1+n_2\mathbf a_2+n_3\mathbf a_3. \]
Born-von Karman boundary conditions require
\[ \mathbf u(\mathbf R) = \mathbf u(\mathbf R+N_i\mathbf a_i), \qquad i=1,2,3. \]
For a plane-wave displacement,
\[ \mathbf u(\mathbf R) = \mathbf A e^{i(\mathbf q\cdot\mathbf R-\omega t)}, \]
we obtain
\[ e^{i\mathbf q\cdot N_i\mathbf a_i}=1. \]
Therefore
\[ N_i\mathbf a_i\cdot\mathbf q=2\pi p_i, \qquad p_i\in\mathbb Z. \]
Write the wavevector in the reciprocal basis,
\[ \mathbf q=h\mathbf b_1+k\mathbf b_2+\ell\mathbf b_3, \]
where
\[ \mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}. \]
Then
\[ h=\frac{p_1}{N_1}, \qquad k=\frac{p_2}{N_2}, \qquad \ell=\frac{p_3}{N_3}. \]
Thus the allowed wavevectors are
For odd \(N_i\), the integers may be chosen as
\[ p_i=0,\pm 1,\pm 2,\dots,\pm \frac{N_i-1}{2}. \]
There are therefore \(N_1N_2N_3=N\) allowed wavevectors in the first Brillouin zone.
If the primitive basis contains \(r'\) atoms, then each wavevector has \(3r'\) phonon branches in three dimensions. Hence the total number of vibrational modes is
This is the expected result: each atom has three displacement degrees of freedom.
7.3.1 Reciprocal-Space Density of States
The volume of the first Brillouin zone is
\[ \Omega_{\mathrm{BZ}} = \frac{(2\pi)^3}{V_{\mathrm{WS}}}, \]
where
\[ V_{\mathrm{WS}} = \mathbf a_1\cdot(\mathbf a_2\times\mathbf a_3) \]
is the primitive-cell volume. Since the crystal volume is
\[ V=NV_{\mathrm{WS}}, \]
the reciprocal-space volume occupied by one allowed \(\mathbf q\) state is
\[ \frac{\Omega_{\mathrm{BZ}}}{N} =\frac{(2\pi)^3}{NV_{\mathrm{WS}}} =\frac{(2\pi)^3}{V}. \]
Therefore the density of allowed states in reciprocal space is
This is the central replacement rule:
\[ \sum_{\mathbf q} \longrightarrow \frac{V}{(2\pi)^3} \int_{\mathrm{1.\ BZ}} d^3q. \]
A useful sanity check is dimensional. The factor \(V/(2\pi)^3\) has dimensions of reciprocal-space inverse volume, so multiplying it by \(d^3q\) gives a dimensionless number of states.
7.4 Density of States in Frequency Space
The reciprocal-space density \(Z(q)\) counts allowed wavevectors. But many physical quantities depend on frequency rather than directly on \(\mathbf q\). We therefore introduce a density of states in frequency space.
Suppose we need a sum of the form
\[ \sum_{\mathbf q,r} F[\omega_r(\mathbf q)], \]
where \(r\) labels the phonon branch. For a macroscopic crystal, the allowed wavevectors are dense, and we write
\[ \sum_{\mathbf q,r} F[\omega_r(\mathbf q)] = \sum_r \int_{\mathrm{1.\ BZ}} d^3q Z(q) F[\omega_r(\mathbf q)]. \]
We now define \(D(\omega)\) by requiring that
\[ \sum_r \int_{\mathrm{1.\ BZ}} d^3q Z(q) F[\omega_r(\mathbf q)] = \int_{\omega_{\min}}^{\omega_{\max}} d\omega D(\omega)F(\omega). \]
This is possible for any function \(F\). The density of states is therefore
The interpretation is direct:
The normalization condition is
\[ \int_{\omega_{\min}}^{\omega_{\max}}D(\omega)d\omega =3r'N. \]
Thus the area under the full density-of-states curve counts all vibrational modes.
7.4.1 Physical Meaning: A Histogram of Frequencies
The density of states is the frequency histogram of the phonon spectrum.
If the dispersion \(\omega_r(\mathbf q)\) changes rapidly with \(\mathbf q\), then only a small number of nearby wavevectors fall into a given frequency interval. The density of states is small.
If the dispersion is flat, many different wavevectors have nearly the same frequency. The density of states is large.
This is the main intuition:
In one dimension, this can be seen from
\[ Z(q)dq=D(\omega)d\omega. \]
Therefore
\[ D(\omega) = Z(q) \left| \frac{dq}{d\omega} \right|. \]
If \(d\omega/dq\) is small, then \(dq/d\omega\) is large, and the density of states is enhanced.
7.5 General Geometric Formula for \(D(\omega)\)
We now derive a general expression for the density of states of one branch \(\omega(\mathbf q)\) in three dimensions.
Consider all points in reciprocal space with frequencies between \(\omega\) and \(\omega+\Delta\omega\). These points form a thin shell between two constant-frequency surfaces.
The number of states in this shell is
\[ D(\omega)\Delta\omega \simeq \frac{V}{(2\pi)^3} \int_{\omega}^{\omega+\Delta\omega} d^3q. \]
Use local coordinates adapted to the constant-frequency surface. The reciprocal-space volume element is
\[ d^3q=dS_q dq_\perp, \]
where \(dS_q\) is a surface element on \(\omega(\mathbf q)=\mathrm{const}\) and \(dq_\perp\) is the distance perpendicular to that surface.
The frequency change perpendicular to the surface is
\[ \Delta\omega = |\nabla_{\mathbf q}\omega(\mathbf q)| dq_\perp. \]
Therefore
\[ dq_\perp = \frac{\Delta\omega}{|\nabla_{\mathbf q}\omega(\mathbf q)|}. \]
Substitution gives
\[ D(\omega)\Delta\omega = \frac{V}{(2\pi)^3} \int_{\omega=\mathrm{const}} dS_q \frac{\Delta\omega}{|\nabla_{\mathbf q}\omega(\mathbf q)|}. \]
Canceling \(\Delta\omega\) yields
For several branches, one sums this expression over all branches \(r\).
7.5.1 Relation to Group Velocity and van Hove Singularities
The group velocity of a phonon wave packet is
\[ \mathbf v_g(\mathbf q) = \nabla_{\mathbf q}\omega(\mathbf q). \]
Thus the general density-of-states formula can be read as
\[ D(\omega) \propto \int_{\omega=\mathrm{const}} \frac{dS_q}{|\mathbf v_g(\mathbf q)|}. \]
This shows why slow modes contribute strongly to the density of states. A small group velocity means that the frequency changes only weakly as \(\mathbf q\) changes, so many nearby wavevectors correspond to nearly the same frequency.
If
\[ \nabla_{\mathbf q}\omega(\mathbf q)=0 \]
at special points or regions in reciprocal space, the integrand becomes singular. Such features are called van Hove singularities.
In real crystals, these singularities are often broadened by finite resolution, disorder, anharmonicity, and the fact that experimental measurements average over finite intervals. Nevertheless, the peaks and edges in measured phonon spectra often reflect precisely these flat parts of the dispersion.
7.6 Example: Isotropic Acoustic Medium in Three Dimensions
As an important example, consider an isotropic medium with one longitudinal acoustic branch and two transverse acoustic branches. The long-wavelength dispersions are
\[ \omega_L(\mathbf q)=v_Lq, \]
and
\[ \omega_T(\mathbf q)=v_Tq, \]
where
\[ q=|\mathbf q|. \]
For one branch \(i\), the surface \(\omega_i(\mathbf q)=\omega\) is a sphere with radius
\[ q=\frac{\omega}{v_i}. \]
The surface area is
\[ 4\pi q^2, \]
and
\[ |\nabla_{\mathbf q}\omega_i|=v_i. \]
Therefore the contribution of one branch is
\[ D_i(\omega) = \frac{V}{(2\pi)^3} \frac{4\pi q^2}{v_i}. \]
Using \(q=\omega/v_i\) gives
Adding one longitudinal and two transverse branches gives
This result explains the characteristic low-frequency behavior of three-dimensional acoustic phonons:
\[ D(\omega)\propto \omega^2. \]
The reason is geometric. In three dimensions, the number of allowed wavevectors at radius \(q\) grows like the area of a sphere, namely like \(q^2\). Since acoustic phonons have \(\omega\propto q\) at small \(q\), the density of states grows like \(\omega^2\).
7.7 Dimensionality: One, Two, and Three Dimensions
The same reasoning gives different low-frequency behavior in different dimensions. This is a useful way to remember the result.
7.7.1 Two Dimensions
In two dimensions, the allowed states fill an area in reciprocal space. The density of states in reciprocal space is
\[ Z^{(2)}(q)=\frac{A}{(2\pi)^2}. \]
The constant-frequency set is now a line, not a surface. Therefore
\[ D^{(2)}(\omega) = \frac{A}{(2\pi)^2} \int_{\omega=\mathrm{const}} \frac{dL_q}{|\nabla_{\mathbf q}\omega(\mathbf q)|}. \]
For an isotropic acoustic branch with \(\omega_i=v_iq\), the constant-frequency line is a circle of radius \(q=\omega/v_i\). Its circumference is \(2\pi q\). Hence
\[ D_i^{(2)}(\omega) = \frac{A}{(2\pi)^2} \frac{2\pi q}{v_i} = \frac{A}{2\pi} \frac{\omega}{v_i^2}. \]
For one longitudinal and one transverse acoustic branch,
Thus in two dimensions,
\[ D^{(2)}(\omega)\propto \omega. \]
7.7.2 One Dimension
In one dimension, the allowed states lie along a line in reciprocal space. The density of states in reciprocal space is
\[ Z^{(1)}(q)=\frac{L}{2\pi}. \]
The constant-frequency set consists of points. Thus
\[ D^{(1)}(\omega) = \frac{L}{2\pi} \int_{\omega=\mathrm{const}} \frac{dP_q}{|\nabla_q\omega(q)|}. \]
For an acoustic dispersion \(\omega=v_L|q|\), a positive frequency corresponds to two points, \(+q\) and \(-q\). Therefore
\[ \int_{\omega=\mathrm{const}} dP_q = 2 . \]
Thus
The one-dimensional acoustic density of states is independent of frequency.
7.7.3 Dimensional Trend
For acoustic phonons with linear dispersion, the low-frequency density of states follows the geometry of reciprocal space:
\[ D^{(3)}(\omega)\propto \omega^2, \]
\[ D^{(2)}(\omega)\propto \omega, \]
\[ D^{(1)}(\omega)\propto \omega^0. \]
The lower the dimension, the more strongly low-frequency modes are represented.
7.8 Reading a Phonon Density-of-States Plot
A density-of-states plot should be read as a mode-counting graph.
Key features have direct interpretations:
- The total area counts the total number of phonon modes.
- The low-frequency onset is controlled by acoustic branches.
- A steep dispersion produces a small density of states.
- A flat dispersion produces a large density of states.
- Peaks often indicate van Hove singularities or weakly dispersive branches.
- Optical branches, when present, often produce high-frequency structures because their dispersion can be relatively flat.
A useful practical rule is:
Here “move slowly” refers to the group velocity in reciprocal-space language: a small \(|\nabla_{\mathbf q}\omega|\) means that the frequency changes slowly as the wavevector changes.
7.9 Optional / Appendix: Debye Cutoff as a Mode-Counting Approximation
Open optional appendix
The isotropic acoustic result motivates the Debye approximation used in heat-capacity theory. The idea is to replace the true Brillouin zone by a sphere in reciprocal space and the true acoustic dispersions by linear dispersions.
For a monatomic three-dimensional crystal, there are three acoustic branches and \(N\) allowed wavevectors per branch. The cutoff wavevector \(q_D\) is chosen so that the Debye sphere contains \(N\) allowed wavevectors:
\[ \frac{V}{(2\pi)^3} \frac{4\pi q_D^3}{3} = N . \]
With three acoustic branches, this gives \(3N\) modes in total.
The corresponding Debye frequency is
\[ \omega_D=v_Dq_D, \]
where \(v_D\) is an effective sound velocity. This approximation preserves the correct number of modes and the correct low-frequency power law, but it does not reproduce detailed van Hove features or optical branches.
For crystals with a multi-atom basis, the acoustic Debye model accounts naturally for the acoustic branches. Optical branches require additional modeling if one wants to represent the entire phonon spectrum.
7.10 Connection to the Next Lecture: From Mode Counting to Phonons
This lecture counted the normal modes of a harmonic crystal. The next step is to quantize these normal modes.
Once the normal modes are quantized, each mode behaves like a harmonic oscillator with discrete energy levels. The density of states then becomes the connection between microscopic lattice dynamics and macroscopic thermal properties.
The main replacement will be
\[ \sum_{\mathbf q,r} F[\omega_r(\mathbf q)] \longrightarrow \int d\omega,D(\omega)F(\omega). \]
In the next lecture, \(F(\omega)\) will contain the thermal occupation and energy of a phonon mode. This will lead directly to the lattice internal energy and then to the heat capacity.
- A finite crystal allows only a discrete set of phonon wavevectors.
- Born-von Karman boundary conditions give a convenient bulk mode-counting scheme.
- Each phonon branch contains one allowed wavevector state per primitive cell.
- The reciprocal-space density of states is uniform for a macroscopic periodic crystal.
- The frequency-space density of states is a histogram of phonon frequencies.
- Flat dispersions produce large density-of-states features.
- Van Hove singularities occur where the phonon group velocity vanishes.
- The low-frequency acoustic density of states depends strongly on dimensionality.
Problem Set
Fixed-end chain. Consider a one-dimensional chain with equilibrium positions \(x_n=na\), length \(L=Na\), and fixed endpoints at \(n=0\) and \(n=N\). Starting from \[ u_n = A_1 e^{i(qna-\omega t)} + A_2 e^{-i(qna+\omega t)}, \] derive the allowed values of \(q\). How many nontrivial vibrational modes are obtained?
Periodic boundary conditions. For a one-dimensional chain with \(N\) atoms and periodic boundary condition \(u_n=u_{n+N}\), use \[ u_n=Ae^{i(qna-\omega t)} \] to derive the allowed values of \(q\). Show that the spacing between neighboring allowed wavevectors is \(2\pi/L\) and obtain \(Z(q)\).
Constant-frequency surfaces. For a single branch \(\omega(\mathbf q)\) in three dimensions, derive \[ D(\omega) = \frac{V}{(2\pi)^3} \int_{\omega=\mathrm{const}} \frac{dS_q}{|\nabla_{\mathbf q}\omega(\mathbf q)|}. \] Why is the density of states large where the dispersion is flat?
Isotropic acoustic phonons. For a three-dimensional isotropic medium with one longitudinal branch \(\omega_L=v_Lq\) and two transverse branches \(\omega_T=v_Tq\), derive \[ D(\omega) = \frac{V}{2\pi^2} \left( \frac{1}{v_L^3} + \frac{2}{v_T^3} \right) \omega^2. \] What is the corresponding low-frequency power law in two and one dimensions?