Gapless Quantum Matter: Phonons and the Debye Model, Goldstone’s Theorem

Phonons and the Debye Model

Previously, we examined systems with a small number of degrees of freedom, where the quantum Hamiltonian exhibits an energy gap, such as \(\Delta E = \hbar \omega\) or \(\Delta E = \frac{\hbar^{2}}{I}\). This leads to specific heat behavior like \(C_V \sim e^{-\Delta E / (k_B T)}\).

However, as \(N \rightarrow \infty\), more intriguing behavior emerges if the Hamiltonian becomes gapless. To illustrate this, we now explore the vibrations in a crystal as an example of a gapless system.

Consider a crystal consisting of a regular lattice of atoms, where the equilibrium positions are described by:

\[\begin{split} \begin{aligned} \vec{q}_{l, m, n}^{*} = l \vec{R}_{1} + m \vec{R}_{2} + n \vec{R}_{3}, \\ l, m, n \in \mathbb{Z}. \end{aligned} \end{split}\]

Here, \(\vec{R}_{i}\) are the “Bravais vectors.” For a simple cubic lattice, these vectors are given by:

\[ \vec{R}_{1}=a \hat{x}, \quad \vec{R}_{2}=a \hat{y}, \quad \vec{R}_{3}=a \hat{z}. \]

Other lattice structures, such as hexagonal or orthorhombic, are also possible, leading to 14 distinct Bravais lattices in 3D. Additionally, there may be multiple atoms within a unit cell. However, for simplicity, we will focus on the simple cubic case. The equilibrium positions of the atoms, \(\vec{q}^{*}_{l, m, n}\), are maintained by various chemical or electrical forces. However, due to thermal and quantum effects, the atoms can deviate from these positions:

\[ \vec{q}_{l, m, n} = \vec{q}_{l, m, n}^{*} + \vec{u}_{l, m, n}, \]

where \(\vec{u}_{l, m, n}\) represents the displacement of the atom at \((l, m, n)\) from its equilibrium position.

To simplify notation, it is convenient to label the atom at \((l, m, n)\) by its equilibrium position \(\vec{r} \equiv \vec{q}_{l, m, n}^{*}\). This leads to:

\[ \vec{q}_{\vec{r}} = \vec{r} + \vec{u}_{\vec{r}}. \]

The potential energy of the system can be expressed as:

\[ V(\{\vec q_{\vec r}\})=V^*+\frac 12 \sum_{\vec r, \vec r', \alpha, \beta} \frac{\partial^2 V}{\partial q_{\alpha,r} \partial q_{\beta, r'}}u_{\alpha, r}u_{\beta, r'}, \]

where \(\alpha, \beta\) are the vector indices (\(\{x, y, z\}\)) of the displacement vector. This expression assumes \(|\vec{u}| \ll a\), which is valid when \(T \rightarrow 0\) and the atomic mass \(\rightarrow \infty\) (since \(\Delta v \Delta u \geq \hbar / \text{mass}\)). Outside this regime, the crystal would melt. Note that at equilibrium, \(\left.\frac{\partial V}{\partial q}\right|_{q^{*}}=0\).

Using translation invariance, \(V\left(\left\{\vec{q}_{\vec{r}}+\vec{c}\right\}\right)=V\left(\left\{\vec{q}_{\vec{r}}\right\}\right)\), we deduce:

\[ \frac{\partial^{2} V}{\partial q_{\alpha}(r) \partial q_{\beta}\left(r^{\prime}\right)} \equiv K_{\alpha \beta}\left(\vec{r}-\vec{r}^{\prime}\right). \]

The total Hamiltonian of the system is then given by:

\[ \hat{H}=V^{*}+\sum_{\vec{r}} \frac{\vec{p}_{\vec{r}}^{2}}{2 m}+\frac{1}{2} \sum_{\vec{r}, \vec{r}^{\prime}} u_{\alpha, \vec{r}} K_{\alpha \beta}\left(\vec{r}-\vec{r}^{\prime}\right) u_{\beta, \vec{r}^{\prime}}, \]

where \(\left[p_{\alpha, \vec{r}}, u_{\beta, \vec{r}^{\prime}}\right]=-i \delta_{\alpha \beta} \delta_{\vec{r} \vec{r}^{\prime}}\).

To simplify \(\hat{H}\), we use the discrete Fourier transform:

\[\begin{split} \begin{aligned} & u_{\alpha}(\vec{k}) \equiv \frac{1}{\sqrt{N}} \int \sum_{\vec{r}} e^{-i \vec{k} \cdot \vec{r}} u_{\alpha, r} \\ & p_{\alpha}(\vec{k}) \equiv \frac{1}{\sqrt{N}} \sum_{\vec{r}} e^{-i \vec{k} \cdot \vec{r}} p_{\alpha, r} \\ & K_{\alpha \beta}(\vec{k}) \equiv \sum_{\vec{r}} e^{i \vec{k} \cdot \vec{r}} K_{\alpha, \beta}(\vec{r}) \end{aligned} \end{split}\]

For simplicity, assume \(K_{\alpha \beta}(\vec{k}) = \delta_{\alpha \beta} \tilde{K}(\vec{k})\). The Hamiltonian then becomes:

\[ \hat{H} = V^* + \sum_{\vec{k}} \left[ \frac{\vec{p}(\vec{k}) \cdot \vec{p}(-\vec{k})}{2m} + \frac{1}{2} \tilde{K}(\vec{k}) \vec{u}(\vec{k}) \cdot \vec{u}(-\vec{k}) \right]. \]

Discrete Values of \(\vec{k}\)

1. Periodicity in \(k\)-Space

Recall that \(\vec{r} = a(l \hat{x} + m \hat{y} + n \hat{z})\). This implies:

\[ e^{i \vec{k} \cdot \vec{r}} = e^{i \left( \vec{k} + \frac{2\pi}{a} \hat{e}_i \right) \cdot \vec{r}}, \quad \hat{e}_i = \hat{x}, \hat{y}, \hat{z}. \]

Thus, \(u_{\alpha}(\vec{k}) = u_{\alpha}\left(\vec{k} + \frac{2\pi}{a} \hat{e}_i\right)\), meaning \(\vec{k}\) is periodic in \(k\)-space. This periodicity defines the “Brillouin Zone” (BZ), a toroidal region in \(k\)-space.

2. Periodic Boundary Conditions in Real Space

Assuming periodic boundary conditions in real space, \(\vec{r} \sim \vec{r} + L \cdot \hat{e}_i\), we find that \(\vec{k}\) must satisfy:

\[\begin{split} \begin{aligned} e^{i \vec{k} \cdot \vec{r}} &= e^{i \vec{k} \cdot (\vec{r} + L \cdot \hat{e}_i)} \\ \Rightarrow \vec{k} &\in \frac{2\pi}{L} \cdot (i, j, k), \quad i, j, k \in \mathbb{Z}, \\ -\frac{L}{2} &\leq i, j, k \leq \frac{L}{2}. \end{aligned} \end{split}\]

This establishes a one-to-one correspondence between the \((L/a)^3\) points in real space and the \((L/a)^3\) discrete \(\vec{k}\) points in \(k\)-space.

3. Original variables are real-valued

Since \(\vec{p}_{\vec{r}}\) and \(\vec{u}_{\vec{r}}\) are real-valued, we have:

\[ \vec{u}_{\alpha, \vec{k}} = \vec{u}_{\alpha, -\vec{k}}^*, \quad \vec{p}_{\alpha, \vec{k}} = \vec{p}_{\alpha, -\vec{k}}^*. \]

The Hamiltonian can then be rewritten as:

\[ \hat{H} = V^* + \sum_{\vec{k}} \left[ \frac{|\vec{p}(\vec{k})|^2}{2m} + \frac{1}{2} \tilde{K}(\vec{k}) |\vec{u}(\vec{k})|^2 \right]. \]

Interpretation in \(k\)-Space

In \(k\)-space, the system decomposes into \(3N = 3(L/a)^3\) independent harmonic oscillators, each with frequency:

\[ \omega(\vec{k}) = \sqrt{\frac{\tilde{K}(\vec{k})}{m}}. \]

These oscillations correspond to waves in the atomic positions, which are the sound waves of the crystal. The group velocity of these waves is given by:

\[ \vec{v}_{\vec{k}} = \frac{\partial \omega(\vec{k})}{\partial \vec{k}}, \]

and depends on the dispersion relation \(\omega(\vec{k})\).

QM description

Quantum mechanically, we define \(3N\) raising and lowering operators, \(\hat{a}_{\alpha, \vec{k}}^{+}\) and \(\hat{a}_{\alpha, \vec{k}}\), with the number operator \(\hat{n}_{\alpha, \vec{k}} = \hat{a}_{\alpha, \vec{k}}^{+} \hat{a}_{\alpha, \vec{k}}\). These operators satisfy the commutation relations:

\[ \left[\hat{a}_{\alpha, \vec{k}}, \hat{a}_{\beta, \vec{k}'}^{+}\right] = \delta_{\alpha \beta} \delta_{\vec{k}, \vec{k}'}, \quad \left[\hat{a}_{\alpha, \vec{k}}, \hat{a}_{\beta, \vec{k}'}\right] = \left[\hat{a}_{\alpha, \vec{k}}^{+}, \hat{a}_{\beta, \vec{k}'}^{+}\right] = 0. \]

Using the ansatz:

\[\begin{split} \begin{aligned} & \hat{u}_{\alpha, \vec{k}} = \sqrt{\frac{\hbar}{2m\omega(\vec{k})}} \left(\hat{a}_{\alpha, \vec{k}}^{+} + \hat{a}_{\alpha, -\vec{k}}\right), \\ & \hat{p}_{\alpha, \vec{k}} = i \sqrt{\frac{\hbar m \omega(\vec{k})}{2}} \left(\hat{a}_{\alpha, \vec{k}}^{+} - \hat{a}_{\alpha, -\vec{k}}\right), \end{aligned} \end{split}\]

we can verify that the canonical commutation relation holds:

\[ \left[\hat{u}_{\alpha, \vec{k}}, \hat{p}_{\beta, -\vec{k}'}\right] = i\hbar \delta_{\alpha \beta} \delta_{\vec{k}, \vec{k}'}. \]

we find the Hamiltonian becomes:

\[ \hat{H} = V^{*} + \sum_{\vec{k}, \alpha} \hbar \omega(\vec{k}) \left(\hat{n}_{\alpha, \vec{k}} + \frac{1}{2}\right). \]

These quantized sound wave excitations are called “phonons,” analogous to photons in quantum electrodynamics (QED). Phonons are bosonic quasiparticles, even though the underlying atoms may be fermions. This exemplifies a recurring theme in condensed matter and quantum many-body physics: the emergence of low-energy quasiparticle excitations distinct from the system’s constituent particles, often referred to as “collective modes.”

The Hilbert space is spanned by specifying the occupation numbers of the \(3N\) modes, \(n_{\alpha, \vec{k}} = 0, 1, 2, \ldots\), with basis states:

\[ \left|\{n_{\alpha, \vec{k}}\}\right\rangle = \left|0, 3, 1, 2, \ldots\right\rangle. \]

The quantum mechanical partition function is given by:

\[\begin{split} \begin{aligned} Z &= \sum_{\{n_{\alpha, \vec{k}}\}} e^{-\beta \hat{H}} = e^{-\beta E_{0}} \sum_{\{n_{\alpha, \vec{k}}\}} e^{-\beta \sum_{\vec{k}, \alpha} \hbar \omega(\vec{k}) n_{\alpha, \vec{k}}} \\ &= e^{-\beta E_{0}} \prod_{\vec{k}, \alpha} \left(\sum_{n_{\alpha, \vec{k}}=0}^{\infty} e^{-\beta \hbar \omega(\vec{k}) n_{\alpha, \vec{k}}}\right) \\ &= e^{-\beta E_{0}} \prod_{\vec{k}, \alpha} \left(\frac{1}{1 - e^{-\beta \hbar \omega(\vec{k})}}\right). \end{aligned} \end{split}\]

The free energy is then:

\[ F = -\frac{1}{\beta} \ln(Z) = E_{0} + \frac{1}{\beta} \sum_{\vec{k}, \alpha} \ln\left(1 - e^{-\beta \hbar \omega(\vec{k})}\right). \]

Key physical observables include the average energy:

\[ \langle E \rangle = E_{0} + \sum_{\alpha, \vec{k}} \langle n_{\alpha, \vec{k}} \rangle \hbar \omega(\vec{k}), \]

and the heat capacity:

\[ C = \frac{\partial \langle E \rangle}{\partial T}. \]

For a single quantum mechanical degree of freedom, we previously found that the heat capacity scales as \(C \sim e^{-\Delta E / (k_B T)}\). However, as we will now demonstrate, this behavior does not hold for phonons.

To begin, we calculate the average occupation number of a phonon mode:

\[ \begin{aligned} \left\langle n_{\alpha, \vec{k}} \right\rangle &= \frac{\sum_{n=0}^{\infty} n e^{-\beta \hbar \omega(\vec{k}) n}}{\sum_{n=0}^{\infty} e^{-\beta \hbar \omega(\vec{k}) n}} = \frac{1}{e^{\beta \hbar \omega(\vec{k})} - 1}. \end{aligned} \]

This expression describes the average number of phonons in a given mode, and is called Bose-Einstein statistics.

The total energy of the system is then given by:

\[ \begin{aligned} \langle E \rangle &= E_0 + \sum_{\alpha, \vec{k}} \left\langle n_{\alpha, \vec{k}} \right\rangle \hbar \omega(\vec{k}), \end{aligned} \]

where \(E_0\) is the zero-point energy. The heat capacity can be derived by differentiating the energy with respect to temperature:

\[ \begin{aligned} C &= \frac{\partial \langle E \rangle}{\partial T} = k_B \sum_{\alpha, \vec{k}} \frac{1}{\left(e^{\beta \hbar \omega(\vec{k})} - 1\right)^2} \left(\frac{\hbar \omega(\vec{k})}{k_B T}\right)^2. \end{aligned} \]

To proceed further, we need to determine the specific form of the phonon dispersion relation \(\omega(\vec{k})\).

Debye Model

The exact form of \(\omega(k)\) depends on the detailed chemistry of the system, which determines \(V(\{\vec{q}_{\vec{r}}\})\) and, consequently, \(K_{\alpha \beta}(\vec{r})\). However, translation invariance ensures that \(V(\{\vec{q}_{\vec{r}} + \vec{c}\}) = V(\{\vec{q}_{\vec{r}}\})\), implying that a rigid shift of the lattice, \(\vec{u}_{\vec{r}} = \vec{c}\), results in no change in energy (\(\Delta E = 0\)). This corresponds to the \(\vec{u}(\vec{k} = 0)\) mode, leading to:

\[ \tilde{K}_{\alpha \beta}(\vec{k} = 0) = 0. \]

If \(K_{\alpha \beta}(\vec{r})\) is local, meaning it decays exponentially as \(\vec{r} \to \infty\), its Fourier transform is analytic and can be expanded as:

\[ \tilde{K}(\vec{k}) = 0 + \tilde{\kappa}_{\alpha} k_{\alpha} + \frac{1}{2} \tilde{\kappa}_{\alpha \beta} k_{\alpha} k_{\beta} + \frac{1}{3!} \tilde{\kappa}_{\alpha \beta \gamma} k_{\alpha} k_{\beta} k_{\gamma} + \cdots \]

With inversion symmetry, \(K(\vec{r}) = K(-\vec{r})\), the odd-\(k\) terms vanish, leaving:

\[ \tilde{K}(\vec{k}) = \frac{1}{2} \tilde{\kappa}_{\alpha \beta} k_{\alpha} k_{\beta} + O(k^4). \]

For an isotropic system, where \(x \sim y \sim z\), this simplifies to:

\[ \tilde{K}(\vec{k}) = B |\vec{k}|^2 + O(k^4), \]

resulting in the dispersion relation:

\[ \omega(\vec{k}) = \sqrt{\frac{B |\vec{k}|^2}{m}} = v |\vec{k}| + O(k^3), \]

where \(v = \sqrt{B / m}\) is the speed of sound. This approximation breaks down as \(k \to \pi / a\).

Under this approximation, the total energy is:

\[ \langle E \rangle = E_0 + \sum_{\alpha, \vec{k}} \frac{\hbar v |\vec{k}|}{e^{\beta \hbar v |\vec{k}|} - 1}. \]

In the limit \(L \to \infty\), where \(\vec{k} = \frac{2\pi}{L}(i, j, k)\), the sum over \(\vec{k}\) becomes an integral:

\[ \sum_{\vec{k}} \to \left(\frac{L}{2\pi}\right)^3 \int_{BZ} d^3k. \]

Thus:

\[ \langle E \rangle \approx E_0 + 3 V \int_{BZ} \frac{d^3k}{(2\pi)^3} \frac{\hbar v |\vec{k}|}{e^{\beta \hbar v |\vec{k}|} - 1}. \]

For \(T \ll T_D\), where \(T_D = \frac{\hbar v \pi}{k_B a}\), the boundary of the Brillouin Zone, with energy \(E \sim \hbar v \frac{\pi}{a} \gg T\), does not contribute significantly. Extending the integration domain to infinity, we find:

\[ \langle E \rangle = E_0 + 3 V \int_0^\infty dk \frac{4\pi k^2}{(2\pi)^3} \frac{\hbar v k}{e^{\beta \hbar v k} - 1}. \]

Evaluating this integral yields:

\[ \langle E \rangle = E_0 + \frac{\pi^2}{10} V \frac{(k_B T)^4}{(\hbar v)^3}. \]

The heat capacity per unit volume is then:

\[ \frac{C}{V} = k_B \frac{2\pi^2}{5} \left(\frac{k_B T}{\hbar v}\right)^3. \]

The key result is that \(E \sim T^4\) and \(C_V \sim T^3\), contrasting with the exponential behavior \(E, C \sim e^{-\Delta E / T}\). This difference arises because phonons are gapless: \(\hbar \omega(k) \sim v |\vec{k}|\) permits excitations with arbitrarily low energy. In high-energy physics terms, phonons are “massless,” with dispersion \(E(k) = v |\vec{k}|\), unlike systems with a mass gap where \(E(p) = \sqrt{m_0^2 v^4 + p^2 v^2}\).

Note

This highlights a profound principle: low-temperature behavior depends only on a few “relevant” parameters (e.g., \(v\)) and not on the detailed microscopic structure.

Relation to spontaneous symmetry breaking: “Goldstone modes.”

For \(T < T_c\), the Ising model spontaneously breaks time-reversal symmetry, where \(E(\{\sigma\}) = E(\{-\sigma\})\).

Similarly, the phonon model exhibits a symmetry: \(E\left(\left\{q_{\vec{r}}\right\}\right) = E\left(\left\{\vec{q}_{\vec{r}} + \vec{c}\right\}\right)\).

This means that any realized equilibrium position \(\left\langle \vec{q}_{\vec{r}} \right\rangle = \vec{r}\) spontaneously breaks translation symmetry.

Note that \(\left\langle \vec{q}_{\vec{r}} \right\rangle = \vec{r} + \left\langle \vec{u}_{\vec{r}} \right\rangle\), and:

\[\begin{split} \begin{aligned} \left\langle \vec{u}_{\vec{r}} \right\rangle &= \sum_{\vec{k}} \frac{e^{i \vec{k} \cdot \vec{r}}}{\sqrt{N}} \sqrt{\frac{1}{2 m \omega_{\vec{k}}}} \left\langle \hat{a}_{\vec{k}}^{+} + \hat{a}_{-\vec{k}} \right\rangle \\ &= 0. \end{aligned} \end{split}\]

(Caveat: In \(D = 2\), this is more subtle due to the “Mermin-Wagner theorem,” which states that continuous symmetry cannot be spontaneously broken. This is covered in Physics 212.)

The existence of translation symmetry ensures \(\hbar \omega(k = 0) = 0\), leading to \(\omega(k) \sim v \cdot k\), which implies gaplessness.

This is a specific case of a broader result: Goldstone’s Theorem.

“A system with a spontaneously broken continuous global symmetry has a gapless mode (the Goldstone boson).”

[There are subtleties depending on whether the symmetry is Lorentz-invariant or non-relativistic, and whether it is internal or spacetime-related.]

In this context, the phonon (sound wave) is the “Goldstone mode” associated with the spontaneous breaking of translation symmetry.