Phonons and Second Quantization

Recommended resources

The freely available resources this lesson will direct you to.

Hello! Welcome back to our course.

In the last lesson, we extended our model of lattice dynamics from a simple 1D chain to a full 3D crystal. We saw that by solving the eigenvalue problem for the 3D dynamical matrix, we obtain the dispersion relations $ω_{j} (q)$ for the normal modes of vibration. We also learned to classify these modes as longitudinal or transverse based on their polarization. However, our entire discussion was based on classical mechanics.

Today, we will make the crucial leap to a quantum mechanical description. Our goal is to apply second quantization to express the lattice Hamiltonian in terms of phonon creation and annihilation operators. This process will formally introduce the phonon, the fundamental quantum of lattice vibration. Your familiarity with quantizing fields in QFT and using ladder operators in quantum mechanics will be a significant asset here, as the underlying concepts are very similar.

1. From Normal Modes to a Collection of Oscillators

The first key insight is that the classical Hamiltonian of the crystal lattice, which is a complicated function of the positions and momenta of all $N$ atoms, can be simplified dramatically. When expressed in terms of the normal mode coordinates corresponding to the vibrations we found in the last lesson, the Hamiltonian transforms into a sum of independent simple harmonic oscillators (SHOs), one for each mode $(q, j)$ .

The total Hamiltonian becomes:
$H_{lattice} = \sum_{q, j} (\frac{1}{2 M} ∣ P_{q, j} ∣^{2} + \frac{1}{2} M ω_{j}^{2} (q) ∣ Q_{q, j} ∣^{2})$
where $Q_{q, j}$ and $P_{q, j}$ are the canonical position and momentum for the normal mode with wavevector $q$ and branch index $j$ . The system of $N$ coupled atoms has been mathematically transformed into a system of $3 p_{N}$ uncoupled harmonic oscillators ( $p_{N}$ is the number of atoms in the basis).

2. Quantizing the Lattice

Our task now is to quantize this collection of harmonic oscillators. This is the heart of "second quantization" in the context of lattice vibrations. We will promote the classical displacement and momentum variables to quantum operators and find the energy spectrum.

The following text provides a clear and self-contained derivation of this process for a 1D monatomic chain, which is easily generalized to 3D.

4. Phonons

These lecture notes, 'Phonons' by David Tong, provide an excellent walkthrough of the quantization of lattice vibrations. This section parallels the standard quantization of the simple harmonic oscillator, but for all the normal modes of the lattice at once.

Please read the subsection 'Quantum Vibrations' (pages 111-113, starting from section 4.1.4). Focus on these key steps: Promoting the classical displacement u_n and momentum p_n to operators. Inverting the classical solution to express the normal mode amplitudes (α_l) in terms of u_n and p_n operators. Deriving the commutation relations for α_l and α_l^†. Rescaling these to define the standard bosonic creation (a_l^†) and annihilation (a_l) operators. Seeing how the Hamiltonian takes its final, diagonal form.

As derived in the notes, by defining the creation and annihilation operators for each mode, we can rewrite the lattice Hamiltonian. Generalizing to 3D, with wavevector $q$ and branch index $j$ , the Hamiltonian becomes:

$H = \sum_{q, j} ℏ ω_{j} (q) (a_{q, j}^{†} a_{q, j} + \frac{1}{2})$

Here, the operators $a_{q, j}$ and $a_{q, j}^{†}$ are bosonic operators satisfying the commutation relations:
$[a_{q, j}, a_{q^{'}, j^{'}}^{†}] = δ_{q, q^{'}} δ_{j, j^{'}}$
$[a_{q, j}, a_{q^{'}, j^{'}}] = [a_{q, j}^{†}, a_{q^{'}, j^{'}}^{†}] = 0$

This is a profound result. The complex, interacting system of atoms is now described as a simple collection of non-interacting quantum particles.

3. The Phonon: A Bosonic Quasiparticle

This quantum description gives rise to a new "particle" concept.

The operator $a_{q, j}^{†}$ creates a quantum of excitation in the mode $(q, j)$ . This quantum carries energy $ℏ ω_{j} (q)$ and crystal momentum $ℏ q$ . This quantum of lattice vibration is called a phonon.
The operator $a_{q, j}$ annihilates a phonon in that same mode.
The operator $n^_{q, j} = a_{q, j}^{†} a_{q, j}$ is the number operator; its eigenvalues tell us how many phonons exist in the mode $(q, j)$ .
The ground state of the crystal, or the vacuum state $∣0 ⟩$ , is the state with no phonons, defined by $a_{q, j} ∣0 ⟩ = 0$ for all modes. Its energy is the zero-point energy $E_{0} = \sum_{q, j} \frac{1}{2} ℏ ω_{j} (q)$ .
Excited states are built by applying creation operators to the vacuum, forming a Fock space. For example, $a_{q_{1}, j_{1}}^{†} a_{q_{2}, j_{2}}^{†} ∣0 ⟩$ represents a state with two phonons.
Since the creation operators commute, you can place an arbitrary number of phonons into the same quantum state. This, by definition, means that phonons are bosons.

This general formalism for expressing a non-interacting Hamiltonian is a cornerstone of many-body physics. The video below provides a broader perspective on representing single-particle operators in second quantization, which you can see is perfectly analogous to what we've just done for phonons.

4- Operators in Second Quantization - Course on Quantum Many-Body Physics

This video from the 'Course on Quantum Many-Body Physics' by Luis Gregorio Dias explains how a general non-interacting many-body Hamiltonian is expressed in the second quantization formalism. This provides a great parallel to our phonon Hamiltonian.

Watch the segment from 37:57 to 42:44. The key idea is to see how a Hamiltonian that is a sum of single-particle Hamiltonians (H = Σ h_n) becomes diagonal in the number operator representation (H = Σ ε_i a_i^† a_i). Notice the direct structural similarity to our phonon Hamiltonian.

Test your understanding!

Consider a crystal lattice in an excited state described by the vector $∣ ψ ⟩ = \frac{1}{2} (a_{q_{1}, LA}^{†})^{2} ∣0 ⟩$ , where LA denotes the longitudinal acoustic branch. What is the total energy (above the ground state) and total crystal momentum of this state?

Show answer

The state $∣ ψ ⟩$ contains two phonons, both in the same mode $(q_{1}, LA)$ .

The number operator for this mode gives $n^_{q_{1}, LA} ∣ ψ ⟩ = 2∣ ψ ⟩$ .
The total energy of the state is given by applying the Hamiltonian:
$H ∣ ψ ⟩ = (\sum_{q, j} \frac{1}{2} ℏ ω_{j} (q) + 2ℏ ω_{LA} (q_{1})) ∣ ψ ⟩$
The energy above the ground state is therefore $E = 2ℏ ω_{LA} (q_{1})$ .
The total crystal momentum is the sum of the momenta of the individual phonons: $P = ℏ q_{1} + ℏ q_{1} = 2ℏ q_{1}$ .

4. Connection to Quantum Field Theory

The procedure we just followed is a specific example of a more general concept: quantizing a classical field. In the long-wavelength limit ( $∣ q ∣ \to 0$ ), we can ignore the discrete atoms and treat the atomic displacement as a continuous field, $u (x, t)$ . The quantization of this displacement field is what gives rise to phonons.

This is perfectly analogous to how other particles arise in QFT:

Photons are the quanta of the quantized electromagnetic field.
Klein-Gordon particles are the quanta of the quantized scalar Klein-Gordon field.
Phonons are the quanta of the quantized lattice displacement field.

The David Tong notes briefly discuss this powerful connection, which should resonate strongly with your studies in QFT.

4. Phonons

This brief follow-up reading from David Tong's notes formalizes the connection between phonons and quantum field theory.

Now, please read the subsection 'From Fields to Phonons' (pages 116-118). Notice how the discrete index n is replaced by the continuous position x, and the sum over k becomes an integral. The final Hamiltonian is the continuum version of the one we just derived.

5. Seeing Phonons Experimentally

How do we know this quantum picture is correct? We can't "see" a phonon directly, but we can see its effects. Inelastic scattering experiments, particularly with neutrons, provide direct evidence for phonons.

In these experiments, a neutron with known momentum $k_{i}$ and energy $E_{i}$ scatters off the crystal. It emerges with a different momentum $k_{f}$ and energy $E_{f}$ . The energy and momentum lost or gained by the neutron has been transferred to or from the crystal lattice.

Phonon Creation (Stokes Process): If the neutron loses energy ( $E_{f} < E_{i}$ ), it must have created one or more phonons. The energy of the created phonon is $ℏ ω = E_{i} - E_{f}$ . This process is governed by the creation operator $a^{†}$ .
Phonon Absorption (Anti-Stokes Process): If the neutron gains energy ( $E_{f} > E_{i}$ ), it must have absorbed a pre-existing phonon from the thermally vibrating lattice. The energy of the absorbed phonon is $ℏ ω = E_{f} - E_{i}$ . This process is governed by the annihilation operator $a$ .

By measuring the energy and momentum change of many scattered neutrons, we can map out the phonon dispersion relation $ω (q)$ that we've been discussing. The Feynman-like diagrams below provide a nice visual for these quantum processes.

Inelastic Neutron Scattering: Scattering Function and Feynman Diagrams for Phonon Processes — Diagrams illustrating inelastic neutron scattering. The left (Stokes) process shows an incoming neutron emitting a phonon and scattering away. The right (Anti-Stokes) process shows an incoming neutron absorbing a phonon. These correspond to the action of the creation and annihilation operators, respectively.

Conclusion

In this lesson, we have taken the crucial step from a classical to a quantum description of lattice vibrations. By quantizing the normal modes of the lattice as a collection of harmonic oscillators, we arrived at a beautifully simple picture.

Key Takeaways:

The classical Hamiltonian for lattice vibrations can be decomposed into a sum of independent harmonic oscillators, one for each normal mode $(q, j)$ .
Quantizing these oscillators leads to the introduction of bosonic creation ( $a_{q, j}^{†}$ ) and annihilation ( $a_{q, j}$ ) operators.
The lattice Hamiltonian takes the diagonal form $H = \sum_{q, j} ℏ ω_{j} (q) (a_{q, j}^{†} a_{q, j} + 1/2)$ .
The quantum of lattice vibration is the phonon, a bosonic quasiparticle with energy $ℏ ω_{j} (q)$ and crystal momentum $ℏ q$ .
Phonon creation and annihilation are not just theoretical constructs; they correspond to measurable processes in experiments like inelastic neutron scattering.

Preview of Next Lesson:
Now that we have the quantized energy levels of the lattice vibrations, we can apply the tools of statistical mechanics. In the next lesson, we will use our second-quantized Hamiltonian to calculate the phonon density of states and derive the general expression for the lattice specific heat, one of the most fundamental thermal properties of a solid.