The textbook says that: if the vibration of the lattice were harmonic, then the states of phonons would be orthogonal to each other. As a result, there would be no interaction between phonons, hence no phonon-phonon scattering, and the thermal conductivity would be infinite. In other words, finite thermal conductivity is an anharmonic effect.

My mind is not so well satisfied with this explanation. Shouldn't the eigenstates of a Hamiltonian be orthogonal to each other? And what is the QM formalization of the phonon-phonon scattering? Is the perturbation theory suitable for this problem? Thanks in advance :)

============== EDIT ==============

After reviewing classical mechanics I have realized the meaning of "orthogonality of modes". And the source of my misunderstanding lies in my confusion between "orthogonality of modes" and "orthogonality of eigenstates".

In a multi-degree-of-freedom system, one can expand the Lagrangian near the equilibrium point:

$$ L(\vec{q},\dot{\vec{q}})=\frac{1}{2}m_{\alpha\beta}\dot{q_\alpha}\dot{q_\beta}-\frac{1}{2}k_{\alpha\beta}{q_\alpha}{q_\beta}, $$

where Einstein convention is used.

The theory of "normal modes" says that the Lagrangian can be written as:

$$ \begin{aligned} L(\vec{Q},\dot{\vec{Q})}&=(\frac{1}{2}M_1\dot{Q}_1^2-\frac{1}{2}K_1{Q}_1^2)+(\frac{1}{2}M_2\dot{Q}_2^2-\frac{1}{2}K_2{Q}_2^2)+\cdots \\ &=L_1(\dot{Q}_1,{Q}_1)+L_2(\dot{Q}_2,{Q}_2)+\cdots \end{aligned} $$

where $Q_\alpha=A_{\alpha\beta}q_\beta$ and $A_{\alpha\beta}$ is an orthogonal matrix.

Therefore, different modes can be seen as different non-interacting subsystems. This is very different from the orthogonality of eigenstates of a Hamiltonian.


1 Answer 1


I think your confusion comes from which orthogonality is referred to. Your textbook means that the modes are orthogonal, not the energy eigenstates. This is a general fact for any quadratic system: they can be always diagonalized and treated as non-interacting systems. In a full QM treatment, these modes need to be quantized leading to an infinite dimensional Hilbert space.

Take for example two coupled oscillators $1,2$ (with their associated creation/annihilation operators) described by the Hamiltonian: $$ H = \omega_1 n_1+\omega_2 n_2+I (a_1^\dagger a_2+a_2^\dagger a_1) $$ with the final term being the interaction. You can easily diagonalize it with the new (non-interacting) modes indexed by $\pm$: $$ H = \omega_+ n_++\omega_- n_- $$ If you carried out the calculation, you would find up to reordering that: $$ \omega_\pm = \frac{\omega_1+\omega_2\pm\sqrt{(\omega_1-\omega_2)^2+4|I|^2}}{2} $$

For phonons, you'll have many more modes that you can diagonalize using a Fourier basis, taking advantage of the translation symmetry. Phonon scattering can later be included as an additional non quadratic terms in the Hamiltonian. For example, they could model the interaction between an electron and a phonon (typically of order $3$), or phonons amongst themselves (typically of order $3$ as well). Actually, for thermal conductivity, a naive third order interaction would not be enough, you'll need to include Umklapp processes. Check out Kittel's Introduction to Solid State Physics for more information, for example.

You can then apply perturbation theory in terms of the interaction parameter, and visualize each term of the perturbation series as a Feynman diagram. However, for quantum systems, you'll typically be faced with issues of regularization and renormalization which makes a naive perturbative treatement ill-defined.

Hope this helps and tell me if you need more details.

  • $\begingroup$ Thank you very much lpz. But I still don't quite understand why the non-quadratic terms can break the orthogonality. Isn't the Hamiltonian matrix Hermitian, which means that its eigenvectors (whether they are modes or eigenstates) are orthogonal? Or maybe I have a problem with my understanding of "modes"? You said that it is a general fact that quadratic systems can be always diagonalized and treated as non-interacting systems. Could you please provide more details on that? Thank you in advance. $\endgroup$
    – Godfly666
    May 2, 2022 at 12:55
  • $\begingroup$ The non-quadratic terms are mixing terms that cannot typically be transformed away using a unitary transformation. Once again, the orthogonality of modes isn't restricted in QM, it's a problem that also arises in classical mechanics. After quantization, your Hamiltonian will always be hermitian, so the spectral theorem still applies. When we say orthogonality, we mean that the Hamiltonian is expressed as sums of terms only involving one degree of freedom. $\endgroup$
    – LPZ
    May 2, 2022 at 13:00
  • $\begingroup$ For references around phonon interactions, check out for example introduction to many-body quantum in condensed matter physics, by H Bruus $\endgroup$
    – LPZ
    May 2, 2022 at 13:01
  • $\begingroup$ Thank you again lpz :) I think maybe the scope of my knowledge isn't yet enough for this problem. Have to read more books $\endgroup$
    – Godfly666
    May 2, 2022 at 13:08

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