In my syllabus about solid state physics they state that lattice vibration is quantized, analogous to the harmonic oscillator: $$E = (n+\frac{1}{2})\hbar\omega$$ So the lattice vibration has zero-point energy $\frac{1}{2}\hbar\omega$.

I wonder what this actually means: are all possible vibration modes of the lattice quantized in this way? So do all the vibration modes (acoustical/optical and transversal/longitudinal) have nonzero energy for all possible wavevectors $\vec{k}$ in the Brillouin zone? So If I interpret this quantization in this way, it means that the lattice is at all time vibrating in all possible vibration modes.

In the syllabus they state though (somewhat earlier (and translated to English)) : "The vibration modes are purely longitudinal or transversal only in the case of sufficient symmetry, e.g. in some directions of a cubic crystal structure. Otherwise, the waves consist of a mixture of the two." (I thinks that this refers to f.i. the [100] direction in a cubic crystal)

This seems to be contradicting the idea that longitudinal and transversal modes should have zero point energy. I hope that someone can clarify this.

  • $\begingroup$ I'm sorry, but I can't understand your quotation. Can you try to clear it up? $\endgroup$
    – Danu
    Commented Nov 16, 2013 at 12:21
  • $\begingroup$ I tried to clear it up, I hope it is better now. The original quotation is in dutch, but I see that you live in Amsterdam, so you will probably understand it (excuse me if this is not the case). So I would like to quote literally what the author wrote: "Enkel in een situatie met voldoende hoge symmetrie, zoals in bepaalde richtingen van een kubisch kristal zijn de trillingen zuiver longitudinaal of transversaal; in minder symmetrische situaties treden gemengd longitudinale/transversale golven op" $\endgroup$
    – Rayman
    Commented Nov 16, 2013 at 13:01

1 Answer 1


In classical mechanics, you can describe a crystal (in some approximation) by a Hamiltonian that is a quadratic form in coordinates and momenta of atoms. After you diagonalize this quadratic form, you obtain a Hamiltonian of a set of independent, rather than coupled, oscillators (modes). Then you can quantize this system and you do get zero-point energy for each independent mode (by the way, this energy is $\frac{1}{2}$ℏω, not ℏω). However, not all independent modes are purely longitudinal or transversal. In other words, longitudinal modes are often coupled with transversal modes and, therefore, they are not independent modes that you get as a result of diagonalization. In other words, longitudinal and transverse modes are some linear superpositions of independent modes (which are also called eigen modes, or characteristic modes, or normal modes:-) ) .

  • $\begingroup$ Thank you, I think I begin to understand it a bit. I wonder, if there are $p$ atoms in a primitive cell, then there are $3p$ modes (transversal and longitudinal) for each wavevector $\vec{k}$ that is permitted. Are there also $3p$ normal modes per wavevector? So can I say that for each wavevector there are $3p$ normal modes, that are linear superpositions of the $3p$ transversal and longitudinal modes? So with $N$ atoms in the lattice (which means $N$ permitted wavevectors), there are $3pN$ normal modes? (This may be trivial, but for me it is pretty hard to tell at this moment) $\endgroup$
    – Rayman
    Commented Nov 16, 2013 at 15:46
  • $\begingroup$ I don't quite get it. As far as I know, the number of modes is typically counted for the crystal as a whole. If you have $N$ atoms in the crystal, each atom has 3 translational degrees of freedom, therefore the crystal has $3N$ relevant degrees of freedom, so the number of phonon modes is $3N-6$ (we need to deduct 6 = 3 translations of the crystal as a whole + 3 rotations of the crystal as a whole - pit.physik.uni-tuebingen.de/PIT-II/teaching/… ) I cannot tell you right now if or how rotational degrees of freedom of atoms are relevant. $\endgroup$
    – akhmeteli
    Commented Nov 16, 2013 at 18:16
  • $\begingroup$ I was mistaken: $N$ stands for the amount of primitive cells. My syllabus mentions that there are $N$ permitted wave vectors for oscillation modes in the Brillouinzone. Furthermore if there are $p$ atoms in a primitive cell, there should be $3p$ branches of the dispersion relation (your document also mentions this, at p.92). So there should be $3pN$ modes of oscillation: $2pN$ transversal and $pN$ longitudinal. But if I got it right from your answer, the normal modes are not this kind of modes, because they are superpositions of these? So it would sound logical if there are $3pN$ normal modes. $\endgroup$
    – Rayman
    Commented Nov 16, 2013 at 19:03
  • $\begingroup$ So I am really confused, to make it clear about the following: what exactly is quantized if we are talking about phonons? And if we talk about a phonon with wavevector $\vec{K}$ then what are the possible frequencies of the phonon? Can we use the dispersion relations f.i. shown at p.90 of the document you linked? If I understand it right, we can't, because if we are talking about phonons, we are talking about superpositions of modes with wavevector $\vec{K}$ but different frequencies. So the so called "phonon dispersion relations" are completely different dispersion relations? $\endgroup$
    – Rayman
    Commented Nov 16, 2013 at 19:10
  • $\begingroup$ @Rayman: Unfortunately, I don't have time to sort out your notation, sorry. I told you how one counts degrees of freedom of a crystal (3N) and phohon degrees of freedom (3N-6). So there are (after diagonalization) 3N-6 independent (uncoupled) phonon modes. Some of them are not purely longitudinal or transverse, but superpositions of longitudinal and transverse vibrations. So if you know how to perform quantization for one independent oscillator, you can perform quantization for (3N-6) independent oscillators. $\endgroup$
    – akhmeteli
    Commented Nov 16, 2013 at 19:34

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