Q0: What are normal modes (accoustic/optical modes) in 1d diatomic chain?

According to What are normal modes? and wiki, The most general motion of a system is a superposition of its normal modes. 'The modes are normal in the sense that they can move independently, that is to say that an excitation of one mode will never cause motion of a different mode. In mathematical terms, normal modes are orthogonal to each other.'

So mathematically, 'normal' here = 'independent', 'orthogonal' in the sense of vector spaces. Since vibrations are represented by wave(function)s, they could possibly form a vector space of functions, which makes using concepts of vector spaces here natural.

Update: Q3: why the two modes are 'orthogonal ... with respect to the inner product defined by the mass matrix'?

Are the two modes represented by eigenvectors of the (Hermitian) matrix in equ 10.5, and since 'Hermitian matrix has orthogonal eigenvectors for distinct eigenvalues', we have the above orthogonality?

But below it seems the author uses simply eigenvalues $\omega_{+,-}$ (functions of the momentum $k$) to represent the two modes. Then what is the physical meaning/significance of the eigenvectors of the (Hermitian) matrix?

It is not necessary to read what follows to answer the questions (marked as Q0, Q1, Q2; where Q0 = Q1 + Q2).

The DOF is the same as the size of independent/orthogonal basis vectors, supporting my above thoughts. (References: https://www.physics.rutgers.edu/grad/601/CM601/simon2012.pdf)

we obviously have two degrees of freedom per unit cell

The 1d diatomic chain is described by a 2*2 matrix, with 2 eigenvalues, also supporting my thoughts.

We now proceed by plugging in our ans ̈atze (Eqs. 10.3 and 10.4) into our equations of motion (Eqs. 10.1 and 10.2). We obtain (2 equations containing Ax, Ay)
... a1 a2

Since $\omega$ is a parameter in the dispersion relation $E(k)$, with two values of $\omega$ we obtain two dispersion relations, corresponding to two normal modes in Brillouin zone scheme. (Note that from the figure below we see when $\kappa_1 - \kappa_2 \to 0$, the two curves will be connected; in extended BZ scheme, the two curves will be one, and the 1st and 2nd BZ could be combined, and the model will converges to the monoatomic chain model.)


Q1: But since 'independent' is less restricting than 'orthogonal', is 'normal' here = 'independent' or 'orthogonal'?
Q2: Why the two modes are called accoustic/optical modes?
(I am still confused by the author's energy/momentum conservation explanation of accoustic/optical modes.)

Generally the definition of an acoustic mode is any mode that has linear dispersion as k → 0.

(the author seems to say that because when light interact with phonons, due to momentum and energy conservation, only phonons with higher energy/momentum could be involved. Phonons in accoustic modes have small $k$ in BZ nearest to 0. So only phonons in other modes would be involved and therefore called 'optical'.
(this seems to be relevant to (in)direct band and why when photons require phonons to transit from valence band to conduction band such transition can hardly happen (as discussed here). )

The following is uncessary for the question, which I put here simply to compare mono-/di-atomic chains

In 1d monoatomic chain, we calculate $E(k)$ using an differential equation (named Newton’s equation of motion) involving the frequency $\omega$, and an ansatz too.

b1..b2..b3 Since $E$ is proportional to the frequency $\omega$, we get the $E(k)$ directly from $\omega$ (as a function of $k$). (Correction: $\omega$ is not a parameter in the function $E(k)$)


2 Answers 2


It's hard to find what is the question being asked here.

Firstly this is classical problem, so there are no "wavefunctions". It is a standard classical mechanics small vibrations problem and the normal modes with different frquencies are orthogonal in the usual sense of such problems --- i.e with respect to the inner product defined by the mass matrix. In particular acoustic and optical modes with the same $k$ are orthogonal because the matrix in 10.5 is hermitian.

Physically, in the acoustic modes adjacent different-mass atoms move in the same direction while in the optical modes they move in opposite directions.

The Op has asked for more details of the orthogonality.

A small vibration problem is an eigenproblem of the sort $$ \omega_n^2 {\bf M}{\bf x}_n= {\bf Vx}_n $$ here for $n$ degrees of freedom the mass matrix $M$ is an $n$-by-$n$ symmetric matrix and the displacement vector is $n$ dimensional. The potential matrix $V$ is also $n$-by-$n$ and symmetric. If we have two modes with different frequencies $\omega_n^2$ and displacement eigenmodes ${\bf x}_n$ and $ {\bf x}_m $ then we use the matrix symmetry to see that $$ (\omega^2_n-\omega^2_m) {\bf x}_n^T {\bf M} {\bf x}_m=0. $$ So when $(\omega^2_n-\omega^2_m)\ne 0$ we have orthogonality $$ \langle {\bf x}_n, {\bf x}_m\rangle \stackrel{\rm def}={\bf x}_n^T {\bf M} {\bf x}_m=0. $$

For the diatomic chain with $N$ dimers (i.e $2N$ atoms) the ${\bf x}^T =(x_1,x_2, \ldots x_{2N})$ is a $2N$ dimensional vector. The matrix $M$ is diagonal with alternating masses and $V$ is tridigonal. We try an ansatz $x_{2n} = ae^{ik2n}$ for the even sites and and $x_{2n+1}= b e^{ik(2n+1}$ for the odd numbered sites. We find the 2-by-2 eigenproblem for $(a,b)^T$ like 10.5. (Not exactly the same as they have scaled $a,b$ to make $M$ proprtional to the identity matrix and so make 10.5 hermitian)

  • $\begingroup$ I modified the post and marked the questions as Q1, Q2.// Thanks for putting the question into the context of vibrations in classical mechanics. // Yes, I mixed vibrations (about atoms, or any points) with wavefunctions (about electrons), which I can tell from ch11 of the book (linked). Can we say in this question $E(k)$ or $\omega(k)$ represents standing waves; and (confusingly) are wavefunctions usually only used to describe electron waves/particles in quantum?//According to the 'physical' explanation, in the optical modes the atoms should have more tensions so the energy would be higher. $\endgroup$ Commented Jun 2, 2022 at 14:46
  • $\begingroup$ Would you like to further explain 'orthogonal ... with respect to the inner product defined by the mass matrix' (I know 'orthogonal, inner product, matrix'). 'the mass matrix' is that in 10.5? $\endgroup$ Commented Jun 2, 2022 at 14:50
  • 1
    $\begingroup$ I'll add a section on the orthogonality. $\endgroup$
    – mike stone
    Commented Jun 2, 2022 at 15:07
  • $\begingroup$ mass here means the two masses (particles, atoms, etc.) in the system of vibration? $\endgroup$ Commented Jun 2, 2022 at 15:22
  • 1
    $\begingroup$ You can always change basis to make $M\to A^TMA$ (and $V\to A^TVA$) so that $A^TMA$ is proportional to the identity matrix. Then the modes are usual eigenvectors. $\endgroup$
    – mike stone
    Commented Jun 2, 2022 at 17:41

This is relevant to Q2. https://physics.stackexchange.com/a/635103/273056

'The optical modes in crystal vibrations correspond to modes whose wavelength is smaller than the smallest separation between the atoms

my comment: Q: mathematically why this is true?

..As a result, optical modes "decay" into acoustic modes. This is in accordance with .. that momentum is conserved modulo 𝜋/𝑎
..is also why band structure diagrams can be "reflected" into the first Brillouin zone'

find this sentence and the 'decay' confusing


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