I am reading the The Oxford Solid State Basics and on pages 82-83 it appears the following:

If a classical harmonic system (i.e., any quadratic Hamiltonian) has a normal oscillation mode at frequency ω the corresponding quantum system will have eigenstates with energy: $E_n=\hbar\omega\left(n+\frac{1}{2}\right)$ (9.7)


Each excitation of this “normal mode” by a step up the harmonic oscillator excitation ladder (increasing the quantum number $n$) is known as a “phonon."

If we think about the phonon as being a particle (as with the photon) then we see that we can put many phonons in the same state (ie., the quantum number n in Eq. 9.7 can be increased to any value), thus we conclude that phonons, like photons, are bosons. As with photons, at finite temperature there will be a non-zero number of phonons “occupying” a given mode (i.e., n will be on average non-zero) as described by the Bose occupation factor

Why is each phonon occupying an eigenstate with a different $n$ number? Why is it not possible, for example, to be all on the ground level $n=0$?


3 Answers 3


When the question "what is a quantum particle --- a phonon, a photon, an electron, a Higgs ?" is asked, then the answer should be

--- and here for concreteness I will replace the word "quantum particle" by phonon ---

One phonon is the first and lowest (above the ground state) excitation of the quantum field — here the crystal lattice.

2 phonons are the second excitation of the crystal lattice.

$n$ phonons are the $n^{\text{th}}$ excitation of the crystal lattice.

What does this mean? If there are no phonons there is no excitation, so we are in the ground state. Actually, the crystal lattice does not only contain one type of oscillators, it contains a myriad of different oscillators which could be or not be excited. We will give each oscillator an "index" called "$k$". So taking into account that there are many different oscillators, the total energy of these oscillators (in the ground state) is $\sum_k \frac{\hbar \omega_k}{2}$ (the ground state has a non-zero energy, a phenomenon that is typically quantum.)

But if we have already a non-zero excitation of the lattice in an oscillator, let's say $l$, then the total energy of the system is:

$$E = \hbar\omega_l + \sum_k \frac{\hbar \omega_k}{2}.$$

Let's assume we have a second excitation in the lattice, and even one which happens in the same oscillator $l$ as the first one, we get as total energy:

$$E = 2\hbar\omega_l + \sum_k \frac{\hbar \omega_k}{2},$$

and if we have $n$ excitations in the same oscillator $l$ then the total energy would be:

$$E = n\hbar\omega_l + \sum_k \frac{\hbar \omega_k}{2}.$$

Note as all excitations so far happen in the same oscillator $l$, they are all in the same state which we can call neatly $l$.

But, actually the excitation can happen in any of the many oscillators. In that case if we have 1 excitation in oscillator $l$ and another one in oscillator $m$ then the total energy would be:

$$E = \hbar\omega_l + \hbar\omega_m + \sum_k \frac{\hbar \omega_k}{2}.$$

On the other hand, if we only consider just one oscillator $k$ and only look at its excitations, we can reduce the general formula for the energy to the one cited in the post (for generality we consider $n$ excitations, where $n$ can be zero (ground state, no phonons), 1 (one phonon) or even 2, 3, 4, ... phonons):

$$E = \hbar\omega_k \left(n + \frac{1}{2}\right).$$

We forget about the ground state energy of the other oscillators, because in this particular consideration they don't play any role (and the energy level can be always adapted to our needs (think of the definition of potential energy in the gravitational field)).

So in order to recap:

$k$ indexes or distinguishes the different states inside the crystal, whereas $n$ numbers the level of excitation. Remember, the excitations are quantized. So each time the excitation is increased by 1 quantum (i.e. 1 phonon) the energy increases by $\hbar\omega_x$. $x$ here specifies the oscillator in which the excitation takes place, which could be any oscillator of the many different ones. Increasing the level of excitation means increasing the number of phonons. Increasing the level of (boson) excitation can happen in the same oscillator or different oscillators.

  • 1
    $\begingroup$ What a fantastic explanation! $\endgroup$
    – DamyHao
    Commented Jan 17, 2022 at 10:36

Why is each phonon occupying an eigenstate with a different $n$ number?

No, it's not each phonon that occupies that eigenstate. It's $n$ phonons of the same mode with frequency $\omega$ that occupy this state. It's a single state with $n$ phonons. Add one more phonon, and you increase $n$ by $1$. Remove a phonon, and you decrease $n$ by $1$. The $n$ variable counts the phonons.

The eigenstates being described are the so called number states—the states with a definite number of phonons. There can be superpositions of them, like $(|n=3\rangle+|n=4\rangle)/\sqrt2$, so you can get states with indefinite number of phonons. A useful example of such non-number state is coherent state.

Also note that all this discussion only touches one single mode of the phonon field. More realistic situations, like e.g. wave packets, are described by states where multiple modes are excited. But each of these modes still behaves as your textbook describes.


The phonons in the same mode occupy the same eigenstate, whereas $n$ number refers to the collection of phonons, i.e., their total energy. That is, for any $n$ (non-interacting) bosons occupying the same state of energy $E_0$, their total energy will be $nE_0$.

  • 1
    $\begingroup$ So the eigenstate is not defined by n? $\endgroup$
    – DamyHao
    Commented Jan 17, 2022 at 8:55
  • $\begingroup$ Aigenstate of a single photon or eigenstate of a system of photons? $\endgroup$
    – Roger V.
    Commented Jan 17, 2022 at 8:58
  • $\begingroup$ Not photons. Phonons. $\endgroup$
    – joseph h
    Commented Jan 17, 2022 at 8:58
  • $\begingroup$ @josephh it works in the same way for both. $\endgroup$
    – Roger V.
    Commented Jan 17, 2022 at 9:22
  • 1
    $\begingroup$ Right, but technically we are talking about phonons. $\endgroup$
    – joseph h
    Commented Jan 17, 2022 at 9:40

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