# How can a phonon be localized? A meaningless concept? Yet Kittel...

1. ### nonequilibrium

Hello,

As you all know, if a crystal contains a phonon of frequency $\omega$, it means that the fourier decomposition of the physical vibration of that crystal lattice contains a plane wave of frequency $\omega$, and the more phonons of a given frequency, the bigger the amplitude of that component. Hence, the phonon is a nifty quasi-particle.

But it also seems obvious that the phonon cannot be localized. What would the position of a phonon mean, after all? If a phonon is an indication of the presence of a plane wave (of a certain frequency), then since plane waves are not localized, one cannot sensibly associate a place to the phonon. Sure, a lot of plane waves may make a localized wave, but that's hardly relevant: it's not as if one associates one phonon to that localized wave; no, that localized wave is rather a whole bunch of delocalized phonons, exactly like a localized wave is a whole bunch of delocalized plane waves. (I'm saying this to show that it's not at all analogous to for example a psi-plane wave of an electron: in that case the electron is also not localized, but if we have localized wave packet, we can speak about the position of the electron; this reasoning is obviously irrelevant for phonons.)

But I'm reading Kittel and where it talks about thermal conduction (p121, chapter 5, edition 8), it first mentions a formula of kinetic gas theory (a formula for the thermal conductivity $K=\frac{1}{3}C v l$ with C the heat capacity of the gas, v the average particle velocity and l the mean free path of a particle between collisions). And he then, without justification, applies this formula to phonons. This of course implies that he views phonons as localized particles, using a kind of kinetic theory of phonons. How can this make any sense, in the light of what I said above?

2. ### alemsalem

159
I think, you can apply the same reasoning to phonons as to electrons, a superposition of single phonon momentum eigenstates gives a localized particle.. and is different than having a collection of phonons with different frequencies.

3. ### nonequilibrium

I addressed your vague reasoning in my OP and why it isn't valid.

4. ### alemsalem

159
why do you think we can have a localized wave packet for electrons and not phonons?
a whole bunch of delocalized plane waves is localized wave packet, why can't you think of that as a single particle?

5. ### nonequilibrium

I understand that what you're implying is an intuitive idea, but it seems to me that if you actually look at it more thoroughly, it seems to make no sense. I'll try to convince you. If you're not convinced, please indicate why:

The main realization once has to make, is that a phonon can actually not be localized by virtue of its definition. A phonon is often vaguely called "a quantized vibration", and your suggestion fits that vague description, but the actual meaning of a photon is more precise (and the more precise definition does not fit your suggestion). I'll repeat the definition I gave in the OP with more detail: given a lattice vibration $u(x,t)$ (one dimension will do for now), you can decompose this in a fourier expansion $u(x,t) = \sum_{k=-\infty}^{+\infty} q_k e^{i(kx-\omega(k) t)}$. $u(x,t)$ of course satisfies a wave equation; this can be translated into equations for the $q_k$'s. One can show that the $q_k$'s satisfy a harmonic oscillator equation (at least in the harmonic lattice approximation). Quantum-mechanically, the $q_k$'s are hence described by a quantum harmonic oscillator (of frequency $\omega$, such that $\omega = \omega(k)$). One can show (it's a standard upper undergrad stat mech exercise) that the excitation of a harmonic oscillator (i.e. saying that the harmonic oscillator of frequency $\omega$ is in its $n$th mode/excitation/energy level) is mathematically equivalent to the statistics describing $n$ bosons (i.e. with Boson-Einstein statistics) each occupying an energy state $\hbar \omega$. These pseudo-particles are called phonons.
More shortly: saying a crystal lattice has $n$ photons of frequency $\omega$ means by definition that the fourier decomposition of the crystal lattice vibration has a fourier term with frequency $\omega$ with its amplitude proportional to $n$.

Hence, you directly see that a localized vibration corresponds to a lot of phonons, each delocalized since what location would you give to a plane wave vibration?
And this is quite different from the situation of an electron, which is not at all defined the way a phonon is. The only "connection" there seems to be is that both have something to do with a wave, but I hope it's clear how crucially different both concepts are (unless, of course, my idea of a phonon is wrong, or at least incomplete, but that requires an argument).

6. ### alemsalem

159
I guess I'm not that familiar with phonons,, but what I do know is that when you quantize a field,, the $q_k$ becomes an annihilation operator (its hermitian conjugate is the creation operator) I assume this applies to phonons too. it applies to photons..
the field becomes an operator when you quantize the vibrations it does not describe a state. the states are:
$(q_k)^{*} |0>$ where |0> is the graound state(vacuum) and $(q_k)^{*}$ is the creation operator. this is a 1 particle state, superposition of these still leads to a single particle state.
these are two particle states:
$(q_(k1))^{*} (q_(k2))^{*} |0>$

7. ### nonequilibrium

I'm not sure, but aren't you talking about quantum waves? Id est you're talking about quantum states. Phonons refer to crystal lattice vibrations: a physical wave in a solid. More simply put: superpositions of $\psi$ "things" are different than superpositions of $u(x,t)$ "things". We're talking about different concepts here.

159
9. ### cgk

484
OP, this localization concept makes more sense if you think not about the one-particle wave functions (WFs), but the many-particle WFs. There is not only a single phonon in the entire crystal: There is an entire photonic many-body WF made out of single-particle WFs. It is the many-body WF which is really relevant, not the single-particle ones. And these many-body WFs tend to be invariant to unitary rotations amongst the single-particle WFs they are made of. Due to this, it is perfectly possibly to localize the one-particle states without this having any effect on the many-body WF whatsoever.

Consider the similar case of electrons in a metal: Typically you would describe the electrons with Bloch waves which are delocalized over the entire crystal. In a simple metal, you could approximate the true many-body WF with a single Slater determiant made out of those occupied Bloch waves. However, it is possible to perform a unitary rotation amongst the occupied Bloch waves and transform them into something called "Wannier orbitals", which are localized to the unit cells and still reproduce exactly the same many-body wave function when put into a deterimant.

Now with phonons things are a bit more difficult due to the bosonic instead of fermionic nature, and thus the full many body wave function being decomposed into permanents intead of determinants. However, most of those concepts can still be applied, even though they are not often described in that way.

10. ### nonequilibrium

Hm, thanks for the answers, but we seem to be having a fundamentally different view. What does the "wavefunction of a phonon" even mean? The only thing that is really there is the vibrating crystal, right? The phonon is merely a handy quasi-particle that quantifies the presence of certain terms in the fourier expansion of the aforementioned crystal lattice vibration. I don't see why one would start talking about the wavefunction of a phonon, or what it could even mean.

Can an answer to my question address the way I view phonons (which I didn't invent myself, by the way)?

11. ### alemsalem

159
the wave function of a phonon could mean (I'm not really an expert I'm also trying to understand it) what it normally means: the probability that when you look you find only a localized group of atoms vibrating.,, and the fact that its a single particle means that the wave function can't get localized around two different positions when you make a measurement.

Sorry if thats not what you're looking for

12. ### cgk

484
Sorry, you are right that the term "wave function of a phonon" does not make sense. What I meant was the "wave function of the nuclei". Via the Born-Oppenheimer approximation one can decompose the entire wave function of the crystal into the electronic and nuclear degrees of freedom. I mean the latter: The one in which the (quantum) nuclei sit on a potential energy surface created by the electrons.

It is this nuclear wave function which is related to the "phonons". Due to the nature of the potential energy surface, this wave function describes lattice vibrations --- but still as a many-body wave function (i.e., a single wave function with one coordinate for each nuclues). This many-body wave function again can be decomposed into permanents or Hartree-products (that's usually okay for rovibrational wave functions) of one-particle wave functions. And these one-particle wave functions are the same as the elementary vibrational wave functions you gave in your previous post under a number of assumptions (e.g., harmonic approximation, decoupling etc).

Now if one defines phonons as in your previous post (as normally done), then indeed localizing a phonon makes no sense. But if you define them as "one-particle states from which one can build a nuclear wave function", then it is possibly to apply localizing unitary rotations to them which leave the many body wave function invariant. The resulting localized states would not be uncoupled (by the Hamiltonian) anymore, though.

I am not saying that this is what Kittel had in mind, though. Just that the possiblity of localizing such wave functions makes sense in some context. I think this might be a similar issue as with band theory or molecular orbital as it is usually discussed. Bands and molecular orbitals tend to be described as completely delocalized and various physical phenomena are deduced from that---but this is just disregarding the fact that there are other just as valid descriptions of the entire system which are localized in fact (Wannier and localized orbitals, respectively).

13. ### sam_bell

67
You really can think of free electrons and phonons in an ideal lattice as on the same footing. A wave function of an electron with a definite momentum p is <x|p> = 1/sqrt(V) exp(i p.r/h) inside a box of volume V. The wave function of a phonon with definite momentum h*k is more complicated, though we rarely talk about it; it is a function of all the ionic displacements in the lattice: F[U1,U2,...,UN] for N sites. It is easier to write down in a basis of normal coordinates of the momenta, in which case it looks like F'[Pk1,...,Pk3N] = H0(Pk1) H0(Pk2)...H1(PkI)...H0(Pk3N) where H0(p), H1(p) is the single-particle harmonic oscillator ground and 1st excited state in momentum representation respectively, there is ONE phonon in mode kI, Pk := sum(all sites Rn, Pn exp(-ik.Rn)/sqrt(N)), and similarly for Uk if you need it. (I'm sure I'm screwing something up with complex numbers, but nevermind.) Using the relationships between normal coordinates and displacements, you could determine the function F_kI[U1,...,UN]. You would find it has significant amplitude even if you played with ionic coordinates far from the origin. It's not localized. (I'm much too lazy to do this, but YOU could.) For simplicity, we can write |kI> := F_kI[U1,...,UN]

Now, what do you do when you want to build a localized electron wave packet with approximate momentum p? You take a Gaussian superposition of eigenstates |p'> centered about |p>. There is a tradeoff between the localization and the spread in |p'> of course. What if you want to build a phonon wave packet? Same thing. Take superpositions of |k> about |kI>. Again, it would take a lot of work to verify that the resulting G_kI[U1,...,UN] would have significant amplitude only for tweaking the few ionic coordinates near the origin. (If you want the packet away from the region you have to introduce an additional phase factor.) This is still ONE phonon because it is a superposition of ONE phonon wave functions. It also happens to be localized.

You could take localized Gaussian packets |p> as a (non-orthogonal) basis to solve problems in for phonons (as well as electrons) if you wanted. Probably it would be a nightmare.

Last edited: Mar 17, 2012
14. ### Cthugha

1,682
One could use exactly the same argument to argue that photons cannot be localized -as single mode "building blocks" of the em field they are usually pictured monochromatic and delocalized - but still no one has problems in imagining localized photons. The issue is pretty much the same. It is convention to call the single frequency delocalized eigenmode as well as localized bullet-like superposition entities phonon/photon.

15. ### sam_bell

67

To establish further correspondence between free electrons and phonons, you can construct extremely localized phonon states |x> such that <x|x'> --> dirac_delta(x-x'). Remember, we're still thinking of |x> as a ket in a Hilbert space of ionic displacements. Having constructed such states for all x, we can use them as an basis expansion of the ONE phonon subspace. Then we can expand our one delocalized phonon wave as |k> = sum( |x><x|k> ). Now I'm just sure that <x|k> = integral[ all ionic displacements U_n, x({U_n})* k({U_n}) ] =(?) A exp(ik.x). Perhaps it would be worth going through for a 1D chain lattice. Once you've established such a relation, all the ionic coordinates recede from view and you can speak of a phonon wave functios |psi> = sum( psi(x) |x> ) = sum( psi(p) |p> ), just as for free electrons.

You can also turn all this on it's head. In QFT, |x> for an electron can be thought of as a localized excitation in the quantized Dirac Field. In that sense, the object |x> is much more complicated than it looks. (Now, my QFT understanding amounts to a chapter or two. I'm making an educated guess here, but I'm sure they would bear out.)

Last edited: Mar 18, 2012
16. ### Hurkyl

16,090
Staff Emeritus
Aside: |x> and |x'> can't be kets in a Hilbert space. If they were, <x|x'> would be an ordinary function of x,x', rather than a distributional function. They can be distributions themselves -- e.g. kets in a rigged Hilbert space -- for which the statements of your paragraph make sense.

17. ### chrisbaird

617
Quantum objects are weird. They can be waves and particles at the same time. Consider a single electron bound to an atom, and therefore localized to it. This electron makes a level transition where it looses some of its energy in the process of creating a phonon. We say that the electron has emitted a phonon. Since the electron is localized, then due to causality, the phonon it emits must be in some sense localized, even if only for a brief moment. If you need a picture in your head, think of the electron as knocking against a nearby atom, but because the atom is bound in a crystal lattice, it's temporary displacement generates a ripple that propagates away. And because the whole system is quantum, the electron is not a hard ball knocking against a hard atom, rather it is a waving thing that resonates against the crystal lattice as a whole.

18. ### zhanhai

50
We shall first determine the origin of phonon.

Lattice waves exist in crystal as lattice modes, and each mode corresponds to phonons of the same frequency and wavevector. My interpretation is that phonon is NOT the lattice mode itself. Rather, phonon is the quanta of interaction between its lattice mode and another object (electron, neutron, ......)

A non-quantized version of equations describing the energy and wavevector relations in lattice mode-electron (neutron....) interaction, partly corresponding to those in Appendix J of Kittel (8th Edition), could be taken as the origin of phonons. Indeed, the energy and wavevector relations of phonon can be derived WITHOUT application of second quantization.

Moreover, I would like to say that second quantization is "source of evils"; it is often introduced prematurely, as it is in some description of eletron-phonon interaction, which should be electron-lattice wave interaction actually.

This issue is far-reaching.

BTW, the same arguments would apply to interaction of light with other objects. What takes part in the interaction, photon or EM wave?

Last edited: Mar 23, 2012
19. ### sam_bell

67
Zhanhai, second quantization is just notation. It makes life simpler; .. or perhaps confusing to those who aren't comfortable with it. And talking about phonons is just a descriptor. But it's a very useful one that gives us a good mental picture that unites it with electrons, photons, etc. and confirms our understanding of wave-particle duality. You don't have to ascribe any deep physical significance to them if you don't want to. Mind you, there are a few people who might wonder if EM fields are _really there_ because we only perceive them through their interactions with charged particles. It doesn't matter. We like to talk about them.

20. ### chrisbaird

617
While I agree with what you are saying, we should be careful not to draw too close the similarities of photons and phonons. Photons are real particles that can exist by themselves as single entities in a vacuum, whereas phonons are not.