bare Phonon and Symmetry Breaking

In condensed matter physics, the phonon is considered as a quasiparticle which is a result of the quantization of lattice vibrations. In many textbooks on solid state physics, it can be done if we consider a harmonic approximation, which means that if we consider the ions form a periodic lattice, and are moving around this position, the restoring force is proportional to the square of the displacement, we can get acoustic phonon, whose frequency $\omega_q$ goes to 0 as $q\rightarrow0$, and this can be understand as an example of Goldstone mode;

However, as mentioned in the book by Mahan, this acoustic phonon is a result of electron screening of the bare phonon, where the bare phonon is just a plasma mode of ions, whose frequency is finite as $q\rightarrow0$. Seems plasma is related to a continuum model of ions.

So I wonder how to connect these two ideas together and give a unified description of phonon. PS: can the ions form a periodic lattice without electrons?

@GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics.

To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is utterly unstable except in the 1-d case depicted in the previous answer. In 2 or more dimensions and in the absence of the electron background, ion-ion repulsion instantly flings the ions to whatever the boundary of the spatial domain is, just as electrostatic repulsion flings free charges from the bulk volume to the surface in a typical conductor.

To answer the other question: I think you are misunderstanding the concept of "bare phonon Hamiltonian" and "electron screening". If your question is based on the description in Sec. 7.4.1 of Mahan, "Phonons in Metals", pgs. 482-483, please keep in mind that the model considered therein is that

"the ions are point charges and the electron gas is jellium".

When Mahan says that

"The Hamiltonian of the phonons is first solved without reference to the electrons" (pg. 483, paragraph before Eq.(7.244))

and that

"These phonons are calculated by ignoring electron-electron and electron-phonon interactions. The ion motions are calculated using only the direct ion-ion interaction potential V_{ii}" (pg. 483, paragraph after Eq.(7.244))

he means that in a zero order approximation the phonons and their "bare frequencies" are calculated for ion-ion interaction in a static background of electron gas (or jellium) that does not adjust to the motions of the ions. This is nothing but the Hartree-Fock approximation for the motion of ions or nuclei in the field of the electrons as determined by the instantaneous electronic state - when the electron state is considered as time-independent. As further explained, this is only a convenient first approximation, since in reality the electrons are expected to follow the ions motion rather swiftly, and so their state and the background charge distribution it generates will vary in time. Accounting for such time variation in the electron configuration involves

a) accounting for perturbations in the electron state wrt the electronic zero order approximation of fixed nuclei, leading to electron-electron interaction terms, and

b) accounting for dynamical changes in the ion-ion interaction potential wrt to the nuclear/ionic zero order approximation of fixed background electronic configuration, which reflects in electron-phonon interactions.

As for the acoustic branch of the "bare phonons", it refers to in-phase oscillations of the ions similar to sound waves or charge oscillations in a plasma, wherefrom the analogy. The long wavelength limit $q\rightarrow 0$ corresponds to ions moving coherently in the same direction (whole lattice displacement), and so the oscillation frequency vanishes $\omega_q \rightarrow 0$. The optical branch corresponds to out-of-phase oscillations in which neighboring ions move in opposite/different directions, see here.

• I want to ask that if we just consider the zeroth order, which is, as you mentioned, the ions are point charges and electron gas is jellium which doesn't response to the motion of ions, does the ions form a crystal like structure? I would expect not since the elementary excitation frequency does not goes to $0$ as $q\rightarrow 0$, so there is no Goldstone mode and no symmetry breaking. In other words, to get the symmetry breaking (periodic lattice), or say acoustic phonon(Goldstone mode), we need to consider the "dynamical screen" from electrons. May 27 '16 at 14:56
• Sorry about the delay, for whatever reason your comment didn't show up in my queue. It happens, occasionally. The zeroth order approximation is in principle the result of a self-consistent Hartree-Fock calculation. This means that the electrons' state is determined for any given configuration of the nuclei and the resulting electronic ground state energy functions as an effective potential energy for the nuclei. The equilibrium configuration for the latter is given by the minimum of the effective potential (electronic ground state energy) as a function of the positions of the nuclei,
– udrv
May 29 '16 at 6:55
• and then the phonons are determined from the harmonic approximation of the effective potential around this equilibrium configuration. So yes, the nuclei (ions) do form a crystal equilibrium structure, but it is not determined by "dynamical screening" from the electrons. It is actually determined by the electronic energy for static configurations. On the other hand, phonons are determined by the behavior of the electronic ground state energy when the nuclear configuration is varied around its equilibrium position, so in principle they do arise from dynamical adjustment of the electron state
– udrv
May 29 '16 at 6:56
• to the displacement of the nuclei (ions). Metaphorically you can look eventually at the nuclei's crystal structure as "breaking" the continuous translational symmetry of free particles (but we do not have free particles in this case, which is why this holds only "metaphorically") , in which case phonons can be likened to the associated Goldstone mode(s).
– udrv
May 29 '16 at 6:56
• I think the whole Goldstone idea is related to these notes, lassp.cornell.edu/sethna/OrderParameters/…, where it was used to provide an intuitive analogy of sorts. But imho it's a bit of stretch too take it as the fundamental point of view for regular acoustic phonons.
– udrv
May 29 '16 at 6:57

Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons.

Now consider three protons A, B, C. If you displace proton B from its central position toward proton C and away from proton A, then B will experience a net force pushing it back to the center. Meanwhile, B exerts opposing forces on A and C. By the time B gets back to zero, A and C are moving, and they perturb the positions of the protons on either side of them, which starts a propagating wave. The frequency of this wave has a minimum, it is the frequency of a harmonic oscillator with the mass of a proton and a spring constant derived from the repulsion of protons on either side of it.

But a lattice of bare protons would want to explode. So lets pour electrons into the lattice until we achieve charge neutrality on average. The protons still want to be spaced at regular intervals in the ground state. Now when you move proton B, protons A and C are given a push, but not so strongly as before because the electrons slosh from the larger gap (A-B) to the smaller gap (B-C). This cancels some but not all of the restoring force. The slower you move B, the better job the electrons can do of screening out the force. So the spring constant depends on how quickly you move or rather on the size of q. As q goes to zero, so does the frequency of vibration.

• I don't quiet understand the last sentence:"So the spring constant depends on how quickly you move or rather on the size of q. As q goes to zero, so does the frequency of vibration." what's the $q$ there? is it reciprocal momentum? May 27 '16 at 15:01