# Why zero chemical potential does not allow the Bose-Einstein Condensation of Phonons?

Here I report the reasoning from which my question comes. According to:

O.V. Misochko, Muneaki Hase, K. Ishioka, and M. Kitajima. Transient bose–einstein condensation of phonons. Physics Letters A, 321(5–6):381 – 387, 2004.

one cannot have the phonon Bose-Einstein condensation at equilibrium due to zero chemical potential.

Physically I expect that as $T\to0$ the vibrational energy of a Crystal tends to its zero point energy (considering the nuclei of the crystal as quantum particles and according to the uncertainty principle). So at $T=0$ the number of phonons should be the smallest possible and all should be in the ground state. As far as I know the number of phonons in a crystal depends on the Temperature through the Bose-Einstein distribution:

$$n_{({\bf q},s)}=\frac{1}{e^\frac{\hbar \omega({\bf q})}{k_B T}-1} \tag{1}$$

The reason for which the Phonons have zero chemical potential, $\mu=0$, is that they do not have to be fixed in number, so when finding the phonon distribution function we have only one Lagrange multiplier for the energy which turns out to be related to $k_B T$.

The phrase remarked above made me think that $\mu=0$ should imply that it is more convenient to destroy phonons at $T=0$ instead of having them condensed in the ground state.

I have also noticed that an acoustic phonon at $\Gamma$ has zero energy and so the occupation number from Eq. (1) should diverge... concerning this, I have read the last answer to this question:

Could the chemical potential of a Bose gas be zero?

and since the dispersion relation of an acoustic phonon in gamma has zero second derivative, its mass is zero and it seems me that phonons in $\Gamma$ can be seen as analogous to blackbody photon mentioned in that answer. Anyway that answer is still too qualitative to be satisfactory to me, for this reason I am still asking this question.

I studied the Bose-Einstein condensation a gas of a fixed number, $N$, of Bosons. In that case I had $\mu <0$ at finite $T$ and $\mu \to 0$ in presence of the condensate. This is also confirmed in the article

G. Cook and R. H. Dickerson. Understanding the chemical potential. American Journal of Physics, 63(8):737–742, 1995.

which treats also the case of photons but unfortunately it does not explain why for photons the BEC does not occur.

Summing up, I have some ideas why BEC does not happen for phonons with $\mu=0$ but I am still not satisfied. I would appreciate any help and some good references for this question.

I think you pretty much have all the right ideas in your question.

In a conventional BEC, particle number is conserved. Therefore, as T goes down towards 0, $\mu$ goes up towards the lowest state to conserve particle number. So the number of particles in the lowest state goes up ... eventually becoming macroscopically large, and that's the BEC (in the non-interacting situation).

For phonons, particle number is not conserved, and $\mu$ is always 0 (in equilibrium). Therefore, as T gets lower but $\mu$ stays the same, the occupation of every single mode goes down. You can see that directly from the formula.

PS: You referred to the zero-frequency acoustic phonon. Well, it's only zero-frequency in the textbook calculation for an infinitely-large crystal. Real objects have finite size, and the lowest phonon frequency corresponds to the fundamental acoustic mode of the object. In everyday terms, try whacking your crystal with a stick, and listen to the ringing sound it makes. What frequency is it? Maybe 300Hz for a small object? Well then, 300Hz is the lowest-frequency phonon. The occupation of this lowest-frequency 300Hz phonon mode decreases with temperature, just like all the other phonon modes.

There is an important distinction between Bose-Einstein condensation of atoms and quasiparticles, which I will describe below.

Bose-Einstein condensation of an ideal (atomic) Bose gas

For bosons, chemical potential $\mu$ has an important restriction: $\mu<E_{0} \equiv 0$ where $E_{0}$ is the ground state energy. Otherwise, the occupation number $\mathcal{N}(E)$ will have a negative value, which is not physical. The total particle density $n$ is calculated using the density of states per volume $\mathcal{D}(E)$ and the occupation number $\mathcal{N}(E)$ following the Bose-Einstein distribution: \begin{align} n &= \int_{0}^{\infty} \mathcal{D}(E)\mathcal{N}(E)dE\\ &= \frac{1}{V}\frac{1}{e^{-\mu/k_{\textrm{B}}T}-1} + \int_{0}^{\infty} \frac{1}{(2\pi)^{2}}\left(\frac{2m}{\hbar^{2}}\right)^{3/2}\frac{\sqrt{E}}{e^{(E-\mu)/k_{\textrm{B}}T}-1}dE \end{align} Suppose we hold $T$ constant and increase the particle density $n$ by adding particles to the system (note that $n$ is an increasing function of $\mu$). For density to increase, we must correspondingly raise the value of $\mu$. In the upper limit $\mu \rightarrow 0$, the particle density becomes $$n \simeq \underbrace{-\frac{k_{\textrm{B}}T}{\mu V}}_{n_{0}} + \underbrace{\frac{2.612}{\lambda_{\textrm{th}}^{3}}}_{n_{\textrm{th}}}$$ where $\lambda_{\textrm{th}}$ is the thermal de Broglie wavelength. The first term (ground-state density $n_{0}$) increases indefinitely as $\mu\rightarrow 0$, while the second term (excited-state density $n_{\textrm{th}}$) approaches the finite limit. Therefore, the ground state can accommodate excess particles that all the excited states cannot take. After this point, any added particles must go into the ground state and form a Bose-Einstein condensate. The chemical potential is then expressed as $$\mu = -\frac{k_{\textrm{B}}T}{(n-n_{\textrm{th}})V} = -\frac{k_{\textrm{B}}T}{N_{0}}$$ where $N_{0}$ is the number of particles in the ground state. Note that the chemical potential is exact zero only in the thermodynamic limit, where the total number of particles $N$ is infinite. In a realistic system, the chemical potential is always non-zero.

Bose-Einstein condensation of photons and phonons

However, there are systems where the chemical potential is intrinsically zero even at high temperatures and low densities. The zero chemical potential $\mu = 0$ means that its conjugate variable, number of particles $N$, is not conserved. In other words, if particle-number-changing interactions are dominant, there is no free energy accompanied by adding or removing a particle. The most common example is photons, where they can be absorbed and emitted by matter; hence, the number of photons can vary without the cost of energy. The number of photons is constantly and automatically being adjusted to the Planck distribution $$\mathcal{N}(E) = \frac{1}{e^{E/k_{\textrm{B}}T}-1}$$ which is the Bose-Einstein distribution with zero chemical potential. Using the density of states $\mathcal{D}(E)$ for massless particles, the photon density becomes $$n = \int_{0}^{\infty}\mathcal{D}(E)\mathcal{N}(E)dE = \int_{0}^{\infty}\frac{E^{2}}{\pi^{2}\hbar^{3}c^{3}}\frac{1}{e^{E/k_{\textrm{B}}T}-1}dE \simeq 2.404\frac{(k_{\textrm{B}}T)^{3}}{\pi\hbar^{3}c^{3}}$$ The same holds for acoustic phonons with linear dispersion, except for the light speed $c$ being replaced by the sound velocity.

Note that this satisfies most of the conditions of Bose-Einstein condensation. The density of states vanishes for the ground state $\mathcal{D}(E\rightarrow 0) = 0$, so the number of particles has a finite limit given by the above integration. Also, the ground-state occupation $\mathcal{N}(E=0)$ is infinite. However, the particle density in the ground state is $$n_{0} \propto \lim_{E\rightarrow 0}\frac{E^{2}}{e^{E/k_{\textrm{B}}T}-1} = 0$$ unlike the previous case for atoms where the ground state density diverges as $$n_{0} \propto \lim_{E\rightarrow 0}\frac{\sqrt{E}}{e^{E/k_{\textrm{B}}T}-1} = \infty.$$ The ground-state density $n_{0}$ for photons is never comparable to the thermal photon density $n_{\textrm{th}}$, even though the ground-state occupancy is infinite. Furthermore, any added (removed) particle will disappear (reappear) so that the free energy is minimized, i.e., the chemical potential is zero. The only way to change the average number of particles is to change temperature. Therefore, Bose-Einstein condensation cannot occur in systems without a particle-number conservation.

Furthermore, as noted by the other answer, the ground state of a photon (or an acoustic phonon at $\Gamma$ point) has zero energy except for vacuum fluctuation—there is no excitation at all. The zero energy $\hbar \omega = 0$ essentially means that there is no photon (lattice vibration), and the number of photons (phonons) in the ground state is zero by definition. This comes from the shape of the dispersion.

But what if you engineer a photonic (acoustic) band gap structure so that there is a finite minimum of the photon (phonon) dispersion? Even in this case, and even if the density of states scales as $\mathcal{D}(E)\propto\sqrt{E}$ as in atomic case, another important point should be addressed: quasiparticle timescales.

Bose-Einstein condensation of quasiparticles

Quasiparticles, in general, will disappear after a long time and the number of quasiparticles is conserved only at a certain timescale. As mentioned above, the particle-number conservation is necessary to have a non-zero chemical potential and Bose-Einstein condensation. Therefore, the comparison between the lifetime and other timescales is important.

Quasiparticle lifetime ($\tau_{\textrm{l}}$) has two components: radiative lifetime ($\tau_{\textrm{r}}$) and non-radiative lifetime ($\tau_{\textrm{nr}}$). The former is related to the interaction with photonic modes, while the latter is related to the interaction with other particles such as phonons or defects. On the other hand, interaction time ($\tau_{\textrm{int}}$) can be classified in several ways. One way is based on the type of particles that scatter: self-interaction ($\tau_{\mathrm{self}}$) and interactions with bath ($\tau_{\mathrm{bath}}$). Another way is to use the number conservation as a criterion: number-conserving interactions ($\tau_{\mathrm{c}}$) and non-number-conserving interactions ($\tau_{\mathrm{nc}}$).

These timescales are typically inter-related. For example, $\tau_{\mathrm{self}}$ can include both $\tau_{\mathrm{c}}$ (e.g. elastic two-magnon collision) and $\tau_{\mathrm{nc}}$ (e.g. second-harmonic generation of photons). As another example, exciton-phonon interactions can take away the energy of excitons but do not destroy excitons ($\tau_{\mathrm{c}}$), while magnon-phonon interactions can destroy magnons by emitting phonons ($\tau_{\mathrm{nc}}$ and $\tau_{\mathrm{nr}}$). For that reason, it is not trivial to make a single statement for the condition of quasiparticle Bose-Einstein condensation. Nevertheless, the following condition must be satisfied: $$\tau_{\mathrm{c}} < \tau_{\mathrm{l}} < \tau_{\mathrm{nc}}$$ The first inequality is the thermalization condition: quasiparticles need long enough lifetime to thermalize themselves ($\tau_{\mathrm{self}}$) or to thermalize to bath ($\tau_{\mathrm{bath}}$). The second inequality implies that the quasiparticle can be treated as a thermodynamic particle if the non-number-conserving scattering time ($\tau_{\mathrm{nc}}$) is longer than the lifetime.

For example, consider the case where some non-thermal quasiparticles are added to an equilibrium system. The system is perturbed from the original equilibrium Bose-Einstein distribution. After $\tau_{\mathrm{c}}$, quasiparticles will have a new distribution with non-zero chemical potential and well-defined temperature, which can be equal to or different from the bath temperature. It is determined by whether the nature of the number-conserving interactions is dominated by $\tau_{\mathrm{self}}$ or $\tau_{\mathrm{bath}}$. If the quasiparticles decay after this timescale but before $\tau_{\mathrm{nc}}$, then the system can be steadily pumped to replenish quasiparticles—called a driven-dissipative system. However, if $\tau_{\mathrm{nc}}$ is shorter than $\tau_{\mathrm{l}}$, the distribution is adjusted to the equilibrium distribution with zero chemical potential, and the quasiparticle will always, at the end, have zero chemical potential. For photons, $\tau_{\mathrm{nc}}$ is typically much shorter than $\tau_{\mathrm{l}}$ so that the number of photons is quickly adjusted to the Planck distribution. For atomic condensates, three-body collision time $\tau_{\mathrm{nc}}$ sets the upper limit of the particle density.