Solid State and Condensed Matter

(Part of the Wolverhampton Lectures of Physics's Quantum Physics Course)

By Quantum Physics—the title of this course—one can understand the application to Physics as a whole of the frame of mind of quantum mechanics. That is, one can apply its principles (quantization, wave-particle duality, indistinguishability, non-commutativity of the observables, wavefunctions, superpositions, entanglement, etc.) to describe well-known topics such as optics, acoustic, electromagnetism, fluid dynamics, thermodynamics, crystallography, metallurgy, material sciences, etc. In many cases, that leads to a new discipline, often with the adjective "quantum" added to it, such as quantum optics, quantum electrodynamics (with electrodynamics the dynamic of electromagnetic fields), etc. As a whole, this body of new topics constitutes Quantum Physics and this breaks into as many separate fields as there exist in classical physics, sometimes even going beyond that, as exemplified by quantum information science, quantum chemistry or quantum technology.

We will illustrate how one particular aspect, namely, the particle statistics—boson or fermions (introduced in Lecture XI)—lays the foundations for the cases of solid-state and condensed-matter physics. These particular examples will illustrate the trend of revisiting well-established topics through the angle of quantum mechanics, since one has theories for the possible states of matter since before Physics itself (e.g., the four elements earth, water, air, fire). Modern versions, for instance the kinetic theory of gases or the Drude solid were actually very much advanced. We will see how the basic principle of symmetry of the wavefunction however constitutes not a correction or refinement to the existing theories but actually provide the foundations for them. We refer you to other courses to become a specialist of particular topics and to cover a much broader breadth of the details.

Atoms (from the previous lectures) can form molecules, from the simplest one (the hydrogen molecule-ion $H_2^+$, i.e., the di-Hydrogen which lost an electron, making it a one-body chemical problem) to the most complicated one (DNA). This is described by chemistry. To make a substance, various such blocks (atoms or molecules) must assemble into another structure, a gas, a fluid or a solid.

We start with the gas. We consider that particles (still of any types, bosons or fermions) are confined in a box of size~$L$ (volume~$V=L^3$). They are free to roam inside, with kinetic energy $E=\hbar^2k^2/(2m)$. To describe the gas as a whole, we will average over particles in the box, and with $k$ as the quantum number, we will want to make a sum of the type:


Further assuming isotropy and that the system is large enough so that the $\Delta k$ granularity washes out and one can integrate instead of summing, we have:

$$\sum_{\mathbf{k}}=\left({L\over2\pi}\right)^3\sum_{\mathbf{k}}{(\Delta k)^3}\rightarrow\left({L\over2\pi}\right)^3\iiint d\mathbf{k}=\left({L\over2\pi}\right)^3\int 4\pi k^2dk$$

and shifting over the more natural integration over energy, since, for a free gas

$$\begin{align} k&={\sqrt{2mE}\over\hbar^2}\,,\\ dk&=\sqrt{m\over2}{1\over\hbar}{dE\over\sqrt{E}}\,, \end{align}$$

we have


Show that in $D$ dimension, the sum over $\mathbf{k}$ reads, more generally: $$\sum_\mathbf{k}\rightarrow\left({L\over2\pi}\right)^D\int{\Omega_D\over D}\left({2m\over\hbar^2}\right)^{D\over2}E^{{D\over 2}-1}\,dE\,.$$ where $\Omega_D=4\pi$ in 3D, $2\pi$ in 2D and $2$ in 1D.

The integrand is an important dimension-dependent quantity known as the density of states:

$$\rho_D(E)\equiv E^{{D\over 2}-1}\,.$$

It grows in 3D, is constant in 2D and decreases in 1D. It is singular, or a Dirac distribution, in 0D as the sum simply disappears. This describes how many states are available at a given energy.

Now enters the statistical argument. It can be shown (we will demonstrate this result ourselves in Modern Physics for the quantum particles) that the distributions for the various types of particles read, for the effective (dimensionless) inverse temperature $\beta\equiv 1/(k_\mathrm{B}T)$ (with $T$ the real temperature and $k_\mathrm{B}$ the Boltzmann constant) and so-called "chemical potential $\mu$":

$$f_{\varsigma,\sigma}(E)\equiv{1\over e^{\beta (E-\mu)}-\varsigma}$$

where $\sigma$ is the spin (what determines the statistics in the first place), and where we remind that $\varsigma=-1$ for fermions (antisymmetric wavefunctions), $0$ for classical (distinguishable) particles $+1$ for bosons (symmetric wavefunctions). The total number of particle in the system $N$ is thus distributed according to:

$$N=\int f_{\varsigma,\sigma}(E)\rho_D(E)\,dE\,.\tag{1}$$

We can now compute for the various particular cases.

The cases of low temperatures will be more interesting, this is where quantum statistics play a role. In the high-temperature limit, i.e., when $\beta\to0$, then $\beta E$ is small so the probability that any energy state in particular is populated is small, meaning that a lot of them will be populated, which makes physical sense: temperature distributes things around. But an integral of the type (1) diverges unless if the exponential term is large, while $\beta E$ would end to make it small. Here the chemical potential comes to the rescue and helps further boosting the exponential, that, despite a small $\beta$, becomes correspondingly large so that it overtakes the $\varsigma$ term. This also shows that the chemical potential has to be temperature dependent, otherwise the integral would not converge. Neglecting $\varsigma$, one recovers the Boltzmann (classical) case, meaning that at high temperatures, particles loose their bosonic/fermionic character: they do not populate the available quantum states in sufficient quantitu to feel an effect of the symmetry of their wavefunctions. So we now turn to the case of low temperatures.

Starting with the electron gas, we have:

$$N={L^3\over2\pi^2}\left({2m\over\hbar^2}\right)^{3\over2}\int_0^\infty{\sqrt{E}\over e^{\beta (E-\mu)}+1}\,dE\,.\tag{2}$$

Note a factor $2$ for the two states of spin possible for electrons, as well as the 3D element $\Omega_D=2\pi$ that both contrive to make the prefactor $L^3/(2\pi^2))$. This integral has no known analytical expression. Approximation are thus made for low-enough and very high temperatures. First for small temperatures, i.e., $\beta\to\infty$, then the expression actually becomes very simple, since either $E>\mu$ in which case $e^{\beta(E-\mu)}\gg 1$ and $1/(e^{\beta(E-\mu)}+1)\approx 0$ or $E<\mu$ and, on the opposite, $e^{\beta(E-\mu)}\ll 1$ and $1/(e^{\beta(E-\mu)}+1)\approx 1$, in which case the distribution simply behaves as a cutoff and

$$N={L^3\over 2\pi^2}\left({2m\over\hbar^2}\right)^{3\over2}\int_0^{E=\mu}\sqrt{E}dE\,.$$

This makes an equation for $\mu$, and since $\int_0^{E=\mu}\sqrt{E}dE={2\over 3}\mu^{3\over 2}$, we find:

$$\mu={\hbar^2\over 2m}\left({3\pi^2 N\over L^3}\right)^{2\over 3}\,.$$

The cubic root of the quantity in parenthesis has units of a wavector ($\mu$ itself has units of energy). It is an important parameter of the electron gas, that we call $k_\mathrm{F}$ (F is for Fermi and we now have the Fermi wavevector, or momentum, for an electron gas:

$$k_\mathrm{F}\equiv\left({3\pi^2n}\right)^{1\over 3}$$

where we also introduced the volume density of electrons (that is a well-defined extensive quantity, independent from $N$ or $L$ in isolation):

$$n\equiv{N\over L^3}\,.$$

Its relation to $k_\mathrm{F}$ is thus:

$$n={k_\mathrm{F}^3\over 3\pi^2}\,.$$

The corresponding energy is known as the Fermi energy $E_\mathrm{F}\equiv\hbar^2 k_\mathrm{F}^2/(2m)$. The total energy of this electron gas is obtained, similarly, as:

$$E_\mathrm{tot}={L^3\over 2\pi^2}\left({2m\over\hbar^2}\right)^{3\over2}\int_0^{E_\mathrm{F}}E\sqrt{E}dE\,,$$

and since $\int_0^{E_\mathrm{F}}E\sqrt{E}dE={2\over 5}E_\mathrm{tot}^{5\over 2}$, we find:

$$E_\mathrm{tot}={E_\mathrm{F}^{5\over 2}L^3\over5\pi^2}\left({2m\over\hbar^2}\right)^{3\over 2}={\hbar^2k_\mathrm{F}^5\over10\pi^2}{L^3\over m}\,.$$

Calling $V$ the volume $V\equiv L^3$, one can compute how the energy of a fixed number of electrons in the gas changes with the volume, which relates to a pressure. For this we need to express the energy in terms of the $N$ and $V$:

$$E_\mathrm{tot}={\hbar^2(3\pi^2 N)^{5\over3}\over10\pi^2 m}{V^{-{2\over 3}}}$$

so that

$${dE_\mathrm{tot}\over dV}=-{2\over 3}{E_\mathrm{tot}\over V}\,.$$

Now moving to the solid, a particular simple model for matter is the crystal, and the simplest description of the crystal just take into account Fermi statistics. The electron gas is confined: if one reduces its volume, its energy increases, i.e., a pressure is exerted against this. Alternatively, the system tries to reduce its energy by increasing the available volume. This is known as as the electron degenracy pressure. Note, in particular, that it has nothing to do with Coulomb repulsion between the charges, and is only due to the fermionic repulsion. One cannot compress all the electrons together because of this exclusion, which is for a gas the counterpart of the atomic growth with the number of electrons.

Still at low temperature, we now turn to the Bose gas, in which case, instead of Eq. (3), we now have (we now assume only one state of spin, so the factor 4 is back):

$$N={L^3\over4\pi^2}\left({2m\over\hbar^2}\right)^{3\over2}\int_0^\infty{\sqrt{E}\over e^{\beta (E-\mu)}-1}\,dE\,.\tag{3}$$

The probability distribution has to remain positive, which implies that $e^{\beta (E-\mu)}>1$ which in turns demand that $\mu\le 0$ (for positive energies). The closed-form expressions are again cumbersome when they exist, so this could be solved numerically. But consider instead one solution, i.e., we have $\mu(\beta)$ for a given temperature. Now we decrease the temperature, i.e., increase $\beta$, therefore, to keep the integral the same (equal to the number of particles), we must compensate with a decrease of $E-\mu$, or, since $\mu$ is negative, to bring $\mu$ closer (from below) to zero. In all cases, $\mu$ must remain negative, or zero. Let us therefore consider what happens when it is zero. We compute:

$$N\equiv{L^3\over4\pi^2}\left({2m\over\hbar^2}\right)^{3\over2}\int_0^\infty{\sqrt{E}\over e^{\beta E}-1}\,dE\,.\tag{4}$$

This time the integral can be solved and is found to be:

$$N\equiv{L^3\over4\pi^2}\left({2m\over\hbar^2}\right)^{3\over2}{\sqrt{\pi}\zeta({3\over 2})\over 2\beta^{3\over 2}}$$

in term of the Riemann zeta function $\zeta (s)\equiv\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots$ so that $\zeta(3/2)\approx 2.61238$. So for a fixed $N$ or density $n$, we can lower the temperature down to:

$$k_\mathrm{B}T_\mathrm{C}\equiv{2\pi\hbar^2\over\lambda_\mathrm{C}^2m}\approx 3.3125{n_{\mu=0}^{2/3}\hbar^2\over m}$$

where $T_\mathrm{C}$ is the so-called critical temperature, for which we also introduced the critical wavelength

$$\lambda_\mathrm{C}\equiv\sqrt[3]{\zeta({3\over2})\over n}\,.$$

If we further lower the temperature, then there is no solution for Eq. (3): whatever $\mu$ we chose (and $\mu=0$ was our most extreme case accommodating the most particles from $N$ into $N_{\mu=0}$), the integral spills over. Einstein understood that in such a case, what happens is that bosons would do what they are good at: they would simply start to superimpose together to accumulate into the ground state (state of lowest energy). They can do that. Therefore, the excess particles that cannot fit the excited state of the thermal distribution go into a separate, condensed state (so-called Bose-Einstein condensate or "BEC") with population $N_{\mathrm{BEC}}$ such that:

$$N=N(T\le T_\mathrm{C})+N_{\mathrm{BEC}}$$

where $N$ is the fixed, initial number of particles we have, $N(T\le T_\mathrm{C})$ is the expression (4) with $\mu$ clamped to zero and $N_{\mathrm{BEC}}$ is the difference between these, i.e.,

$$N_{\mathrm{BEC}}=N-L^3\left({mk_\mathrm{B}T\over\hbar^2}\right)^{3\over2}{\zeta({3\over 2})\over2\sqrt{2}\pi^{3\over2}}$$

where the $T$ here is a temperature below $T_\mathrm{C}$ (otherwise $\mu>0$) or, diviving both side by $N$:

$${N_{\mathrm{BEC}}\over N}=1-\left(T\over T_\mathrm{C}\right)^{3/2}\,.$$

For $T>T_\mathrm{C}$, then all particles can be accommodated by the Bose-Einstein distribution and $N_\mathrm{BEC}=0$.

One sees that at the critical temperature, bosons start to accumulate in a single state until, at $T=0$, all of them are there. Interestingly, one can check that in 2D, the arguments above show that BEC should not be possible as there is always enough room for bosons to fit in the Bose-Einstein distribution. BEC can also exist in 2D but as a result of more refined models.

This constitutes, in any dimensions, a new phase of matter, and an important one as corresponding to a macroscopic quantum state, where all particles are described by the same wavefunction (that of the ground state). This was postulated by Einstein who knew of De Broglie's hypothesis of matter waves, so the idea that all atoms would condense into a wave-looking single state looked plausible to Einstein. This is how, reading Einstein's paper on the statistics of bosons, Schrödinger learned about De Broglie's idea, which prompted him to look for a wave equation, that became Schrödinger's equation. So we have, at an historical level, the birth of modern quantum theory rooted in such arguments.

Despite its importance and central role, this phase of matter was realized in 1995 only, with cold atoms. One can see that the mass, in the denominator, makes the critical temperature very small for possible densities. This improved in the solid-state where excitons (electron-hole pairs, with the correct Bose statistics from adding two spin 1/2) and/or polaritons (excitons further coupled to light), with a weight of the order of the electron mass as opposed to atomic ones, made it possible to realize such quantum states at much higher temperatures. The fact that one can describe a gas of such excitons as if they were free particles in vacuum as opposed to excitations in a crystal is a leading idea of condensed matter: such complicated objects behave, in many respects, as independent particles indeed. In fact, the free electron gas above better describes electrons in a solid (this is known as the free electron model or Sommerfeld, or Drude-Sommerfeld model), because their Coulomb repulsion can be assumed to be canceled by the atomic crystal itself.