20Ge 131Chapter ThreeBeyond the Fermi Gas; Application to Hydrogen & More ComplexMaterialsClearly there is something that makes the material more compressible than ifit were just a free electron gas. That something also provides stability (aFermi gas with no attractive forces always has positive pressure and sowants to expand no matter what the density). The competing effect (the“something else”) comes from Coulomb forces, but is affected by thedictates of quantum mechanics (exchange and correlation effects).Coulomb Effects of the NucleiIn reality, there is no uniform background positive charge. Rather this chargeis in the form of roughly uniformly spaced nuclei of charge Z. The problemof the total electrostatic energy of a distribution of charges is difficult tosolve because of the long-ranged nature of the Coulomb force, but we canemploy a trick which works very well..... place about each nucleus a sphereof electronic charge, uniformly distributed, such that the entity as a whole isneutral. The radisu if this sphere is Z1/3rs in atomic units. Then think of theentire system as an ensemble of these “atoms”. The error incurred has to dowith the overlapping regions of these spheres, and is small in a close packedsystem. The electrostatic energy of one of these “atoms” is the sum of theterm arising from the attraction between the electron cloud and the nucleusand the repulsive interaction of the electrons with themselves. (This is allpurely classical electrostatics). The electron-nucleus term is:Eel−nucl= −Zr0Z13rs∫.143rs3.4 r2dr = −3Z532rs (atomic units)[Note: The atomic unit of energy is 27.2 eV = 2 Rydbergs]. The repulsiveinteraction between the electrons is the classical resultEel−el=3Z25.(Z13rs)=3Z535rs21(Note: The factor of 3/5 is exactly the same as you get for the gravitationalenergy of a uniform density sphere, where the result is –3GM2/5R). To getthe energy per electron, we must divide by Z, and to get the energy inRydbergs we must multiply by 2. So we find thatECoulomb= −1.8Z23rsIf this calculation is done for a specific lattice then instead of 1.8 you getsomething else. It is about 1.76 for simple cubic, 1.79.. for fcc (face-centeredcubic) and bcc (body centered cubic) and hcp (hexagonal close packed).This dimensionless constant is often referred to as the Madelung constant.The Exchange EnergyBecause electrons are Fermions, the total wave function of the system mustbe antisymmetric (i.e. the wave function changes sign when two electronsare interchanged). What this does is prevent two electrons of the same spinfrom being close to each other. This has nothing to do with Coulombrepulsion, it is a purely quantum mechanical effect.To illustrate this, consider just a two electron wave function (where theelectrons have the same spin). Then by definition, antisymmetry of the wavefunction says thatΨ(r1,r2)= -Ψ(r2,r1)where r1 and r2 and the positions of electrons 1 and 2, respectively. But thatmeans the wavefunction is identically zero when r1 = r2 (i.e., the electronsare on top of each other). Since the probability is proportional to |Ψ|2, andsince the wavefunction is a smooth function (i.e., it will be small even whenr1 and r2 are merely similar), this is equivalent to “keeping the electronsapart”. But because this quantum effect keeps the electrons apart, we haveoverestimated the repulsive electron-electron interaction term in theCoulomb energy above. The correction is called the exchange energy and ithas the valueEexchange= −0.916rs Ryd/eln22The Total EnergyThere is one other term, even when the electron gas is uniform. This arisesfrom the fact that even electrons of opposite spin avoid each other, becauseof the Coulomb force. This is called a correlation energy for obvious reasons(the electron states are correlated to avoid each other). This is also a negativeenergy (because we can lower the energy by arranging to have the electronsavoid getting near each other). It goes to a finite value as the electronspacing parameter goes to zero (infinite density limit). And then there is anenergy that arises from the non-uniformity of the electron gas. This is calledband-structure energy, and it is small (bounded like the correlation energy),provided the density is sufficiently high. But for many purposes, it sufficesto retain just the Coulomb and exchange energies for determining pressure:E ≈2.21rs2−(1.8Z23+ 0.916)rsP = −dEdV=51.6rs5[1− (0.407Z23+ 0.207)rs]Notice now that here is a state corresponding to zero pressure. It is given by:rs=10.407Z23+ 0.207⇒ rs= 1.63 ( = 0.62g/cc) for hydrogenThe energy for this state (when computed as carefully as possible; i.e.,including correlation and band structure) is about -1.04 Rydbergs, so it isbound relative to hydrogen atoms (which lie at exactly –1.00 Rydberg). It isnot lower than the energy per proton of molecular hydrogen gas or liquid orsolid however, where the energy is about -1.15 Ryd. (It takes about 4 eV orso to bust up a hydrogen molecule). So hydrogen prefers to form moleculesat zero pressure. (This is not true of other first column elements in theperiodic table, i.e., Li, Na, K...)The Molecular-Metallic Transition in Hydrogen; A Traditional ViewWe have seen that hydrogen could exist in a metallic state (free-electron-likestate) analogous to alkali metals. Recall that hydrogen is in column one ofthe periodic table. But we know from everyday experience that