Past exam of the condensed matter physics course of the University of Oxford / 2013 / 2

Put $C=\alpha q^2/(4\pi {ϵ}_0)$ with $\alpha =2\ln 2$. Relative to a chosen positive ion, successive neighbours have alternating charge. The sum of its interactions with all other ions is $-C/R$. Summing this over all $2N$ ions would count every pair twice, so the total electrostatic energy is $-NC/R$, or $-C/R$ per ion pair.

There are $2N$ nearest-neighbour bonds in a periodic chain of $N$ pairs; boundary corrections are negligible for large $N$. If $A/R^n$ is the energy of each such bond, the repulsive contribution per ion pair is $2A/R^n$. Therefore the alternating-charge chain cohesive energy isThe factor two in the short-range term counts two bonds per ion pair. If a repulsion parameter were instead defined as the whole pair's repulsive coefficient, it would equal $2A$ in this convention; changing that definition must not silently change the equilibrium formula. The absolute zero of energy is separated ions for a genuinely decaying repulsion.

For a repulsive short-range potential and a stable finite minimum, assume $A>0$ and integer $n>1$. Setting $u^{\prime }(R)=C/R^2-2nA/R^{n+1}=0$ givesAt this point $2A/R_0^n=C/(n{R}_0)$, soHere $u^{\prime \prime }$ denotes the second derivative of the energy $u$, not a displacement variable. The negative equilibrium energy is the cohesive energy per pair relative to infinite separation.

The qualifiers on $A,n$ matter: an arbitrary real coefficient and integer exponent do not automatically describe stabilizing repulsion. For instance $A<0,n>1$ gives collapse as $R\to 0$, whereas $n=1$ gives $u=(2A-C)/R$ and has no isolated finite equilibrium. The displayed minimum therefore applies to the physically repulsive, faster-than-Coulomb branch of the stated model.

For a one-dimensional free-electron gas of length $L$, allowed wave vectors have spacing $2\pi /L$. Including two spin states and occupied wave vectors from $-k_F$ to $+k_F$,The ground-state kinetic energy is found by integrating rather than assigning every electron the Fermi energy:Charge neutrality gives $N_e=ZN$ because each positive ion has charge $+Ze$. Thus $K_e=ZN{E}_F/3$. Since the ion separation is $2R$, $L\simeq 2RN$ and $n_e=Z/(2R)$. These are one-dimensional number densities and state counts; using a three-dimensional $k_F\propto n^{1/3}$ relation here would change the model. The results concern the ground-state Fermi-gas contribution, neglecting electron interactions beyond the supplied electrostatic energy.

Use the supplied effective electrostatic term and enforce charge neutrality while varying the chain length: $R=Z/(2n_e)$, with $N$ and $Z$ fixed. The total energy per positive ion becomesThe positive quadratic term supplies degeneracy pressure against compression. Differentiating gives the neutral one-dimensional metal equilibrium density,where $a_B=4\pi {ϵ}_0{\hslash }^2/(m_e{e}^2)$ is the Bohr radius. The second derivative $Z{\pi }^2{\hslash }^2/(12m_e)>0$ verifies a minimum. Within this specified toy energy, the minimizing electron density is independent of $Z$; instead the equilibrium ion spacing is $2R^{\ast }=Z/n_e^{\ast }$.

At the minimum the electrostatic contribution is minus twice the kinetic contribution, so ${\epsilon }_{\ast }=-Z{E}_F^{\ast }/3$. These follow from the given electrostatic approximation, not from a full self-consistent calculation of a real one-dimensional metal. In particular, additional Coulomb terms cannot be appended without the corresponding neutral-background subtraction that defines the supplied finite energy.