Quantum matter course of the University of Oxford (MMathPhys/MSc paper)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a1853381.pdf

TODO: establish the year coverage of the lecturer’s prior-exam collection: www-thphys.physics.ox.ac.uk/people/SteveSimon/QCM2025/quantummatter.html.

The Hartree-Fock method approximates the ground state wavefunction by a single Slater determinant, minimizing the expectation value of the Hamiltonian over its occupied orbitals. It retains fermionic exchange symmetry exactly within that determinant but omits electron correlation beyond a single Slater determinant.

The periodic boundary condition requires $e^{ikM}=1$, so there are exactly $M$ independent wavevectors in a Brillouin zone:The inverse discrete Fourier transform, with the sign convention of the paper, isFourier orthogonality gives the tight binding dispersion relation ${ϵ}_k=-2t\cos k$. For the interaction, first use the canonical anticommutation relations at distinct sites to write $n_{\sigma ,j}{n}_{\sigma ,j+1}=c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}^{†}{c}_{\sigma ,j+1}{c}_{\sigma ,j}$. Summing its discrete Fourier transform over $j$ enforces crystal momentum conservation, givingEvery wavevector is reduced modulo $2\pi$. The phase $e^{iq}$ comes from the displacement of the second site by one lattice spacing. Equivalently replace $e^{iq}$ by $\cos q$: exchanging the two incoming and the two outgoing fermions changes $q$ to $-q$ with two cancelling fermionic signs. Thus the usual interaction vertex is $2V\cos q$ accompanied by the factor $1/(2M)$.

For each spin let ${\mathcal{F}}_{\sigma }$ contain the $N_{\sigma }$ lowest tight binding energy levels, and set $n_{\sigma }=N_{\sigma }/M$ and $C_{\sigma }=M^{-1}{\sum }_{k\in {\mathcal{F}}_{\sigma }}{e}^{ik}$. The normalized trial Slater determinant isIts nearest-neighbour one-body density matrix is $⟨c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}⟩=C_{\sigma }$. By Wick's theorem, fermionic exchange symmetry subtracts the squared off-diagonal one-body density matrix from the direct contributionThis formula also specifies the finite-$M$ answer exactly, including partially filled terminal energy levels through the chosen occupied set. For $N_{\sigma }$ consecutive allowed wavevectors, $|C_{\sigma }|=|\sin (\pi N_{\sigma }/M)|/[M\sin (\pi /M)]$; its phase locates the centre of that occupied interval.

Write $\tau =|t|$. In the large-$M$ limit the Fermi sea is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with Fermi wavevector $k_{F,\sigma }=\pi n_{\sigma }$ measured from $k_0$. ConsequentlyThe subtraction due to fermionic exchange symmetry is essential: replacing $⟨n_j{n}_{j+1}⟩$ by $n^2$ would discard the suppression of neighbouring equal-spin occupation. A completely filled spin band has zero kinetic energy and interaction energy $MV$, as the formula also shows.

At $N=M$, the two spin occupations per site are $x$ and $1-x$. With $\tau =|t|>0$, the trial energy per site isIts derivative is $e^{\prime }(x)=-4\tau \cos (\pi x)+V[4x-2-(2/\pi )\sin (2\pi x)]$. Since $x=0$ is a boundary point, its local minimum test is one-sided: $e^{\prime }(0)=-4\tau -2V$. At $V=-2\tau$, where this derivative vanishes, its Taylor series begins $e(x)-e(0)=(2{\pi }^2\tau /3)x^3+O(x^4)>0$ for small positive $x$. HenceThe same applies to $x=1$ by spin interchange.

The symmetry $e(x)=e(1-x)$ makes $x=1/2$ a stationary point for every $V$. Here $e^{\prime \prime }(1/2)=4\pi \tau +8V$, so the second-derivative test settles all cases except equality. Writing $y=x-1/2$, at $V=-\pi \tau /2$ the Taylor series is $e(1/2+y)-e(1/2)=({\pi }^3\tau /6)y^4+O(y^6)$. ThusThe minimizing branches interpolate continuously. To find all of them, put $u=\pi (1/2-x)$ on $0<x<1/2$. An interior stationary point obeysThe right side decreases strictly from $-\pi /2$ to $-2$, sinceThe positivity follows from $u-\sin u\cos u={\int }_0^{u}2{\sin }^{2}vdv$. There is therefore exactly one such stationary point on each side of $1/2$ when $-2<V/\tau <-\pi /2$. The signs of $e^{\prime }$ show that both are global minima, whereas $1/2$ is then a local maximum. The complete answer isThe pair merges through a supercritical pitchfork bifurcation at $V/\tau =-\pi /2$. Comparing only $e(0)$ and $e(1/2)$ would miss the intervening ground states with partial spin polarization.

For $V\gg |t|$, the leading objective is to avoid adjacent electrons of the same spin. The unpolarized Fermi sea, although optimal within the preceding restricted family, has interaction energy per site $V(1/2-2/{\pi }^2)>0$. Its cost grows linearly with $V$, while states with no forbidden neighbouring pairs have zero interaction energy.

There are two independent spinless fermion t-V chains, one for each spin. On an even ring, avoiding all neighbouring pairs allows at most $M/2$ particles in each chain. Since the total is $M$, both bounds must be saturated, and each chain must alternate occupied and empty sites. Their two independent choices give four limiting occupation patterns:The first two are Néel states with one electron per site; the last two are charge density waves alternating doubly occupied and empty sites. Opposite-spin double occupation costs nothing in this particular Hamiltonian. Each species has alternating occupation in all four patterns.

These are the exact zero-interaction ground states at $t=0$, and the limiting ordered configurations as $V/|t|\to \infty$. At large finite $V/|t|$, virtual hopping admixes configurations with forbidden pairs, producing an energy correction of order $-M|t{|}^2/V$. A finite ring can have superpositions of the ordered configurations, with possible quantum tunnelling splittings; four exactly degenerate product ground states are not implied at finite hopping. The four ordered choices describe the strong-coupling thermodynamic limit, with spontaneous symmetry breaking understood.

Measure single-particle energy from the quantum harmonic oscillator ground state energy $\hslash \omega$, and write $\tilde{\mu }={\mu }_{\mathrm{physical}}-\hslash \omega \le 0$. The Bose-Einstein distribution and the supplied density of states giveThis excited-state population increases with the chemical potential and has the finite saturation valueAny additional particles must occupy the ground state: this is two-dimensional harmonic-trap Bose condensation. The critical temperature and the condensate fraction are thereforeHere $N$ is the total particle number, the integral of the inhomogeneous number density over the trap; there is no fixed uniform area in which $N$ itself is a number density. Equivalently $N\kappa ={\pi }^{2}m(k_B{T}_c{)}^2/(6{\hslash }^2)$ relates the trap population to the critical temperature. Harmonic confinement changes the low-energy density of states and permits this saturation in two spatial dimensions.

These formulas are leading large-$N$ results with $k_B{T}_c\gg \hslash \omega$. A sharp transition belongs to the trapped-gas thermodynamic limit $N\to \infty$, $\omega \to 0$ at fixed $N{\omega }^2$; a finite trap has finite-size rounding. In particular, finite $N_0$ requires $\tilde{\mu }=-k_{B}T\log (1+1/N_0)<0$, approaching zero in the condensation limit.

Write $G(r)=⟨{\psi }^{†}(0)\psi (\mathbf{r})⟩$, $r=|\mathbf{r}|$, and ${\ell }^2=\hslash /(m\omega )$. At zero temperature all $N$ particles occupy ${\varphi }_0(\mathbf{r})=(\ell \sqrt{\pi }{)}^{-1}{e}^{-r^2/(2{\ell }^2)}$. Expanding the bosonic annihilation operator in single-particle orbitals givesThis one-body density matrix is a Gaussian function of width $\ell$; it contains one wavefunction evaluated at $r$, so its exponent is half that of the spatial number density $N|{\varphi }_0(r){|}^2$. In particular,the total number density at the trap centre, including condensed and noncondensed particles.

At positive temperature, with occupations ${\overline{n}}_a$ of the trap energy eigenstates, the exact ideal-gas decomposition isFor $\hslash \omega \ll k_{B}T<k_B{T}_c$, the sketch has a broad Bose-Einstein condensate component on scale $\ell$ and a narrower thermal peak near the origin. On approaching $T_c$ the broad component loses its weight $N_0$, while the thermal part remains. Above $T_c$ there is no macroscopic broad Bose-Einstein condensate component; close to $T_c$, low energy levels can still give a shoulder, so a single narrow Gaussian function is only a high-temperature description. At $k_{B}T\lesssim \hslash \omega$ individual excited energy levels matter and the two-scale description need not apply.

For $T\gg T_c$ the Maxwell-Boltzmann distribution applies. The leading off-diagonal one-body density matrix isHere ${\sigma }_r$ is the Gaussian function width defined by $e^{-r^2/(2{\sigma }_r^2)}$, and ${\lambda }_T$ is the thermal de Broglie wavelength. The thermal cloud itself extends over the much larger spatial scale $\sqrt{k_{B}T/\kappa }$; this is a correlation width, not the cloud radius.

For a normalized orbital $\int d^{2}r|\varphi {|}^2=1$, the symmetric Hartree product state and its equivalent Fock space form areThere are exactly $N(N-1)/2$ interacting pairs, so the expectation value of the grand Hamiltonian is the functionalAt fixed $N$ the term $-\mu N$ is constant. Enforce wavefunction normalization separately by varying ${\mathcal{E}}_{\mu }-N\eta (\int |\varphi {|}^2-1)$ with respect to ${\varphi }^{\ast }$. The resulting stationary Gross–Pitaevskii equation for the normalized orbital isThe Lagrange multiplier $\eta$ enforces the norm; fixed-$N$ variation cannot identify it with the independently inserted chemical potential merely by differentiating the constant $-\mu N$.

For the standard large-$N$ convention put $\Psi =\sqrt{N}\varphi$, so $n=|\Psi {|}^2$ and $\int n=N$, and replace $N-1$ by $N$. Varying the resulting grand energy functionalwith unrestricted $\Psi$ givesEquivalently, at fixed $N$ this same $\mu$ is chosen as the normalization Lagrange multiplier. This is the chemical potential convention used below. The contact interaction coupling $g$ has dimensions of energy times area in two dimensions.

In the large-$N$ Gross–Pitaevskii equation, dropping the kinetic energy gives $[\kappa r^2/2+gn(r)-\mu ]\Psi (r)=0$. The physical nonnegative Thomas–Fermi approximation for a condensate is thereforewhere $(z{)}_{+}=\max (z,0)$ and $R$ is the cloud radius. Its particle number isThus the two-dimensional Thomas–Fermi condensate obeysHere $N=\int n{d}^{2}r$ is a total particle number. Keeping the exact pair count in the fixed-$N$ Hartree product state would replace $g$ by $g(1-1/N)$ in these formulas expressed in terms of $n=N|\varphi {|}^2$.

The approximation is best at the centre, $r=0$, and worst near the edge, $r=R$. For example its omitted quantum pressure of a condensate, evaluated on the interior profile, isIt is smallest at the centre and diverges at the artificial sharp edge. The central healing length ${\xi }_0=\hslash /\sqrt{2m\mu }$ is much smaller than $R$ when $\mu \gg \hslash \omega$, since $R/{\xi }_0=2\mu /(\hslash \omega )$. This is the bulk validity condition; the full Gross–Pitaevskii equation smooths the edge and gives a decaying tail outside $R$.

The Hartree product state underlying the Gross–Pitaevskii equation has the factorized one-body coherence of a Hartree condensate. Choosing its nodeless ground state orbital real givesIn particular, the Thomas–Fermi approximation for a condensate from part (d) predictsThe curve starts at $\mu /g$, decreases with zero slope at the origin, and reaches zero with an infinite slope at $R$. Restoring the kinetic energy in the full Gross–Pitaevskii equation rounds that edge and produces a nonzero tail. This corrects the Thomas–Fermi approximation for a condensate, but the resulting one-body density matrix still factorizes.

The additional missing physics is quantum depletion. Interactions scatter pairs of bosons into orbitals outside the Bose-Einstein condensate even at zero temperature. In terms of the true Bose-Einstein condensate orbital ${\phi }_0$ the exact one-body density matrix decomposes asThe single-orbital Hartree product state sets $N_0=N$ and omits ${\rho }_{\mathrm{ex}}$ altogether. In a weakly interacting Bose gas, Bogoliubov theory adds a comparatively short-range noncondensed contribution near the origin, together with reduced weight in the broad Bose-Einstein condensate component. The central value remains the total number density, not just the condensate density. Its numerical shift relative to the original Gross–Pitaevskii equation solution also depends on the interaction-induced redistribution of number density; it is not fixed by the condensate fraction alone.

Source PDF: mmathphys.web.ox.ac.uk/sitefiles/quantum-matter.pdf

TODO: establish the year coverage of the lecturer's prior-exam solutions collection: www-thphys.physics.ox.ac.uk/people/SteveSimon/QCM2025/quantummatter.html.

Write $h_0=-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})$ for the one-particle Hamiltonian operator. The second quantization of the kinetic energy, external potential energy and repulsive contact interaction isThe factor $1/2$ counts each unordered particle pair once. Normal ordering excludes self-interaction: the interaction acts on two distinct bosons. The fixed particle number sector has $\hat{N}=\int d^{3}r{\hat{\psi }}^{†}\hat{\psi }=N$. In three dimensions, this contact interaction is understood as a regulated low-energy interaction with a matched coupling, as in renormalization of a three-dimensional dilute-gas contact coupling; the finite-mode calculation below uses its projected matrix elements.

The quantized field expansion in the complete orthonormal basis of orbitals isThe complex conjugation in the creation operator expansion is essential for complex orbitals. Indeed, ${\hat{a}}_n^{†}=\int d^{3}r{\varphi }_n(\mathbf{r}){\hat{\psi }}^{†}(\mathbf{r})$ creates the wavefunction ${\varphi }_n$. The resolution of the identity in an orthonormal basis givesrecovering the canonical commutation relation and the spatial Dirac delta function.

The two-mode approximation for a Bose gas retains the real localized orbitals:It is valid when the lowest doublet is well separated from higher energy levels, and the temperature and interaction-induced changes of the orbitals are small on that separation. More explicitly, if $\Delta$ is the energy gap to discarded orbitals, sufficient low-energy conditions are $k_{B}T,t,N{U}_{0}I\ll \Delta$, where $I=\int d^{3}r|{\phi }_L{|}^4$. A deep double-well potential suppresses interwell interaction overlaps. Large $N$ alone does not justify the truncation: increasing $N$ at fixed coupling can violate the interaction condition.

Let $ϵ=(E_0+E_1)/2$ and use the even and odd eigenfunctions to evaluate the projected one-particle Hamiltonian operator:ThusThe same quantum tunnelling matrix element is $t=-\int d^{3}r{\phi }_L{h}_0{\phi }_R$. Reflection symmetry makes $\int |{\phi }_L{|}^4=\int |{\phi }_R{|}^4=I$. Neglecting mixed contact interaction integrals leavesThe last step is the normal-ordered boson pair count, following from the canonical commutation relation. Consequently the two-site Bose-Hubbard model has precisely the requested coefficientsandHere $\mu$ is the coefficient obtained by rewriting the fixed-particle number Hamiltonian operator, rather than an independently prescribed thermodynamic chemical potential. Adding a reservoir term $-{\mu }_{\mathrm{res}}\hat{N}$ would replace it by $\mu +{\mu }_{\mathrm{res}}$.

The negligible-overlap instruction applies to the mixed contact interaction integrals while the small quantum tunnelling matrix element is retained. Exactly disjoint smooth orbitals for a local $h_0$ would also give $t=0$; setting every overlap-related matrix element identically to zero would discard the physical splitting one is trying to describe.

The normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

The printed identity has the wrong sign for the specified relative phase. Differentiating the energy found above and evaluating the operator commutator giveThus the phase derivative of a two-mode condensate energy isA concrete counterexample to the printed sign is $N=100$, $\theta =\pi /4$, $\chi =\pi /2$ and $t>0$: its left side is $100t$, whereas its printed right side is $-100t$, independently of $w$ and $\mu$.

There is also a direct derivation without inserting the Hamiltonian operator. Since $|\theta ,\chi ;N⟩=e^{i\chi {\hat{n}}_R}|\theta ,0;N⟩$, differentiating this unitary operator gives ${\partial }_{\chi }E=i⟨[\hat{H},{\hat{n}}_R]⟩$. Conservation of the total particle number, $[\hat{H},{\hat{n}}_L+{\hat{n}}_R]=0$, then supplies the minus sign for the left well.

By the Heisenberg equation of motion, $\hslash d⟨{\hat{n}}_L⟩/dt=i⟨[\hat{H},{\hat{n}}_L]⟩$. Define the bosonic Josephson junction particle current from left to right as $J_{L\to R}=-d⟨{\hat{n}}_L⟩/dt$. The corrected identity meansThe energy slope with respect to the relative phase drives particle transfer; an extremum with respect to that phase has zero instantaneous particle current. The printed sign would work with $e^{-i\chi }$ in the right-well orbital, or with $[{\hat{n}}_L,\hat{H}]$ in place of $[\hat{H},{\hat{n}}_L]$.

For a uniform continuum gas with pair interactions depending on relative positions, a Galilean boost changes every particle momentum by $p_0\hat{\mathbf{x}}$ and leaves the interaction energy unchanged. Write ${\hat{P}}_x={\sum }_j{\hat{p}}_{j,x}$. The transformed many-body Hamiltonian operator isThe resting ground state has $⟨{\hat{P}}_x⟩=0$, so the Galilean boost energy of an interacting gas isThe mass here is the bare particle mass, because this is an exact transformation of the microscopic kinetic energy. Substituting the Fermi-liquid effective mass $m^{\ast }$ would omit the change in interaction energy encoded by the Landau quasiparticle energy functional. This conclusion requires the Principle of Galilean relativity; a gas in a lattice or a fixed external potential energy landscape need not obey it.

Let $A$ be the real-space area and use ${\sum }_{\mathbf{p}}\to C\int d^{2}p$, with $C=A/(2\pi \hslash {)}^2$. There is one state per momentum cell, since the fermions are spinless. An annulus between $p$ and $p+dp$ contains $C2\pi pdp$ states. Near the Fermi surface, the quasiparticle dispersion relation has $dϵ/dp=v_F^{\ast }=p_F/m^{\ast }$, so the total density of states isThe density of states per unit area and the occupied-state count are, respectively,In particular, $D^{\ast }(\mu )=2N{m}^{\ast }/p_F^2$. No factor of two for spin degeneracy is present. This counts Landau quasiparticle levels, so it is not multiplied by the quasiparticle residue of the microscopic particle Green's function.

The shifted Fermi circle satisfies $|\mathbf{p}-p_0\hat{\mathbf{x}}|=p_F$. Its radial boundary at polar angle $\theta$ is the positive solution of $p^2-2p{p}_0\cos \theta +p_0^2=p_F^2$. Thus, for $p_0<p_F$,and its leading Taylor expansion isForward-pointing momenta gain occupied states and backward-pointing momenta lose them. With the paper's explicitly reversed Heaviside step function convention, ${\Theta }^{\prime }(x)=-\delta (x)$; consequentlywith a positive leading sign.

For the later quadratic energy calculation it is useful to distinguish the first-order shape from exact particle number conservation. The particle-conserving displacement of a Fermi circle hasThe second-order inward correction cancels the apparent area increase one would obtain by treating $\nu =p_0\cos \theta$ as exact.

Within a stable normal Fermi liquid, the low-energy states are described by long-lived Landau quasiparticles and holes in a thin shell about the Fermi surface; the Pauli exclusion principle suppresses their decay. Expanding the energy about the equilibrium occupations gives the one-quasiparticle dispersion relation at first order and the interaction between occupation changes at second order, namely the Landau quasiparticle energy functional. For a smooth interaction, radial deviations in that interaction may be neglected at leading order, leaving its angular dependence; higher occupation terms contribute higher powers of the small shell thickness. Collective occupation deformations are included in the same description. The sufficiency claim is conditional on the normal Fermi liquid regime, and on supplementing this static energy functional with dynamics when required: it is not a description of arbitrary low-energy phases such as a superconductor or non-Fermi liquid.

Use the total density of states $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ and $C=A/(2\pi \hslash {)}^2$ from part (b). In the paper's double-sum convention, $f(\varphi )$ has units of energy and scales as $A^{-1}$. Define its first angular moment byThis convention matters: the Landau quasiparticle energy functional written with an explicit $1/A$ instead uses the intensive kernel $\tilde{f}=Af$.

At zero temperature, the displayed functional is the change in $E-\mu N$; a true Galilean boost preserves $N$, so $\Delta \mathcal{F}=\Delta E$. For a radial displacement $\nu (\theta )$, the first term in the Landau quasiparticle energy functional becomesThe oriented radial integral also works for negative $\nu$: removing occupied levels below the chemical potential costs positive energy. Substituting $\nu =p_0\cos \theta +O(p_0^2/p_F)$ and using $N=C\pi p_F^2$ givesThe exact second-order correction to the boundary is unnecessary here because the unperturbed excitation energy vanishes on the Fermi surface; it changes this term only beyond quadratic order. It must nevertheless be understood when identifying $\Delta \mathcal{F}$ with a fixed-particle number energy change.