Past exam of the quantum matter course of the University of Oxford / 2023 (A15758W1)

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.

For the quadratic interaction term, the first-order occupation change alone suffices. The radial Dirac delta function integrals givePut $\varphi =\theta -{\theta }^{\prime }$. The remaining angular integral usesso only the first cosine component of the interaction survives. Since $D^{\ast }=2\pi C{m}^{\ast }$,Adding both contributions and matching the bare-mass result of part (a) yields the two-dimensional Landau effective-mass relation:Equivalently, using the density of states per unit area $g^{\ast }=D^{\ast }/A$,If dimensionless Landau parameters are defined through the Fourier series $D^{\ast }f(\varphi )=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, then $D^{\ast }{\overline{f}}_1=F_1/2$ andDefining the cosine expansion with an extra factor of two would instead absorb that factor into $F_1$; the angular-integral formula above fixes the convention unambiguously. The physical content is that interaction changes accompanying a moving Fermi sea compensate the renormalized quasiparticle velocity, so the total boost energy is governed by the bare mass. For an interaction independent of angle, ${\overline{f}}_1=0$ and this consistency condition requires $m^{\ast }=m$.