Past exam of the quantum matter course of the University of Oxford / 2023 / 1

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 $[\hat{H},{\hat{n}}_L]$.