Past exam of the theoretical physics I course of the University of Cambridge / 2019 / 3

Under $\psi \mapsto e^{i\alpha }\psi$ with constant real $\alpha$, the field and its complex conjugate acquire opposite phases. These cancel in $|\psi {|}^2$, $|\psi {|}^4$, $\nabla {\psi }^{\ast }\cdot \nabla \psi$, and ${\psi }^{\ast }\dot{\psi }-\psi {\dot{\psi }}^{\ast }$. The Lagrangian density therefore has the particle-number phase symmetry, a global circle group $U(1)$.

For a spacetime-dependent real $\alpha (\mathbf{r},t)$,Writing $n=|\psi {|}^2$, the change in the Lagrangian density isThis is not generally zero or a boundary term. Thus the symmetry is global, with no arbitrary local phase symmetry. A compensating gauge field and gauge-covariant derivative of a charged scalar field would be required to promote it to a local redundancy.

Treat $\psi ,{\psi }^{\ast }$ independently in the Legendre transform in mechanics. Their canonical momenta areThe time-derivative terms cancel in ${\pi }_{\psi }\dot{\psi }+{\pi }_{{\psi }^{\ast }}{\dot{\psi }}^{\ast }-\mathcal{L}$, giving the Hamiltonian densityThe first-order time dependence makes the momenta constrained; this direct calculation still gives the canonical density. In physical terms this is the grand-canonical Hamiltonian of a Bose field, $K=H-\mu N$. The physical energy density is obtained by adding $\mu |\psi {|}^2$ back. A stationary field in this chemical-potential frame has the usual factor $e^{-i\mu t/\hslash }$ in the physical frame.

Complete the square:For uniform or periodic boundary conditions, or in the homogeneous bulk, both nonconstant terms are nonnegative and vanish simultaneously atThe ground states form a circle. A particular nonzero ${\psi }_0$ is changed by every nontrivial phase rotation, although the full family is invariant. Selecting one such mean-field theory state therefore exhibits spontaneous symmetry breaking of $U(1)$. This describes phase selection in the thermodynamic limit; an exact finite-system particle-number eigenstate need not itself have a nonzero field expectation.

Choose the real representative $v={\psi }_0=\sqrt{\mu /g}$. For $\psi =v+\chi$,and the fourth power through quadratic order isIn $\mu |\psi {|}^2-g|\psi {|}^4/2$, the linear terms and the $|\chi {|}^2$ terms cancel because $g{v}^2=\mu$. The remaining potential contribution is ${\mu }^2/(2g)-\mu (\chi +{\chi }^{\ast }{)}^2/2$.

The time-derivative contribution isIts first term is a total derivative and contributes only a temporal boundary term for fixed endpoint fields. Since $v$ is uniform, the gradient contribution is simply $-{\hslash }^2|\nabla \chi {|}^2/(2M)$. Dropping that boundary term and the constant, the quadratic condensate fluctuation Lagrangian isThe omitted cubic and quartic interactions describe scattering between these small fluctuations; they do not enter the linear normal mode calculation.

For independent $\chi ,{\chi }^{\ast }$, the derivatives needed in the ${\chi }^{\ast }$ Euler-Lagrange field equation areConsequently the two conjugate field equations areWrite $\chi ={\chi }_1+i{\chi }_2$ with real components, and put $A={\hslash }^2/(2M)$. Separating real and imaginary parts givesFor each real field take the real part of a complex plane wave amplitude, ${\chi }_j=\mathrm{Re}[q_j{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. With ${ϵ}_k={\hslash }^2|\mathbf{k}{|}^2/(2M)$ the two equations becomeA nonzero normal mode requires their determinant to vanish, soThe positive-frequency Bogoliubov quasiparticle dispersion is thereforeThere is also the negative-frequency conjugate solution, as required for real component fields. Since $\mu >0$, ${\omega }^2\ge 0$, consistent with the stability of the chosen uniform ground state.

At long wavelength,This is the sound-like Goldstone mode of the broken global $U(1)$ continuous symmetry. A spatially uniform infinitesimal phase change is $\delta \psi =i{\psi }_0\delta \theta$, entirely in the ${\chi }_2$ direction, and costs no potential energy. Slow spatial variations acquire gradient energy and propagate as sound. Thus the dispersion relation realizes Goldstone's theorem through a gapless excitation; it is not a massive phase oscillation.

The amplitude and phase are canonically paired by the first-order time derivative in the quadratic condensate fluctuation Lagrangian. There is one positive-frequency branch, rather than an additional independent amplitude branch of a relativistic second-order-time theory. At high momentum,so the same branch becomes particle-like. The healing length $\xi =\hslash /\sqrt{2M\mu }$ sets the crossover scale.