The conserved <number operator> is
$$
N=\int d^dx\,\Psi^\dagger\Psi=\int d^dx\,\rho.
$$
Substitution of $\Psi=\sqrt\rho e^{-i\theta}$ into the <Landau-Ginzburg theory> gives
$$
i\Psi^\dagger\dot\Psi=\rho\dot\theta+\frac i2\dot\rho,
\qquad
|\nabla\Psi|^2=\frac{(\nabla\rho)^2}{4\rho}+\rho(\nabla\theta)^2.
$$
The imaginary term is a <total derivative>, so
$$
\mathcal L=\rho\dot\theta-
\frac{(\nabla\rho)^2}{8m\rho}-
\frac{\rho}{2m}(\nabla\theta)^2-r\rho-\lambda\rho^2+\cdots.
$$
Thus $\rho$ is the <number density> and is the <canonical momentum> conjugate to $\theta$. The <canonical commutation relation> is $[\theta(\mathbf x),\rho(\mathbf y)]=i\delta^{(d)}(\mathbf x-\mathbf y)$. Consequently, for $\theta_0=\int d^dx\,\theta$ in volume $V$,
$$
[\theta_0,N]=iV,
\qquad
\Delta\theta_0\,\Delta N\geq\frac V2.
$$
Equivalently, the averaged phase $\bar\theta=\theta_0/V$ obeys $\Delta\bar\theta\,\Delta N\geq1/2$. This <number-phase conjugacy> means that a state of sharp $N$ has no sharp phase, whereas a phase-selected state exhibiting <spontaneous symmetry breaking> must superpose different number sectors. Such sectors become effectively degenerate in the <thermodynamic limit>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 337 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-337/1/a/solution