The XY-model on a 2-torus ($L_1,L_2$) has a lagrangian given by $$ L_{XY}[\theta] = \int d^2 x \frac{\chi}{2}\big{(}\dot{\theta}^2 - (\partial_x \theta)^2\big{)} $$ Fourier expanding $\theta$ as $$ \theta (\boldsymbol{x},t) = \theta_0(t) + \frac{2\pi}{L_1}m_1x_1 + \frac{2\pi}{L_2}m_2x_2 + \sum_{\boldsymbol{k}}\lambda_{\boldsymbol{k}}(t) e^{i\boldsymbol{k}\cdot \boldsymbol{x}} $$ Putting this into our Lagrangian, we get $$ L=\frac{\chi}{2}\left(L_{1} L_{2} \dot{\theta}_{0}^{2}-\frac{(2 \pi)^{2} L_{2}}{L_{1}} m_{1}^{2}-\frac{(2 \pi)^{2} L_{1}}{L_{2}} m_{2}^{2}+L_{1} L_{2} \sum_{k}\left(\left|\dot{\lambda}_{k}\right|^{2}-k^{2}\left|\lambda_{k}\right|^{2}\right)\right) $$ We see that we have a collection of oscillators ($\lambda_{\boldsymbol{k}}$), a particle on a circle ($\theta_0$), and two integers ($m_1,m_2$) which are the winding numbers for our $\theta$ field around the torus. The eigenenergies will be $$ E(m_i,n_k\equiv \lambda_k, N\equiv p_{\theta_0}) =\frac{1}{2 \chi L_{1} L_{2}} N^{2}+\frac{\chi(2 \pi)^{2} L_{2}}{2 L_{1}} m_{1}^{2}+\frac{\chi(2 \pi)^{2} L_{1}}{2 L_{2}} m_{2}^{2}+\sum_{k}|k| n_{k} $$ where I have used $N \in \mathbb{Z}$ to label the momentum conjugate to $\theta_0$ and $n_\boldsymbol{k} \in \mathbb{N}$ for the occupation number for the $\lambda_k$ harmonic oscillator.

My doubt is, in Xiao-gang Wen's book Quantum field theory of many body systems (Chapter 6, page 263), he mentions that the label $N$ is physically "the total number of bosons minus the number of bosons at equilibrium, namely $N = N_{tot}-N_0$." How do you arrive at that statement?

My understanding is that $N$ just labels the energy from the motion of all the rotors moving uniformly. If I think analogously to the phonon problem, that term is like the energy due to the center of mass motion of the whole crystal. I don't see how bosons have anything to do with it.

  • $\begingroup$ Any suggestions for editing the question such that it becomes more understandable is also appreciated. $\endgroup$ Jul 27, 2020 at 14:15
  • 1
    $\begingroup$ I think this might be more suitable for math.stackexchange. $\endgroup$
    – Wyuw
    Jul 29, 2020 at 20:53
  • $\begingroup$ Thanks for that suggestion. I was looking to understand the physical picture he had so I thought this might be the better place to do it. $\endgroup$ Jul 31, 2020 at 7:01


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.