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  Physical intuition for spatially constant motion in the XY-model in 2+1D

+ 1 like - 0 dislike
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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. 

asked Jul 8 in Theoretical Physics by Nandagopal Manoj (5 points) [ no revision ]

1 Answer

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The coherent state path integral for the partition function of a theory of bosons is an integral over fields $\phi(\tau,x),\overline{\phi}(\tau,x)$; these are the classical fields corresponding to the bosonic creation annihilation operators $b(x),b^\dagger (x)$:

$\text{tr}(e^{-\beta \hat{H}}) = \int d[\phi,\overline{\phi}] e^{-S[\phi,\overline{\phi}]}$

Now perform a change of variable in the path integral $\phi(x), \overline{\phi} (x) \rightarrow n(x),\theta(x)$ where $\phi = \sqrt{n} e^{\text{i}\theta}$. This is the "number phase" representation (see Altland and Simons, chapter 6). The change of variables is a canonical mapping i.e., it preserves the classical poisson bracket. [It has some pathologies at $n=0$ which we'll ignore]

So $n,\theta$ are canonically conjugate variables. Now remember that in a theory of bosons, $\hat{b}^\dagger(x) \hat{b}(x)$, or $\overline{\phi}(x) \phi(x)$  in the path integral, is the local particle density. But note too that $\overline{\phi(x)} \phi(x) = n(x)$; hence we can identify $n(x)$ with the local particle density. What you're calling $N$ is the zeroth spatial fourier component of $n(x)$, and what you're calling $\theta_0$ is the zeroth fourier component of $\theta(x)$. That $N,\theta_0$ are canonically conjugate follows from fact $n(x),\theta(x)$ are canonically conjugate. 

answered Aug 5 by Curt vK (40 points) [ revision history ]
edited Aug 6 by Curt vK

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