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This is a translation into math of the following question, posted on PhysicsOverflow.

Let $H:=L^2(\mathbb C)$.
For every $N$, let $\psi_N\in\Lambda^N H\cong (L^2(\mathbb C^N))^{S_N}$ be the function $$ \psi_N(z_1,\ldots,z_N):=c\cdot\prod_{i<j}(z_i-z_j)^3\prod_i e^{-|z_i|^2/4}, $$ normalized so as to have $\|\psi_N\|^2=1$.
(Note that I could have replaced the $3$ above by any other odd integer — those are also called `Laughlin wave functions'.)

Let $D\subset \mathbb C$ be a unit disc, not necessarily centered at the origin. Let $H_{in}=L^2(D)$ and $H_{out}=L^2(\mathbb C\setminus D)$, so that $H=H_{in}\oplus H_{out}$. We then have a decomposition $$ \Lambda^N H=\bigoplus_{k=0}^N (\Lambda^kH_{in})\otimes (\Lambda^{N-k}H_{out}) $$ Write $\psi_{N,k}$ for the $k$-th summand of $\psi_N$ in the above direct sum decomposition.

Question 1: For any given $k$, is the limit $$ \lim_{N\to \infty} \|\psi_{N,k}\|^2 $$ (assuming it exists) independent of $D$?

Definition: Given a unit vector $\zeta=\sum_i\xi_i\otimes \eta_i\in H\otimes K$ in the tensor product of two Hilbert spaces, the `reduced density matrix' is the operator $[\zeta]_H:H\to H$ given by $$[\zeta]_H(x):= \sum_i \langle x,\xi_i\rangle \|\eta_i\|^2 \xi_i.$$ Equivalently, $[\zeta]_H$ is the partial trace of the rank one projection onto the line spanned by $\zeta$. (Sub-question: what's the correct math terminology and notation for $[\zeta]_H$?)

Let us from now on abbreviate $[-]_{H_{in}}$ by $[-]_{in}$.

Question 2 (which implies Question 1): For any given $k$, is the limit $$ \lim_{N\to \infty} [\psi_{N,k}]_{in} $$ (assuming it exists) independent of $D$?

Here, to make sense of Question 2, we need to identify the Hilbert spaces $H_{in}=L^2(D_1)$ and $H_{in}=L^2(D_2)$ associated to any two discs $D_1,D_2\subset \mathbb C$. This is done by the `magnetic translation' operator, which sends $f\in L^2(D_1)$ to $f_\tau\in L^2(D_2)$ defined by $$f_\tau(z):=e^{\mathrm{Im}(\bar \tau z)}f(z-\tau)$$ where $D_2=D_1+\tau$. (and I may be missing some minus sign or $2\pi$ in the exponent.)

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  • $\begingroup$ the physics answer is that $|\psi|^2=\exp[-\beta H(z_1,z_2,\ldots z_N)]$ has the interpretation of a Gibbs distribution of $N$ classical particles in a disc with a parabolic confining potential, interacting with a logarithmic $-\ln|z_i-z_j|$ interaction, at inverse temperature $\beta=q$, where in your case $q=3$. So the question boils down to asking whether this classical statistical mechanical system is a liquid, hence translational invariant, or a crystal without translational invariance. The answer needs numerical calculations: the liquid exists for $q<70$, for larger $q$ it's a crystal. $\endgroup$ Nov 24, 2014 at 21:56
  • $\begingroup$ the point I want to make is this: since for some exponents $q$ in $|z_i-z_j|^q$ this $\psi_N$ is translationally invariant and for other exponents $q$ it is not, I don't think you can expect a proof (without numerical calculations) that will tell you the system is translationally invariant for $q=3$. $\endgroup$ Nov 24, 2014 at 22:10
  • $\begingroup$ The Laughlin wave function is certainly (and visibly) rotationally invariant, and I've never heard of rotationally invariant crystals... so I'm a bit confused by what you mean by "crystal". Could you maybe elaborate? $\endgroup$ Nov 25, 2014 at 14:20
  • $\begingroup$ the "crystal" refers to the marginal distribution $N^{-1}\sum_{n}P(|z_n|)$, the probability to find a particle at distance $r=|z|$ from the origin of the potential well; in the "liquid" phase this distribution decays smoothly from the origin, and is constant on the scale $1/\sqrt N$ of the average interparticle separation. In the "crystal" phase this distribution has peaks on this scale: the first particle will prefer to set at the center of the well, and then the next particle will be at some fixed radial separation from it, and so on. $\endgroup$ Nov 25, 2014 at 14:49
  • $\begingroup$ figure 3 of this reference arxiv.org/pdf/0709.2320v1.pdf shows the electron density in the liquid phase, so for $\nu=3$; the bulk density is uniform, but in the crystal phase, $\nu>70$, the oscillations persist in the bulk. $\endgroup$ Nov 25, 2014 at 14:55

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