What is off-diagonal long range order in superfluid? From Wikipedia:

[...]Off-diagonal long-range order (ODLRO) [...] exists whenever there is a macroscopically large factored component (eigenvalue) in a reduced density matrix of any order.

How to understand the ODLRO in superfluidity? 
 A: The one-body density matrix is defined by
$\rho(r,r')=\langle \hat\psi^\dagger (r) \hat\psi (r')\rangle$. 
ODLRO is equivalent to say that $\lim_{|r-r'|\to \infty}  \rho(r,r') \neq 0$ and in the case of (bosonic) superfluids this corresponds to 
$\lim_{|r-r'|\to \infty}  \rho(r,r')=\langle \hat\psi^\dagger (r) \rangle\langle\hat\psi (r')\rangle$.
You can see that the $U(1)$ symmetry $\hat\psi (r)\to e^{i\theta} \hat\psi (r)$ is then spontaneously broken by the anomalous average $\langle\hat\psi (r)\rangle \neq 0$.
In a homogeneous system $\langle\hat\psi (r)\rangle$ is independent of $r$ by translation symmetry and $n_0=\langle \hat\psi^\dagger (r) \rangle\langle\hat\psi (r')\rangle$ defines the condensate density of the Bose-Einstein condensate.
A: @Adam. I am studying this ODLRO and just wanted to ask a question about your notation and to get a handle of the density matrices and reduced density matrices. Is it correct that by one-body density matrix you mean the density matrix of a pure state so that $\rho = |\psi><\psi|$ and the average being over the reduced density matrix? From this and Yang's definition (https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.34.694) (I would love to know a textbook that went more in depth with the use of these density matrices and reduced ones) we would get:
$$ <x|\rho_1|x'> = \text{Tr}\left(\hat{\Psi}(x)\rho\hat{\Psi}^{\dagger}(x')\right)  = \sum_n < n |\hat{\Psi}(x)\rho\hat{\Psi}^{\dagger}(x') |n> .$$
Where n denotes a complete set. So on rearranging and putting in $\rho$ we'll get:
$$= <\psi|\hat{\Psi}^{\dagger}(x')\left(\sum_n |n>< n |\right)\hat{\Psi}(x)|\psi>,$$
using $\sum_n |n>< n | = 1$, we get:
$$= <\hat{\Psi}^{\dagger}(x')\hat{\Psi}(x)>_{\psi},$$
with the average being a quantum mechanical average over the pure state $\psi$ and not a thermal average (although I guess it could easily be by letting $\rho = \sum_n e^{-\beta(\epsilon_n-\mu)}|n><n|$?). And then you define this as your $\rho(x',x)$?.
