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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?

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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.

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    $\begingroup$ Thanks! Can you please explain in more detail why is $U(1)$ symmetry broken by the anomalous average $\langle\hat\psi (r)\rangle \neq 0$. $\endgroup$ – Timothy Oct 7 '13 at 3:43
  • $\begingroup$ The Hamiltonian is invariant under the transformation $\hat \psi^{(\dagger)}\to e^{(-)i\theta} \hat \psi^{(\dagger)}$, so one would expect that all averages implying only annihilation operators should vanish $\langle \hat \psi\rangle=\langle \hat \psi^2\rangle=...=0$ by symmetry. Therefore, if $\langle \hat \psi\rangle\neq 0$, this mean that the symmetry is broken. Its called spontaneously, because the Hamiltonian does not break the symmetry, only the ground-state of the system. $\endgroup$ – Adam Oct 7 '13 at 3:51
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    $\begingroup$ Can you please write down the Hamiltonian? $\endgroup$ – Timothy Oct 7 '13 at 3:57
  • $\begingroup$ You can have a look at wikipedia en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensate $\endgroup$ – Adam Oct 7 '13 at 13:12
  • $\begingroup$ @Adam: ODLRO, as you say, implies a non-vanishing value of the one-particle density matrix as $r \rightarrow \infty$; the value it takes is the condensate density. So in a superfluid without Bose-Einstein condensation, there should be no ODLRO. $\endgroup$ – MaviPranav Oct 21 '13 at 17:39

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