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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?

Qmechanic
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Timothy
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1 Answers1

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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.

Adam
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