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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2 Answers
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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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4$\begingroup$ Thanks! Can you please explain in more detail why is $U(1)$ symmetry broken by the anomalous average $\langle\hat\psi (r)\rangle \neq 0$. $\endgroup$– TimothyCommented Oct 7, 2013 at 3:43
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$\begingroup$ 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. $\endgroup$– AdamCommented Oct 7, 2013 at 3:51
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2$\begingroup$ Can you please write down the Hamiltonian? $\endgroup$– TimothyCommented Oct 7, 2013 at 3:57
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$\begingroup$ You can have a look at wikipedia en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensate $\endgroup$– AdamCommented Oct 7, 2013 at 13:12
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1$\begingroup$ @Adam Adam, thanks for your answer. However, it is not obvious to me why $G_{AB}(r) = \langle(A(r)-\langle A\rangle)(B(0)-\langle B\rangle)\rangle\rightarrow 0$ if $r\rightarrow0$. I thought long range order means $G_{AB}\neq 0$ previously. $\endgroup$– BlueCommented Jul 25, 2015 at 21:05
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@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 $\rho(x',x)$?.
answered Feb 3, 2022 at 14:42
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$\begingroup$ If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review $\endgroup$ Commented Feb 3, 2022 at 15:11