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It is defined that a system of identical bosons is said to be condensed, if the largest eigenvalue of the first-order reduced density matrix (1-RDM) is of the order of the number of particles in the system. Check Phys. Rev. 104, 576 (1956) and chapter 2 of Many-Body Schrödinger Dynamics of Bose-Einstein Condensates.

For a given wavefunction $\psi(r_{1},...,r_{N};t)$ of $\mathcal{N}$ identical spinless bosons with spatial coordinates $r_{i}$, the first-order reduced density matrix (1-RDM) is defined as $$ \rho^{(1)}(r_{1},r^{'}_{1};t)=N\int\psi(r_{1},r_{2},...,r_{N};t)\psi^{*}(r^{'}_{1},r_{2},...,r_{N};t)dr_{2}dr_{3}...dr_{N}=\langle\psi(t)|\hat{\Psi}^{\dagger}(r{'}_{1})\hat{\Psi}(r_{1})|\psi(t)\rangle $$

where $\hat{\Psi}(r)$ is the bosonic field operator and the wavefunctions are properly normalized $\langle\psi(t)|\psi(t)\rangle=1$. The first-order reduced density matrix can be regarded as the kernal of the operator $$ \hat{\rho}^{(1)}=\mathcal{N}.{Tr}_{\mathcal{N}-1}[|\psi(t)\rangle\langle\psi(t)|] $$

What exactly are the 1-RDM and its eigenvalues and how do I get a physical picture of it?

What is the physical meaning of the above statement based on the definition of the 1-RDM?

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  • $\begingroup$ Note that for indistinguishable particles, the "first-order reduced density matrix" typically refers to the matrix $M$ with elements $M_{ij}=\langle a^\dagger_i a_j\rangle$. $\endgroup$ – Norbert Schuch Jan 26 '16 at 9:09
  • $\begingroup$ BTW, would help if you could focus your question more. As it stands, it is very broad (and not very clear). $\endgroup$ – Norbert Schuch Jan 26 '16 at 9:10
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    $\begingroup$ Closely related (?) question (though with no answer): physics.stackexchange.com/questions/110371 $\endgroup$ – Norbert Schuch Jan 26 '16 at 9:17
  • $\begingroup$ @NorbertSchuch I have edited the actual question.please check. $\endgroup$ – ss1729 Jan 27 '16 at 10:00
  • $\begingroup$ I'm afraid that while you have added definitions, the question as such is still quite unclear. $\endgroup$ – Norbert Schuch Jan 27 '16 at 10:11
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The 1st order density matrix, $$ \rho({\bf x}, {\bf x'}) = \langle \hat\psi^\dagger({\bf x'})\hat\psi({\bf x}) \rangle $$ is a Hermitian operator since $\rho({\bf x}, {\bf x'}) = \rho^*({\bf x'}, {\bf x})$ and its diagonal entries, $$ \rho({\bf x}, {\bf x}) = \langle \hat\psi^\dagger({\bf x})\hat\psi({\bf x}) \rangle = n({\bf x}) $$ give the number density of particles at location ${\bf x}$, such that $$ \int{d{\bf x} \;\rho({\bf x}, {\bf x})} \equiv \int{d{\bf x} \;n({\bf x})} = N $$ is the (average) total number of particles in the system. Its orthonormal eigenfunctions, defined by $$ \int{d{\bf x'} \rho({\bf x}, {\bf x'}) \phi_j({\bf x'})} = n_j \;\phi_j({\bf x})\\ \int{d{\bf x} \;\phi^*_j({\bf x})\phi_k({\bf x})} = \delta_{jk}\\ \sum_j{\phi^*_j({\bf x})\phi_j({\bf x'})} = \delta({\bf x} - {\bf x'}) $$ represent natural orbitals. The eigenvalues $$ n_j = \int{\int{d{\bf x}d{\bf x'} \;\phi^*_j({\bf x})\; \rho({\bf x}, {\bf x'})\; \phi_j({\bf x'})}} $$ satisfy $$ \sum_k{n_k} = \int{d{\bf x} \;n({\bf x})} = N $$ and represent the occupation numbers of the natural orbitals.

In fact, if $a^\dagger_j$, $a_j$ are creation and annihilation operators of these orbitals, $[a_j, a^\dagger_k]=\delta_{jk}$, the field operators are expanded as $$ \hat\psi({\bf x}) = \sum_j{\phi_j({\bf x})\;a_j} $$ and we have $$ \rho({\bf x}, {\bf x'}) = \sum_{jk}{\phi^*_j({\bf x'})\phi_k({\bf x}) \langle a^\dagger_j a_k \rangle} $$ and $$ \rho_{jk} \equiv \langle a^\dagger_j a_k \rangle = \int{\int{d{\bf x}d{\bf x'} \phi^*_k({\bf x}) \rho({\bf x}, {\bf x'}) \phi_j({\bf x'})}} = \rho_{jj}\delta_{jk}\\ \rho_{jj} \equiv \langle a^\dagger_j a_j \rangle = n_j $$ In the condensate state one orbital achieves a high occupation number $n_0 \sim N$, such that $$ N = n_0 + \sum_{j\neq 0}{n_j} $$ and the largest eigenvalue of $\rho({\bf x}, {\bf x'})$ becomes $$ \rho_{00} = n_0 $$

See for example these lecture notes: "Dynamics of Bose-Einstein Condensates in Trapped Atomic Gases at Finite Temperature"

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    $\begingroup$ Good answer. It might be worth adding (w.r.t. to BEC) that if the largest eigenvalue $\eta_0$ is close (or proportional) to $N$, this means that the mode $a_0$ is occupied by $\eta_0$ particles, which corresponds to Bose-Einstein condensation. $\endgroup$ – Norbert Schuch Jan 27 '16 at 11:23
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    $\begingroup$ @NorbertSchuch Oy! I actually had in mind to add that and got distracted at the last moment. Thanks for bringing it up. $\endgroup$ – udrv Jan 27 '16 at 13:09

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