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I am reading Girvin's Modern Condensed Matter Physics and I have a question about the off-diagonal long-range order (ODLRO). In chapter 18.1, he starts from \begin{align} \rho(\vec{r},\vec{r}') = \left\langle \Psi^{\dagger}(\vec{r}) \Psi(\vec{r}') \right\rangle \end{align} where $\Psi^{\dagger}(\vec{r})$ is the bosonic field operator at $\vec{r}$. He then said that since $\rho(\vec{r},\vec{r}')$ is Hermitian we are allowed to do this decomposition \begin{align} \rho(\vec{r},\vec{r}') = \sum_{i} n_{i} \chi(\vec{r}) \chi^{*}(\vec{r}') \end{align} where $\chi(\vec{r})$ is an eigenfunction of $\rho(\vec{r},\vec{r}')$ satisfying \begin{align} \int d\vec{r}' \rho(\vec{r},\vec{r}') \chi(\vec{r}') = n_{i} \chi(\vec{r}). \end{align} We interpret $n_{i}$ as the occupation number of the eigenfunction $\chi(\vec{r})$. I am totally fine with these since these are all mathematically rigorous. Now we try to do some physical interpretations. When one $n_{i} \sim O(N)$ where $N$ is the scale of the total particle number, we say that this $\rho(\vec{r},\vec{r}')$ will signal an ODLRO \begin{align} \rho(\vec{r},\vec{r}') \sim n_{i} \chi(\vec{r}) \chi^{*}(\vec{r}') \end{align} such that \begin{align} \lim_{|\vec{r}-\vec{r}'|\to \infty }\rho(\vec{r},\vec{r}') \neq 0. \end{align} and in the book it is said that $ \chi(\vec{r})$ is the condensate wave function. I am not totally sure what does it mean by condensate wave function.

Let's say we have a non-interacting Bose gas. Then I will say that this condensate wave function is probably just the single particle eigenstate where nearly all the bosons condense into. However, when we have a (weakly) interacting Bose system, the concept of a single particle eigenstate breaks down since we can't even solve for a single particle eigenstate. Yet, the decomposition of $\rho(\vec{r},\vec{r}')$ should nevertheless work for both non-interacting and interacting Bose system.

My questions are then

  • Is there any physical meaning of the condensate wave function for an interacting Bose system, besides the mathematical fact that the condensate wave function must be an eigenfunction of $\rho(\vec{r},\vec{r}')$?

  • Can we compute this condensate wave function?

  • If yes, which method can we use?

  • Or are we just saying that, there is a condensate wave function $\chi(\vec{r})$, and we will guess one to explain the experimental results of a superfluid?

I know that we can use this condensate wave function to do a lot of meaningful calculation, such as calculating the superfluid velocity field and show that superfluid is an irrotational fluid, but I am not entirely sure what exactly is a condensate wave function. Maybe some suggestions on the questions listed above can help me understand this more.

Thanks!

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$\rho(\vec r, \vec r')$ is the one-body reduced density matrix. If you know that quantity, then you can calculate the expectation value of any 1-body operator. For one-body operators it looks like the density matrix that would be prepared by placing $n_i$ particles in the wave function $\chi_i(\vec r)$, where in your expression you have a typo where you dropped the $i$ subscript. In chemistry the $\chi_i(\vec r) are called natural orbitals. In any case, the single particle state with the macroscopic occupation is often called the condensate wave function.

For example in liquid $^4$He, the ground-state condensate fraction is around 10 percent, and in the bulk the "wave function" is just a constant since it is the momentum zero plane-wave state. If you remove a He atom from some place and put it back in another place, the probability of remaining in the ground state is $\rho(\vec r,\vec r')$ and you could, in principle, measure $\rho(\vec r,\vec r')$ this way.

If you know the ground-state wave function, $\Psi_0(\vec r_1,...,\vec r_N)$ for the state for $N$ identical particles, then you can calculate $\rho(\vec r,\vec r')$ from $\int d^3r_2,...,d^3r_N \Psi^*(\vec r',\vec r_2,...,\vec r_N)\Psi(\vec r,\vec r_2,...\vec r_N)$ which is just the first quantized version of your expectation value of $\langle\psi^\dagger(\vec r')\vec \psi(\vec r)\rangle$. For example for $^4$He, this is readily done exactly for the ground state using quantum Monte Carlo methods.

Easier systems to see this stuff are cold atom experiments. There, the atoms can be cooled to Bose condense in a trap. The interaction can be tuned with a magnetic field. If the interaction is turned off and the trap released, the atom cloud will expand and the atom density at later times can be measuted to give the condensate density directly. Quantum Monte Carlo and other methods can also be used to calculate the one-body density matrix and the theoretical condensate fraction and wave function.

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The first thing that you need to understand is that a superfluid and a Bose-Einstein condensate (BEC) are two different phases: the former is driven by interactions, the latter is driven by statistics.

  • A superfluid is a phase matter where, for flow velocities less than a critical value $v_c$, there is no viscosity. This effect is solely driven by the dispersion relation, i.e. the $\omega$ vs $k$ plot, which usually shows a straight line up to some $k_c$ so that $v_c = \partial \omega / \partial k \,\vert_{k<k_c}$. A straight line means that there cannot possibly be velocities $< v_c$ and hence why there cannot be dissipation (like in a superconductor for resistance).
    Within the Bogoliubov approximation, the dispersion relationship is given by the Bogolibuov dispersion relationship: $$ E(p) = \sqrt{ \frac{g n}{m}p^2 + \left ( \frac{p^2}{2m} \right )^2}, $$ where $g$ quantifies the interaction strength. You can see that for $g=0$ (non-interacting case), the dispersion relation is just the single particle one with $v \propto \partial E / \partial p \propto p $ and hence $v \rightarrow 0$ as $p\rightarrow 0$: there is no (finite) critical velocity $v_c$ such that $v \not \leq v_c$. You can also deduce that the value of the critical velocity goes as the interaction strength $v_c \propto g$. So, if you want, a non-interacting superfluid (which happens to be a fit description for a BEC, but not its defining feature) is a "boring" superfluid, that is one with critical velocity $v_c = 0$.

  • A Bose-Einstein Condensate is a purely non-interacting phenomenon, driven solely by quantum statistics. It triggers the "condensation" of a macroscopic number of particles to the single-particle ground state. Macroscopic here means on the order of the total number of particles. Its definition is when the off-diagonal one-body density matrix $$ n^{(1)}(\mathbf{r}, \mathbf{r}') = \sum_i n_i\, \psi_i^\ast(\mathbf{r}) \psi_i(\mathbf{r}') $$ tends to a constant for large separations: $$ \lim_{|\mathbf{r} - \mathbf{r}'| \rightarrow \infty} n^{(1)}(\mathbf{r}, \mathbf{r}')\rightarrow n_0 \neq 0, $$ which implies phase coherence, only possible if only one energy eigenstate is present (or at least dominant). This is because each energy eigenstate has a phase factor that goes as $\exp(-\mathrm{i} E t/\hbar)$ for each energy $E$, so having multiple ones would lead to destructive interference and decoherence.

You will notice that the above definition is also the definition Off-Diagonal Long-Range Order (ODLRO). Which is really what you get when having a BEC.

Now:
Can you have a BEC without superfluidity?
Yes. Superfluidity is triggered by interactions, and a BEC is by definition a non-interacting phenomenon so it doesn't even need interactions to be considered. It's a "superfluid" if you want but with $v_c = 0$.

Can you have a superfluid without BEC?
Yes. For example, the BKT phase is a supefluid phase that occurs in 2D, where BEC is forbidden by the Mermin-Wagner theorem. The superfluidity is allowed by the energy dispersion relation of vortex/anti-vortex pairs. It has $v_c \neq 0$ but no ODLRO.

Can you have a superfluid and BEC?
BEC by definition is non-interacting. And it means occupation of a single-particle state and ODLRO. If your superfluid has a state that is macroscopically occupied, and it happens to be the "lowest energy one" (ground state - as it usually is), people usually "extend" the term BEC to mean that as well. An "interacting BEC" would have $v_c \neq 0$ but also ODLRO. Because of interactions, however, the Hamiltonian does not commute with the particle number operator so you cannot say that a single-particle state is macroscopically occupied. Usually you do a basis transformation and find a way to express in terms of at least one thing that is macroscopically occupied, and that's what you call your "condensate". This e.g. would be one of the two fluids in the Landay two-fluid model for superfluidity.

Your questions

Is there any physical meaning of the condensate wave function for an interacting Bose system, besides the mathematical fact that the condensate wave function must be an eigenfunction of $\rho(\vec{r},\vec{r}')$?

For an interacting system, $n_i$ as the number of particle is not defined anymore. So yes, there is a physical meaning of the condensate wavefunction but it's not the 'lowest single-particle state'. It's just a state (in a different basis, the one that diagonalises the interacting Hamiltonian) that is macroscopically occupied.

Can we compute this condensate wave function? If yes, which method can we use?

It depends on your theoretical/numerical approach. If you have the Hamiltonian, diagonalise it to find the eigenstates $v_k$. Then find the occupation of $v_0$, the lowest one. See its occupation compared to the others.

Or are we just saying that, there is a condensate wave function $\chi(\vec{r})$, and we will guess one to explain the experimental results of a superfluid?

Hopefully you can answer this from my long discussion above.

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  • $\begingroup$ What do you mean by 'A Bose-Einstein Condensate is a purely non-interacting phenomenon' exactly? You can have BEC in interacting as well as non-interacting systems. Further, the definition of ODLR only applies to homogeneous systems. A more general definition would be the Penrose-Onsager criterion. Moreover, the basis in which the 1-RDM is diagonal is in general not the basis in which the Hamiltonian is diagonal! $\endgroup$ Feb 27, 2021 at 11:40
  • $\begingroup$ You have to diagonalise the Hamiltonian and find its ground state and from that calculate the 1-RDM, diagonalise it and find its lowest eigenvalue and corresponding eigenvector. Note that the eigenvector of the 1-RDM is a single-particle state, while the eigenstates of the Hamiltonian are $N$ particle states, no?! $\endgroup$ Feb 27, 2021 at 11:53
  • $\begingroup$ Thanks for the reply. Could you elaborate more on "This effect is solely driven by the dispersion relation, i.e. the $\omega$ vs $k$ plot, which usually shows a straight line up to some $k_{c}$ so that $v_{c} = \partial{\omega}/\partial k |_{k < k_{c}}$. A straight line means that there cannot possibly be velocities $<v_{c}$ and hence why there cannot be dissipation (like in a superconductor for resistance)."? How does the $E(k)$ tell us the dissipation? Also I am not quite sure, so from $E(k)$ it seems like at low $k$ the $E(k)$ will be linear in $k$, why this mean anything about dissipation? $\endgroup$
    – ocf001497
    Feb 27, 2021 at 18:51
  • $\begingroup$ @Jakob At T=0 every bosonic system is in the many-body ground state. Is that a condensate? Is that a BEC? No; a BEC by definition is the (forced) macroscopic occupation of the lowest single-particle state. You do not need interactions to explain BEC. Though you could BEC survives with interactions, and introduce quantmu depletion. $\endgroup$
    – SuperCiocia
    Feb 27, 2021 at 23:25
  • $\begingroup$ @ocf001497 The $E$ vs $k$ plot just tells you what states are available to dissipate to. It's the same thing for superconductivity (which is just a charged superfluid). If there are no states with $v<v_c$, then the system has to keep whatever velocity it had at beginning, and does not slow down. For an interacting system, $v$ near $k=0$ is not going to be $0$ but a constant, whereas for a non-interacting system it reaches $0$. So, in the latter, you can always find states at lower $k$ that have less $v$, but for the interacting case you are going to keep the same $v$ even at lower $k$. $\endgroup$
    – SuperCiocia
    Feb 27, 2021 at 23:28

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