# Superfluid condensate wave function

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!