Interpretation of the Bose-Einstein Condensate Wavefunction squared and Density Matrix

Question

I'm currently trying to get some background information about the theoretical treatment of Bose-Einstein Condensate, and I'm reading through this paper on the topic by J. Rogel Salazar. I have also been reading through Bose-Einstein Condensation by L. Pitaevskii, S. Stringari, as well as Quantum Liquids by A. Legget.

In each of these sources, the authors use the Bose-Einstein Field operator, $\hat{\Psi}(r)$, which they then use to derive the single-particle density matrix $$n^{(1)}(\mathbf{r}, \mathbf{r'}) = \langle \hat{\Psi}(\mathbf{r}), \hat{\Psi}(\mathbf{r'})\rangle$$ although they never explain quite exactly;y what this means or the interpretation of this is. In Legget it is roughly defined, as a ratio of probability amplitudes but, I'm not really sure, the ratio of what amplitudes.

Now, in Pitaevskii and Stringari's treatment, they are able to separate the field operator into the following terms via the Bougloibov Ansatz:

$$\hat{\Psi}(\mathbf{r}) = \Psi_{0}(\mathbf{r}) + \delta \hat{\Psi}(\mathbf{r})$$
Where $\Psi_{0}(\mathbf{r})$ is known as the condensate wavefunction. I believe that the condensate wavefunction (from my previous readings into the subject) that the condensate wavefunction represents a single particle state for a specific boson in the condensate. While the overall state of the $N$ particle condensate is given by the tensor product of the $N$ states $\Psi_0$.

In every textbook that I've read and every source that I've consumed, they note that the condensate wavefunction squared is normalized to have $\int |\Psi_{0}(\mathbf{r})|^2 = N$. More pertinently the quantity $|\Psi_{0}(\mathbf{r})|^2$ gives the number density of the condensate. I'm still not sure why this is true, where this comes from, and what the difference between the number density and the density matrix is. Is there a nice connection between the interpretation of the regular wavefunction norm squared as the probability density to find the particle to the condensate wavefunction squared being the number density?

Answer 1

The field operator is normalised to the number of particles, $$ \int d^3r \; \langle \Psi^\dagger(\vec r) \Psi(\vec r) \rangle = N $$ because we have a $N$-particle state. This is just like in classical physics: If you integrate the (particle) density $n=\frac{dN}{dV}$ over a volume $V$, you get the number of particles inside that volume, $\int dV \; n = N$.

Bogoliubov realised that if we have many particles in the ground state the quantum correlations inside the ground state are negligible. Hence, he treated the condensate as a classical field $\Psi_0$, by arguing that the BEC wave function is not altered, if a single particle is added or removed. This leads to the perturbation ansatz you wrote in your question, $$ \Psi = \Psi_0 + \delta \Psi$$

In order to understand the difference between a wave function and the density matrix I would advice you to take an other look into an introductory quantum mechanics textbook. Personally, I don't feel it is worth putting to much effort into understanding the theory which goes beyond the mean-field ansatz. If you need this later on and you like to write your master thesis of dissertation about this kind of stuff, you will have to understand it. However, till then there are plenty of interesting phenomena, which are explained using simpler models.

Answer 2

Some notes on things I believe you are misinterpreting:

In BECs, the wavefunctions (or filed operators,a s you have called them) do not represent single particle states, quite on the contrary. For BECs the wave function is macroscopic and phase coherent, describing not the probability density (as in the wave function of a single quantum particle) but the spatial distribution of the BEC at some instant. It's the difference between describing an atom and a cloud of atoms...

Where Ψ0(𝐫) is known as the condensate wavefunction. I believe that the condensate wavefunction (from my previous readings into the subject) that the condensate wavefunction represents a single particle state for a specific boson in the condensate. While the overall state of the 𝑁 particle condensate is given by the tensor product of the 𝑁 states Ψ0.

NO! the state of the N-particle condensate cannot be properly descibed by a tensor product of single-particle states (even after symmetrisation). A new wave function must be built, that is macroscopic because quantum phenomena in BECs are macroscopic. That is why sometimes you might hear, when talking about BECs, the mysterious term huge matter waves! This new wave function is a solution of the Gross-Pitaevskii equation (check Stringari-Pitaevskii for a derivation and further explanation)

Bogoliubov approximation is a mean-filed one. Essentially he considers that

1. the filed operators (essentially boson creation and annihilation operators) can be separated into a condensate and non-condensate component
2. In some regimes the condensate component can be treated as classical field and not an operator!