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I am currently reading trough "Bose-Einstein Condensation and Superfluidity" by Pitaevksii and Stringari and noticed some inconsistencies in my reasoning. In Chapter 5 (Non-uniform Bose gases at zero temperature) the authors introduce the condensate wave function $\Psi$. It is futher stated that the normalization of $\Psi$ is given by $N = \int d\vec{r} |\Psi(\vec{r})|^2$, where N is the total number of atoms in the condensate. Up until this point, I think of $\Psi$ as a probability density, as I have been doing when dealing with Quantum Mechanics for the past few years.

The following sentence then really confuses me:

The modulus $|\Psi(\vec{r})|$ determines the particle density $n(\vec{r}) = |\Psi(\vec{r})|^2$ of the condensate.

My question is: How can something that describes a probability density be a quantity that represents a particle density?

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    $\begingroup$ On the one hand, $n(r) = |\Psi(r)|^2$ treats $\Psi(r)$ as a function of a single variable, yet you're also talking about a multi-particle system of $N$ identical bosons, thus a wave function $\Psi(r)$ of a single variable should be completely irrelevant in general (when $N > 1$), so why are they writing such a thing? Because $\Psi(r)$ is actually a second quantized quantum field operator and $N = \int \Psi^{\dagger} \Psi dV$ is an operator acting on multi-particle states representing some number of identical particles in some volume, the integrand thus an operator giving the particle density. $\endgroup$
    – bolbteppa
    Commented Jan 16, 2022 at 21:11
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    $\begingroup$ When one applies these general 'second quantization' principles to BEC's, there are simplifications one can then make, such as treating the ground state as made from commuting operators due to the large number of particles and so one can again treat it as a literal single particle wave function, but it will look very confusing to normalize it against $N$ or even apply this single particle wave function to a multi-particle system without knowing it's origins in second quantization. See sections 25-26 of Landau 'Statistical Physics' part 2 $\endgroup$
    – bolbteppa
    Commented Jan 16, 2022 at 21:36
  • $\begingroup$ Ok so maybe my misunderstanding lies deeper. In our lecture, we introduced a general correlation function $g(\vec{r}, \vec{r}') = \langle \Psi^{\dagger}(\vec{r}) \Psi (\vec{r'})\rangle$. We then said that for $r=r'$, this reduces to the diagonal density $N(r, r')$ of the system. I see that if one applies the Bogoliubov approximation, this diagonal density is then equal to $|\psi(r)|^2$. So maybe I don't understand how one goes from a probabilistic interpretation of the correlation function to a density interpretation. $\endgroup$
    – LionCereals
    Commented Jan 17, 2022 at 7:48

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The following sentence then really confuses me:

The modulus $|\Psi(\vec{r})|$ determines the particle density $n(\vec{r}) = |\Psi(\vec{r})|^2$ of the condensate.

My question is: How can something that describes a probability density be a quantity that represents a particle density?

For an N-particle system, the particle density operator is: $$ \hat n(\vec r) \equiv \sum_i^N \delta(\vec r - \hat{\vec r_i} ) $$

The expectation value of this is: $$ n(\vec r) = \langle\Phi|\hat n(\vec r)|\Phi\rangle\;, $$ where $\Phi(\vec r_1, \ldots, \vec r_N)$ is the N-body wave function.

The expression for $n$ can be re-written as: $$ n(\vec r) = \sum_i^N\int d^3r_1\ldots d^3r_N |\Phi(\vec r_1, \ldots, \vec r_N)|^2\delta(\vec r - \vec r_i)\;. $$ Due to the symmetry or antisymmetry of the many-body wave fucntion, this can be written more simply as: $$ n(\vec r) = N\int d^3r_2\ldots d^3r_N |\Phi(\vec r, \vec r_2, \ldots, \vec r_N)|^2 $$

You can choose to define: $$ |\Psi(\vec r)|^2 = N\int d^3r_2\ldots d^3r_N |\Phi(\vec r, \vec r_2, \ldots, \vec r_N)|^2 $$

That is all that $|\Psi(\vec r)|^2$ means.

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The wave function describing a BEC is in nature quite different to that of a single quantum particle.

Usually in quantum mechanics you have that $|\psi(t,x)|^2$ is the probability density of the particle being around $x$ at time $t$.

However in BECs the description is different, since quantum phenomena are now apparent macroscopically. This requires us to use a different kind of wave function, describing the condensate as a whole. So in this context, $|\Psi(t,x)|^2$ is the spatial distribution of the BEC cloud at time $t$.

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  • $\begingroup$ To make sure it get this right: Here, $\Psi$ has no connection whatsoever to a probability density? Isn't it defined (in the non-interacting case, Hartree-Fock) as a product of single-particle wavefunctions and as such has to be a probability density? $\endgroup$
    – LionCereals
    Commented Jan 16, 2022 at 17:22
  • $\begingroup$ of course it has a probabilistic nature, everything in quantum mechanics does. Think about what a probability density (from this wave function describing the entire gas) means for a gas $\endgroup$
    – Henrique Calazans Prates
    Commented Jan 16, 2022 at 17:25
  • $\begingroup$ I would have said that $\frac{1}{N}|\Psi(\vec{r})|^2d\vec{r}$ gives the probability of finding a particle in a volume element $\vec{r} + d\vec{r}$. Still, I don't see how that is a physical density. $\endgroup$
    – LionCereals
    Commented Jan 16, 2022 at 17:28
  • $\begingroup$ Doesn't a probability density just tell you the probability of finding a particle being present at an infinitesimal element of space surrounding any given point? $\endgroup$
    – franz
    Commented Jan 16, 2022 at 17:58
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    $\begingroup$ NO. The condensate has a spatial distribution, it is not all condensed in one point of space. $|\Psi|^2$ gives you the way the cloud of bosons making up the BEC are distributed in space (of course this is also probabilistic) $\endgroup$
    – Henrique Calazans Prates
    Commented Jan 16, 2022 at 19:35
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How can something that describes a probability density be a quantity that represents a particle density?

That is open for extensive qm interpretations, and your question is related to "what is the physical property of a qm object before its measured".

I know that in quantum chemistry, that a working assumption is that the probability density also is the particle density, and if relevant also the charge density. This gives a good picture of how things look in average in results, but the assumption is also used in midway calculations - contradicting the Copenhagen interpretation that the physical properties arises at measurement.

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