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We have a system of two bosons particles and we are interested in calculating the one-particle density and two-particle-density when both are in different states.

So, to do that, I consider the following:

First, we know that the exchange of any two identical bosons must be symmetric, therefore the that wave function for a two-particle bosons is given by:

$$ \psi^s\left(\vec{r_1},\vec{r_2} \right) = \frac{1}{2}\left[ \psi_a\left(\vec{r_1}\right) \psi_b\left(\vec{r_2}\right) +\psi_a\left(\vec{r_2}\right) \psi_b\left(\vec{r_1}\right) \right] $$

Here is my question: If the system is composed of two bosons how can I calculate the one-particle density, Which density does it refer to? ... $\\$

For two- particles:

$$ \left| \psi^s\left(\vec{r_1},\vec{r_2} \right) \right|^2 $$

In many books, I found the concepts of density operator and density probability? Are they the same?

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    $\begingroup$ Your chosen ansatz for the two boson state is in the form of Slater determinant, except without minus signs. Slater determinants cannot capture correlations, and so cannot that form. i.e. the true wavefunction actually isn't given by that form, even though it is the best that we can do using single particle approximations. The normalisation is not half, but rather its sqrt. $\endgroup$ Commented Apr 17, 2023 at 18:20

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Assuming your single particle states are orthonormal and $a$ and $b$ are different, the prefactor for $\psi^s$ should be $\frac{1}{\sqrt{2}}$ for a normalized state.

The density matrix is the general quantum form for the case where your system can be in a mixed state. That is, it can be thought of as describing an ensemble of systems, each in a pure quantum state with a given probability. For a pure system, like you describe, you can use either normal expectation values or traces over the density matrix and the operator. The diagonal elements of the one-body reduced density matrix in position space will be the one-body density.

Just using expectation values is the simplest way to get the density for your problem. The position probability density is the expectation value of the one-particle density operator. For a system with $N$ particles that would be \begin{equation} \hat \rho (\vec r) = \sum_{i=1}^N \delta^3(\vec r- \hat{\vec r_i}) \end{equation} where $\vec r$ is the position you measure the density, and $\hat{\vec r_i}$ are the position operators. That is, the density operator when integrated over a volume, should count the number of particles in the volume.

The one-particle density is for your case the expectation value \begin{equation} \langle \hat \rho(\vec r)\rangle = \int d^3r_1 d^3r_2 \left [ \delta^3(\vec r-\vec r_1)+\delta^3(\vec r-\vec r_2)\right ] |\psi^s(\vec r_1,\vec r_2)|^2 = |\psi_a(\vec r)|^2+|\psi_b(\vec r)|^2 \end{equation}

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