Is there a relation between the density matrix and the density of the position probability? Are they the same concept?

Question

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?

Answer 1

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}