Density Matrix Theory: An introduction II

We have now introduced the polarization vector and how to calculate its components. To see the relationship between it and the density we shall move as follows:
Remember that if we have  a pure state then we can assign a state vector to our system but if we have a mixture then we instead have to consider a statistical mixture. We can thus write the density matrix for a statistical mixture in the following manner:\chi
\begin{equation} \label{eq.1}
\rho = \sum_k W_k |\chi \rangle  \langle \chi | \hspace{10mm}  \text{eq.1}
\end{equation}
where \(W_k \) is the statistical weight. From this I hope one can see that a statistical mixture is a generalization of a pure state. Let's multiply the above equation by the pauli matrix \( \sigma_i \) to get:

\begin{equation} \label{eq.2}
 \rho \sigma_i =  \sum_k W_k |\chi \rangle  \langle \chi |  \sigma_i   \hspace{10mm}  \text{eq.2}
\end{equation}
We shall now take the trace of each term i.e the \(tr (\rho \sigma_i \) to get:
\begin{equation} \label{eq.3}
tr ( \rho \sigma_i ) = \sum_k W_k \langle \chi |  \sigma_i | \chi \rangle = P_i   \hspace{10mm}  \text{eq.3}
 \end{equation}

Now, the pauli matrices in combination with the identity matrix form a basis for a 4 dimensional space. Thus if we stick with two level systems, then we can write our density matrix as:
\begin{equation}
 \rho = a_o I + \sum_i a_i \sigma_i
\end{equation}
 We can the use the following easily derivable facts to find the \( a_o , a_i  \)constants:
 1) pauli spin algebra- \(\sigma_i \sigma_j = i \sum_i \epsilon_ {ijk} \sigma_k  + \delta_{ij}I\),
 2) the \(tr(\sigma_i) =0 \)
 3)\( tr (\rho ^2 )= \frac{(1+ P^2)}{2}\)

Using  2) and 3) in combination tells us that \(a_o \) is actually 1/2. Multiply our density matrix again by  \(sigma_i \)  and taking the trace and then using 1) and 2) gives us that
\begin{eqnarray}
 tr ( \rho \sigma_ i) &=& 2 \sum_i b_i \delta_{ij} \\
                              &=& 2 b_i
\end{eqnarray}
Now referring back to eq.3  we see that  \( b_i  = \frac{P_i}{2} \)
But now let's take a close look at eq.1 in conjuction with fact 3). In general eq.1 will be less than 1 since the statistical weights themselves are less than one. Which means that in general  \( \rho^2 \) for  a statistical mixture will be less than one while if we have a pure state then \(  \rho^2\) is equal to one. Now we finally have a mathematical way of distinguishing between a density matrix for a pure state and a density matrix for a statistical mixture.

In our discussion of the density matrix we considered states \( | \chi \rangle \) . These states were not necessarily orthonormal. But we could re -write the density matrix in terms of orthornormal states. That is to say we can first re-write the \( | \chi \rangle \) like this:
\begin{equation}
 |\chi  \rangle  =  \sum _n a_n | \psi \rangle
\end{equation}

So if we then re-write the density matrix (which is in general represent a mixture) we have that:
\begin{equation}
 \rho = \sum_k W_k a_k a_k^{*} | \psi \rangle \langle \psi |
\end{equation}

This last equation gives us a hint of a much more general formulation for the density matrix with the eventual goal of considering open quantum systems and the density matrix that may describe them.