The difference between pure and mixed states is the difference in their density matrix structure.
For density matrix $\rho$ of mixed state the trace of $\rho^{2}$ should be less than 1. For pure state corresponding trace $Tr(\rho^{2}) = 1$.
But when I tried to check the Bell two-qubit state, i got:
$$ \rho = \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1\end{pmatrix}$$
$$ \rho^{2} = \frac{1}{4}\begin{pmatrix} 2 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 2 \end{pmatrix}$$
Trace of which is equal to 1.
As I understand, reduced density matrix is the right describing of bell states. But my matrix is not reduced.
Can you explain me how to find reduced matrix of bell state?
Best Answer
The reduced matrix is defined as the partial trace of the density matrix.
Let $A$, $B$ be finite dimensional Hilbert spaces, and let there be a $T$ such that $T \in$ $L(A \otimes B)$ (i.e., $T$ is a linear operator on $A \otimes B$), then the partial trace of T, represented as $\rm{Tr}_B [T]$ in $L(A)$, is defined by:
\begin{equation} \langle a | \rm{Tr}_B [T]| b \rangle = \sum_n \langle a | \langle n | T| n\rangle | b \rangle \end{equation}
where $| n \rangle$ is an orthonormal basis in $B$, and $|a\rangle, |b\rangle$ are vectors in $A$.
Finally, note that the reduced matrix isn't the correct way of describing a quantum state, it is just a way to describe it as seen by looking only at a subsystem. This usually involves ignoring part of the information of the state; therefore, the reduced density matrix of a pure state may be a mixed state. This is spectacular for the Bell states, as their reduced matrix is $\rm{Id}/2$, the most disordered state.