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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?

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?