so-called, Schmidt representation:\
**Theorem 1** *Schmidt decomposition:*\
*Every $|\psi_{AB}\rangle \in {\cal H}_{A} \otimes {\cal H}_{B}$ can be represented in an appropriately chosen basis as $$|\psi_{AB}\rangle = \sum _{i=1}^{M} a_i |e_i\rangle \otimes |f_i\rangle,
\label{eq:schmidt}$$ where $|e_i\rangle$ ($|f_i\rangle$) form a part of an orthonormal basis in ${\cal H}_{A}$ (${\cal H}_{B}$), $a_i > 0$, and $\sum_{i=1}^{M} a_i^2=1$, where $M\le dim{\cal H}_{A}, dim{\cal H}_{B}$*.\
The positive numbers $a_i$ are known as the Schmidt coefficients and the vectors ${\ensuremath{|e_i\rangle}} \otimes {\ensuremath{|f_i\rangle}}$ as the Schmidt vectors of $|\psi_{AB}\rangle$. Note that product pure states correspond to those states, whose Schmidt decomposition has one and only one Schmidt coefficient. If the decomposition has more than one Schmidt coefficients, the state is entangled. Notice that the squares of the Schmidt coefficients of a pure bipartite state $|\psi_{AB}\rangle$ are the eigenvalues of both the reduced density matrices $\rho_A$ ($= \mbox{tr}_B ({\ensuremath{|\psi_{AB}\rangle}} {\ensuremath{\langle\psi_{AB}|}}) $) and $\rho_B$ ($= \mbox{tr}_A ({\ensuremath{|\psi_{AB}\rangle}} {\ensuremath{\langle\psi_{AB}|}})$) of $|\psi_{AB}\rangle$. The last fact gives us an easy method to find the Schmidt coefficients and the Schmidt vectors.

Bipartite mixed states: Separable and entangled states
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As discussed in the last section, the question whether a given pure bipartite state is separable or entangled is straightforward. One has just to 