If you have a 2 qubit system, you represent measurements/observations as the tensor product of the tensor product of two kets. For an entangled system, what's the difference? Is there just some condition in the operators that leads to say, being non-diagonalizable? I have tried looking for an answer online but I haven't been able to.
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Related/possible duplicate: http://physics.stackexchange.com/q/3489/50583 – ACuriousMind Mar 29 '17 at 10:57
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The Hilbert space is the tensor product, and each basis vector will also be a product of vectors, but the state is not necessarily so. For instance, $|\psi\rangle= |+\rangle|-\rangle$ is separable (and thus not entangled), but $|\psi\rangle=( |+\rangle|-\rangle-|-\rangle|+\rangle)$ is not separable (you cannot write it as a product), and thus entangled.