I've been trying to follow Qiskit Global Summer School 2020 and understood that if a pure state $S$ on systems $A$ and $B$ cannot be written as a tensor product of some state from $A$ and some state from $B$ then it is correlated and therefore entangled.

The thing is that on the same slide, the instructor shares a general formula of the Bell states that represents it as a tensor product which was very conflicting.

Course link: https://youtu.be/9MpSQglnqI0?t=327 minute 5:27

Course slide capture


1 Answer 1


The term $$(\mathbb{I} \otimes \sigma_{x}^j\sigma_{z}^i)|\psi^{00}\rangle$$ is the state $|\psi^{00}\rangle$ multiplied by $(\mathbb{I} \otimes \sigma_{x}^j\sigma_{z}^i)$. So, the formula doesn't represent $|\psi^{ij}\rangle$ as a tensor product. It shows how to get any of the four entangled forms starting from $|\psi^{00}\rangle$ by applying local operations.

  • $\begingroup$ mmhhmmm, does it mean that i can't distribute over ket with the tensor product to obtain |πœ“00⟩ βŠ— (πœŽπ‘—π‘₯πœŽπ‘–π‘§) |πœ“00⟩ and |πœ“00⟩ is system A and (πœŽπ‘—π‘₯πœŽπ‘–π‘§) |πœ“00⟩ is system B? $\endgroup$
    – Tawfik
    Jun 25, 2023 at 17:43
  • $\begingroup$ @Tawfik $|\psi^{00}\rangle$ is already a two qubit entangled state $\endgroup$
    – Mauricio
    Jun 25, 2023 at 19:30
  • $\begingroup$ ah that's true! $\endgroup$
    – Tawfik
    Jun 25, 2023 at 19:35

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