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Wondering if it's possible to produce an entangled state by measuring an unentangled state. I tried a few examples, but it seems it's not possible.

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  • $\begingroup$ Can you share what approach you have taken trying to prove/disprove this? $\endgroup$
    – FDGod
    Sep 24, 2023 at 2:26
  • $\begingroup$ @FDGod I tried to measure qubit 0 in {0} basis for a given unentangled state by doing |0> tensor pdt <0|psi> where psi is the unentangled quantum state but it always comes out to be an entangled state! $\endgroup$
    – codeit
    Sep 26, 2023 at 21:54

2 Answers 2

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Depends on what you mean by "measuring". A "measure", in and of itself, produces (classical) outcomes, not necessarily states. But you can consider the quantum states that "remain" after a measurement. Depending on the physical platform these can make sense, and in such contexts the question is sensible.

An easy approach would be to consider projective measurements and assume post-measurement states correspond to the measured outcome. For example, if you're measuring in an orthonormal basis $\{|b\rangle\}_b$, then upon observing the $b$-th outcome the post-measurement state is $|b\rangle$.

In this scenario, any measurement in an entangled basis will "produce" entangled states even if the originally measured state is not entangled. For example, if you measure the two-qubit state $|0,0\rangle$ in the Bell basis, which has elements $$|\Phi^\pm\rangle=\frac1{\sqrt2}(|0,0\rangle\pm|1,1\rangle), \qquad |\Psi^\pm\rangle=\frac1{\sqrt2}(|0,1\rangle\pm|1,0\rangle),$$ then the possible outcomes are those corresponding to $|\Phi^\pm\rangle$, and the post-measurement states are $|\Phi^\pm\rangle$, both of which are maximally entangled.

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If you only consider the measurement to be single-qubit measurement (which is what state-of-the-art quantum hardware can only do now), the answer is no. But in a more general case, measurement can be done collectively between different qubits and the answer would be yes (@glS provides a good answer).

A natural question you may have is: Can we avoid the use of gates to create entanglement in our state by collective measurements? That is true, but in some sense, multi-qubit gates are identical to multi-qubit collective measurements. For example, the bell state measurement given by @glS can be done using A CNOT gate and Hadamard gate, and you then measure the qubits separately.

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