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Imagine for a moment that we could distinguish between arbitrary quantum states. We’ll show that this implies the ability to communicate faster than light, using entan- glement. Suppose Alice and Bob share an entangled pair of qubits in the state (|00⟩ + √ |11⟩)/ 2. Then, if Alice measures in the computational basis, the post-measurement states will be |00⟩ with probability 1/2, and |11⟩ with probability 1/2. Thus Bob’s sys- tem is either in the state |0⟩, with probability 1/2, or in the state |1⟩, with probability 1/2. Suppose, however, that Alice had instead measured in the |+⟩, |−⟩ basis. Recall that √√ |0⟩ = (|+⟩ + |−⟩)/ 2 and |1⟩ = (|+⟩ − |−⟩)/ 2. A little algebra shows that the initial √ state of Alice and Bob’s system may be rewritten as (| + +⟩ + | − −⟩)/ 2. Therefore, if Alice measures in the |+⟩, |−⟩ basis, the state of Bob’s system after the measurement will be |+⟩ or |−⟩ with probability 1/2 each. So far, this is all basic quantum mechanics. But if Bob had access to a device that could distinguish the four states |0⟩, |1⟩, |+⟩, |−⟩ from one another, then he could tell whether Alice had measured in the computational basis, or in the |+⟩, |−⟩ basis.

I thought if Alice measures in +/- state, then Bob would immediately see his qubit to be in +/- state and know that she measured in +/- basis. What exactly is measuring in different bases mean then?

Tristan Nemoz
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Nir Sharma
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If Alice measures in the $|+ \rangle$, $| - \rangle$ basis, then Bob's qubit will indeed be in either the $|+\rangle$ or $|-\rangle$ state depending on the result of Alice's measurement. But your confusion is that Bob doesn't "immediately see his qubit" in that state. Without communication with Alice, Bob doesn't know anything about the state of his qubit. The best Bob can do is measure his qubit, but the result of his measurement is not enough to determine the state.