I'm trying to understand how the Hadamard gate works with entangled pairs.

If I have two particles A and B which are entangled and have gone through a Hadamard gate to become super-positioned Qubits, if I then observe and collapse particle A to a binary bit, has particle B has also collapsed, even though I have not yet observed it, to a binary state?

If so, if I then pass particle B through a Hadamard gate will it behave like a binary bit input rather than a super-positioned qubit input?


1 Answer 1


Yes (with a mild reservation)

It depends on what entangled state you're using. But let's say that you're using $$ \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle). $$ If you apply a Hadamard to the first qubit, you end up with $$ \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle-|11\rangle). $$ Thus, if you measure the first qubit, you get the answers 0 and 1 with 50:50 probability. If you got answer 0, the second qubit is in the state $(|0\rangle+|1\rangle)/\sqrt{2}$. If you apply hadamard to this, you get the outcome $|0\rangle$ with certainty. If you got the answer 1, the second qubit is in the state $(|0\rangle-|1\rangle)/\sqrt{2}$. Applying the Hadamard gives the outcome $|1\rangle$ with certainty, no superposition.


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