Hi I am quite new to Quantum Mechanics and I came across with this confusion.

Let's say there are 2 entangled particles A and B.

1. When A is measured to be state 0 (doesnt measure B) does it automatically collapse B to 1 or does it imply B should be 1 ?

2. If A is measured (result is 1) and doesn't measure B, by any chance if we change the state of A to state 0 will it change the state of B ?

Thank you very much.

  1. If you measure one of the entangled particles, the other one collapses automatically. However, the state to which it collapses depends not only on measurement result of the first particle, but also on the state in which they were entangled.

For example, if the original state was $\frac{1}{\sqrt2}(|01\rangle + |10\rangle)$, measuring the first particle in state $|0\rangle$ will indeed collapse the second one to $|1\rangle$. But if the original state was $\frac{1}{\sqrt2}(|00\rangle + |11\rangle)$, the state of the second particle will be $|0\rangle$, and if it was $\frac{1}{2}|0\rangle\otimes(|0\rangle + |1\rangle) + \frac{1}{\sqrt2}|11\rangle$, the second particle will end up in $\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$.

  1. Once you measure the first particle, it's no longer entangled with the second one, and whatever you do to it doesn't affect the state of the second particle. Unless you entangle them again...
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  • $\begingroup$ Thank you very much @mariia for your concise explanation. That actually helped me a lot to clarify the doubts I had. If you don't mind will it be possible for me to get a reference for the points 1. and 2. for my further readings. $\endgroup$ – Adhisha Gammanpila May 28 at 5:19
  • $\begingroup$ I don't know a perfect source off the top of my head, sorry. $\endgroup$ – Mariia Mykhailova May 28 at 5:39
  • $\begingroup$ Not at all an issue Marria, thank you very much :) $\endgroup$ – Adhisha Gammanpila May 28 at 6:53

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