# How does Bell measurement work in the teleportation?

I'm a complete beginner and one of the first things I was taught was the teleportation protocol. In the protocol, the party sending its state (which we call say $$|\phi\rangle$$) makes a Bell measurement on a Bell state it has from before along with $$|\phi\rangle$$. From this, it finds the indices of its Bell state which it sends to the receiver. But my question is why does the fact that it is measuring $$|\phi\rangle$$ together with its Bell state have any bearing on the indices of the Bell state, considering $$|\phi\rangle$$ is not involved in the Bell measurement?

• Hi and welcome to Quantum Computing SE. Maybe this helps to better understand how the teleported state $|\phi\rangle$ are connected with entagled state shared between parties: quantumcomputing.stackexchange.com/questions/9627/… Jul 21 '20 at 6:23
• Hi, as it is said in the pointed question in the previous comment, I advise you to look at the way quantum teleportation works in this book, and if you want an example of implementation check also this Qiskit textbook. Hope this will help you :)
– Lena
Jul 21 '20 at 8:48

So, you see the controlled-NOT acts as an entangling operator without affecting the first qubit (here $$|\phi\rangle$$) and changing the entangled Bell state and hence its measurement outcome.