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There is a mistake in design of the first circuit. Both $X$ and $Z$ gate work when value 1 is in classical register. Gate $X$ should work in case qubit $q_2$ is in state $|1\rangle$. Similarly $Z$ acts when $q_1$ in state $|1\rangle$. Also you have to deal with state when both $X$ and $Z$ have to act. In your case you conditioned both $X$ and $Z$ on c==1. ...


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1) In this step, you connect a teleported qubit with entangled qubits between Alice and Bob. This means, Bob now has an "access" to the teleported qubit. 2) Here you get some information about the teleported qubit and "partially colapse" Bob's qubit according to a state of the teleported qubit. 3) In this last step you bring information about the ...


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Currently IBM Q does not support IF statement on real quantum processor. The IF can be used on simulator only. However, there is a theorem stating that quantum gates controlled by classical register can be replaced by those controlled by qubits before measurement. So, you can simply replace the statement measure q[1] -> c[1]; if (c == 1) x q[2]; by cx ...


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Thanks to Martin Vesely! Now I modify method 1 by using the correct c values. I think I got the right answer. My goal is to see the 1st bit entangled with the 4th bit. As can be seen below, 1st and 4th bits always have the same values. So they are entangled.


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I contacted @fran-cabrera from the IBMQ team and he could reproduce the bug! The problem is with the visualization of the transpiled circuit, not the execution (the result should be correct). The team is working on solving it and they expect to deploy a fix at the end of the week. I ran you example in Qiskit and it seems to work IBMQ.load_account() ...


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