# Tag Info

1

The whole point of an EPR pair is that you cannot write (without losing some information) "This is what Alice has" and "This is what Bob has". Partial descriptions can be given using reduced density matrices. However, if you want to identify which bits Alice and Bob each have, we often use a notation like  (|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B)/\...

3

I think faster than light communication is more stubborn than that :-) The principle of deferred measurement, to quote Nielsen and Chuang, page 186, is Measurements can always be moved from an intermediate stage of a quantum circuit to the end of the circuit; if the measurement results are used at any stage of the circuit then the classically controlled ...

1

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.

3

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. ...

3

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 ...

1

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() ...

3

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 -> c; if (c == 1) x q; by cx ...

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