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I know that in order to make a two qubit entangled state, this quantum circuit is used:

Hadamard + CNOT gate

But I was wondering if there are any other gate combinations which also create entangled two quit states.

If there are other quantum circuits which achieve this, I was wondering if they would also create any one of the four bell states. Thanks in advance.


Cross-posted on quantumcomputing.SE

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  • $\begingroup$ This is too broad. There's a plethora of circuits creating entangled states, including Bell states. $\endgroup$ Mar 3 at 21:02
  • $\begingroup$ @NorbertSchuch Ok thanks. So if that means that there are other circuits to create Bell states, are there any certain rules that these circuits follow in order for them to create Bell States? $\endgroup$
    – Will
    Mar 3 at 22:19
  • $\begingroup$ Sure. But again, this would be an overly long list and never be exhaustive. That's simply not a good question. E.g., they must use some gate which is not diagonal, or start from a non-classical state. But agian, this is not sufficient. $\endgroup$ Mar 3 at 22:27
  • $\begingroup$ @NorbertSchuch Ok thank you, that makes sense now. So if I understand completely what you said, would it not be plausible to create an algorithm which could determine if two qubits are in a bell state based on the quantum circuit given? $\endgroup$
    – Will
    Mar 3 at 22:42
  • $\begingroup$ It depends. In principle, yes, but it would be exponentially costly in the number of qubits involved, potentially. (Of course, there is no need to create a Bell state using a complicated circuit, but also a complicated circuit can do so, and then it is very hard to determine if it does so.) $\endgroup$ Mar 3 at 23:06

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