Timeline for Can a qubit be entangled with an arbitrary quantum state, without altering it?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2021 at 19:35 | comment | added | NoLongerBreathedIn | Nope, both of the qubits in the middle have had their states modified. A CNOT is a reversed CNOT in the Hadamard basis. | |
| Apr 30, 2021 at 18:17 | comment | added | Quantum Guy 123 | I further clarified my question because I see now that I was saying 'state' in some places where I should've said 'qubit' or 'bit' | |
| Apr 30, 2021 at 18:08 | comment | added | Quantum Guy 123 | $BCNOT$ is like $CNOT$ except it applies the control to the first bit, instead of the second bit. $BCNOT = \left(\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right)$ | |
| Apr 30, 2021 at 18:04 | comment | added | Quantum Guy 123 | "So if the quantum state in question is half of a Bell pair, then it is not possible to entangle it with anything else without affecting it." this is false. taking the state $\left|\Phi^{+}\right\rangle \bigotimes \left|\Phi^{+}\right\rangle$ can give an entangled state with 4 qubits, by apply $I \bigotimes BCNOT \bigotimes I$, where only one out of the 4 qubits has its value modified. | |
| Apr 29, 2021 at 16:28 | comment | added | Rammus | I see, I thought your claim was that your example implies that there are no instances in which this is possible. | |
| Apr 29, 2021 at 14:52 | comment | added | NoLongerBreathedIn | Being able to do it in general means being able to do it no matter the state you have. Therefore if you cannot do it in one specific case it is impossible to do it in general. | |
| Apr 29, 2021 at 12:33 | comment | added | Rammus | It is impossible to do of some very specific example (Bell pairs) therefore it is impossible to do in general? Can you add some more details to back up this claim? | |
| Apr 29, 2021 at 6:21 | comment | added | Nikita Nemkov | Is this really correct? I would say that entangling the transmitted qubit with another qubit basically ruins its state. For the receiver who gets the qubit it is no longer in a pure state, but instead is described by a density matrix. What does this have to do with the no-cloning theorem? | |
| Apr 28, 2021 at 22:42 | history | edited | NoLongerBreathedIn | CC BY-SA 4.0 |
Added more stuff
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| Apr 28, 2021 at 22:04 | comment | added | Quantum Guy 123 | "This is related to the no-cloning theorem." can you elaborate? | |
| Apr 27, 2021 at 21:33 | history | answered | NoLongerBreathedIn | CC BY-SA 4.0 |