Timeline for Say you run qubit $A = (|0⟩ + |1⟩) / √2$ and qubit $B = |0⟩$ through a CNOT gate. What is the state of qubit $B$ afterwards?
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| Mar 14, 2019 at 21:29 | comment | added | jcgrowley | Ugh, got it. There are other situations in this textbook where they insist on treating a mixed-state post-gate 2-qubit system with respect to their individual states (such as equation 1.43 on page 33.) | |
| Mar 14, 2019 at 20:31 | comment | added | Jahan Claes | @jcgrowley Ah, they're being sloppy! In that diagram (and their formula), they are assuming that the states of $A$ and $B$ are only 0 or 1. In other words, they've ruled out the possibility of sending in the state $(|0\rangle+|1\rangle)|0\rangle$ like you want to do. They are describing the action of CNOT on the basis {00,01,10,11}, and expecting you to extrapolate the rest by linearity. | |
| Mar 14, 2019 at 20:27 | vote | accept | jcgrowley | ||
| Mar 14, 2019 at 20:25 | comment | added | jcgrowley | I took that from "Quantum Computation and Quantum Information", page 21, figure 1.6, which shows qubits A and B passing through a CNOT gate and becoming A and B ⊕ A. (mmrc.amss.cas.cn/tlb/201702/W020170224608149940643.pdf) I'll definitely read up on reduced density matrices. Thanks for your help. | |
| Mar 14, 2019 at 20:19 | comment | added | Jahan Claes | @jcgrowley Do a little reading about entanglement in general, and specifically reduced density matrices. Properties of the reduced density matrix will tell you whether or not you can represent a system state as factored into an A- and B- state. I am not sure where exactly you've heard that particle A retains its state after the CNOT gate, but this is emphatically NOT true. It's possible there's something else going on in that statement, so if you have an example I'd be happy to take a look. | |
| Mar 14, 2019 at 20:18 | comment | added | Jahan Claes | @jcgrowley The issue is not that the particle's state cannot be represented as a combination of 0 and 1, it's that there is no such thing as the particle's state; there's only a state of the overall two-particle system. The two particle system can always be written as a combination of 00, 01, 10, 11, but this can't necessarily be factored into an A- state and a B- state. | |
| Mar 14, 2019 at 16:47 | comment | added | jcgrowley | Thank you! Is this idea of a particle’s state not being representable as a linear combination of 0 and 1 formalized somehow? Or any pattern behind what separates the different classes of particles? The literature seems to make occasional references to individual particle states after being operated on by a gate (for example, indicating that after a CNOT gate, particle A retains its state and particle B becomes A XOR B.) So I’m just trying to reconcile that with your assertion that the particles have no definite individual state anymore. | |
| Mar 14, 2019 at 3:46 | history | edited | Jahan Claes | CC BY-SA 4.0 |
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| Mar 14, 2019 at 2:31 | history | answered | Jahan Claes | CC BY-SA 4.0 |