Timeline for How to splice Hamiltonians corresponding to channels $\Phi_1$ and $\Phi_2$ so as to obtain a Hamiltonian corresponding to $\Phi_2\circ\Phi_1$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2023 at 11:57 | vote | accept | Sam Jaques | ||
| Feb 6, 2023 at 11:57 | comment | added | Sam Jaques | The intuition is that quantum mechanics should be capable of describing the universe completely with unitary evolution from time-independent Hamiltonians. And indeed, Adam Zalcman's answer shows that quantum channels and/or time-dependent Hamiltonians can always be approximated by such a theory. | |
| Feb 5, 2023 at 8:48 | comment | added | Norbert Schuch | Your intuition seems wrong. Why should that be the case? | |
| Feb 4, 2023 at 23:47 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
added 3 characters in body; edited tags; edited title
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| Feb 4, 2023 at 23:39 | answer | added | Adam Zalcman | timeline score: 3 | |
| Feb 4, 2023 at 13:11 | comment | added | Sam Jaques | Right, but since these unitaries and Hamiltonians are non-unique, maybe we can find such an $H$ in the set of all possible $V_1$, $V_2$, and $V$. Physically speaking, I could regard the entire process as some much much larger quantum evolution (e.g., start including the experimenter as part of the environment system) and in this view there is some nearly-universal Hamiltonian that would produce the evolution that causes $\Phi_1$ to happen and then $\Phi_2$. | |
| Feb 3, 2023 at 22:46 | comment | added | glS♦ | so you're asking whether there's $H$ such that you get first $\Phi_1$ and then $\Phi_2$ evolving at different times? I don't get your argument for why we should expect this to exist. The Hamiltonian corrisponding to $\Phi$ needs not have anything in common with those of $\Phi_i$. Note that Hamiltonians generating $V_1 V_2$ will be very different than those generating $V_1$ and $V_2$ individually. Also physically speaking doing a time-independent evolution to get $\Phi_1$ and then another for $\Phi_2$ is very different than doing one to get $\Phi_2\circ\Phi_1$ | |
| Feb 3, 2023 at 17:54 | history | asked | Sam Jaques | CC BY-SA 4.0 |