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May 6, 2021 at 21:43 vote accept DJA
Apr 30, 2021 at 9:56 vote accept DJA
Apr 30, 2021 at 9:56
Apr 13, 2021 at 2:49 vote accept DJA
Apr 30, 2021 at 9:53
Apr 13, 2021 at 2:48 vote accept DJA
Apr 13, 2021 at 2:49
Apr 13, 2021 at 1:09 answer added QuantumEngineer timeline score: 1
Apr 12, 2021 at 22:27 comment added DJA Still dont see how to solve the problem above though. Sorry that I cant quite see what to do!
Apr 12, 2021 at 22:02 comment added user87745 Yes, but the point of my comment is that you only need to solve the eigenvalue problem for the Hamiltonian. The $H\vert\psi\rangle = E\vert\psi\rangle$ that you mentioned. Once that is done, it is trivial. You don't need to work with the exponentiated operators because it takes the form of exponentiated scalars in the eigenbasis of the Hamiltonian.
Apr 12, 2021 at 21:59 comment added DJA Thats very true, but normally I am used to working out stuff such as $H |\psi \rangle = E|\psi \rangle$ solving things that are of the form $exp(-iHt /\hbar)|\psi \rangle$ is a bit more difficult for me at the present moment.
Apr 12, 2021 at 21:54 comment added user87745 I would also mention that the Hamiltonian being given as a summation of operators that don't commute isn't anything exotic. Almost all our examples in QM courses involve Hamiltonians of the type $\hat{p}^2 + V(\hat{x})$ and we know that $[\hat{p},\hat{x}]\neq 0$ ;-)
Apr 12, 2021 at 21:48 comment added user87745 Did you try finding the eigenbasis of the Hamiltonian first? Once you know the eigenbasis of the Hamiltonian, you can express the propagator simply as $\sum_n e^{-iE_nt}\vert n\rangle\langle n\vert$ and the state at time $t$ will be simply $\sum_n e^{-iE_nt}\vert n\rangle\langle n\vert\psi(0)\rangle$.
Apr 12, 2021 at 21:43 history asked DJA CC BY-SA 4.0