Timeline for Schrödinger equation in energy basis
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 13, 2022 at 22:32 | comment | added | Sandejo | @ytlu The "time evolution for the coefficients" is the Schrödinger equation expressed in a given basis. For example, in your answer, when you give the Schrödinger equation in the position basis, you are writing the time evolution equation for the coefficients of position eigenstates, denoted by $\Psi(\vec r,t)$. Writing the Schrödinger equation in any basis (including the position basis) requires that you already have the relevant set of basis kets. | |
| Feb 13, 2022 at 16:38 | comment | added | ytlu | To my understanding (maybe I was wrong about this), the OP is asking: can we solve the eigen values and eigen functions using differential equation in time domain, instead of in the spatial domain as we observed from most textbooks did. You answer was saying that we already had a set of eigen functions, then expands the solution in the bases, check the time evolution of the expanding coefficients. | |
| Feb 13, 2022 at 16:07 | comment | added | ytlu | I thought that OP is requesting a "schrodigner equation" for energy space, not the time evolution for the coefficients expanding in the energy bases. | |
| Feb 12, 2022 at 22:01 | comment | added | Sandejo | @ytlu The Schrödinger equation is a basis-independent description of the time evolution of a system, so it is not necessary to use any basis. How you obtain the the eigenstates is irrelevant to that fact that the equation can be expressed in the energy basis. | |
| Feb 12, 2022 at 11:12 | comment | added | ytlu | The eigen states are supposed to comes out from the constructing schrodinger equation. You use the eigen states to construct the generating equation, a circling logic. It supposes to start with an arbitrary bases. | |
| Feb 12, 2022 at 1:48 | comment | added | Sandejo | @ytlu I'm defining $\langle n\rvert$ to be eigenbras of the hamiltonian, so $\langle n\rvert H=E_n\langle n\rvert$. For this reason, your intermediate step is unnecessary here. | |
| Feb 11, 2022 at 23:01 | comment | added | ytlu | Sandejo jumped too quick from Eq.(2) to Eq.(3). An intermediate equation is $ \langle n \rvert H \sum_m \lvert m \rangle \langle m \rvert \lvert \alpha \rangle$. Then you have to discuss on the element $ \langle n \rvert H \lvert m \rangle$, if it is diagonal. | |
| Feb 11, 2022 at 22:18 | vote | accept | Silas | ||
| Feb 11, 2022 at 21:40 | history | answered | Sandejo | CC BY-SA 4.0 |