Timeline for Assigning initial conditions for Schrodinger's equation
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2022 at 2:14 | comment | added | hft | @ZeroTheHero Thanks for catching that. I've corrected it. | |
| Sep 19, 2022 at 2:14 | history | edited | hft | CC BY-SA 4.0 |
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| Sep 19, 2022 at 1:59 | comment | added | ZeroTheHero | Actually there’s a minor bug in your coefficients since their moduli square should sum to 1, whereas yours sum to $1/2$. There could also be a (relative) phases. | |
| Sep 19, 2022 at 1:10 | comment | added | CBBAM | Thank you for all your help, everything is clear now! | |
| Sep 19, 2022 at 1:09 | vote | accept | CBBAM | ||
| Sep 19, 2022 at 0:54 | history | edited | hft | CC BY-SA 4.0 |
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| Sep 18, 2022 at 23:46 | history | edited | hft | CC BY-SA 4.0 |
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| Sep 18, 2022 at 23:37 | history | edited | hft | CC BY-SA 4.0 |
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| Sep 18, 2022 at 23:36 | comment | added | hft | It you have not purchased Griffiths's Quantum Mechanics book, that might be a good thing to buy. It is a good introduction. Also, if you have not learned about ordinary and partial differential equations yet, you should do so since it is a important prerequisite for understanding quantum mechanics. | |
| Sep 18, 2022 at 23:34 | comment | added | hft | We still have to solve the time-independent equation, which itself it usually pretty hard except for toy problem. Once we have those solutions we can expand any other function of space (in this case the single variable x) in terms of those and the coefficients of that expansion are indeed independent of space and time. | |
| Sep 18, 2022 at 23:33 | comment | added | hft | The coefficients $c_i$, as I am using them, are constant in space, but are functions of time. The time-dependent schrodinger equation is harder to solve because we are really solving a partial differential equation in two variables at once (x and t). This is why we decompose it into parts as shown in the answer. The decomposition into the spatial-only dependence in the time-independent Schrödinger equation let's us make progress. The time dependent part then becomes "easy." | |
| Sep 18, 2022 at 23:30 | history | edited | hft | CC BY-SA 4.0 |
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| Sep 18, 2022 at 23:25 | comment | added | CBBAM | I see, so essentially the only difference is that in the time independent version you may express $\varphi$ as a sum of the $\varphi_i$ with constant coefficients $c_i$. (and this is what we mean by solving the time independent Schrodinger equation). In the time dependent case we have the same summation, but now the coefficients $c_i$ are no longer constant, rather they depend on $t$. Is this correct? | |
| Sep 18, 2022 at 23:21 | history | edited | hft | CC BY-SA 4.0 |
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| Sep 18, 2022 at 23:18 | comment | added | hft | Oh, sorry, I just noticed you are using varphi. I'll edit to be consistent with your notation. My $\phi_i$ were your $\varphi_i$. I've now changed to be consistent with your notation, so my $\varphi_i$ mean the same as your $\varphi_i$. | |
| Sep 18, 2022 at 23:16 | comment | added | hft | My $\phi_i$ are the same as yours; solutions to the time-independent Schrödinger equation. I also just fixed a typo in one of the first lines where I had $\psi_i$ instead of $\phi_i$. But anyways, my $\phi_i$ and my $\varphi$ are they same as in your notation. | |
| Sep 18, 2022 at 23:14 | history | edited | hft | CC BY-SA 4.0 |
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| Sep 18, 2022 at 22:45 | comment | added | CBBAM | What is $\phi_i$ in your answer, are they the eigenvectors of Hamiltonian? | |
| Sep 18, 2022 at 22:01 | history | answered | hft | CC BY-SA 4.0 |