Timeline for Justifying separation of variables in solving the Schrödinger equation in 3D
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2022 at 13:17 | comment | added | basics | Let us continue this discussion in chat. | |
| Sep 7, 2022 at 13:14 | comment | added | Chris Yang | I think if we were discussing solving the time-dependent SE, writing solutions as a linear combination of base functions would work, since $\hat H (\Psi_1 + \Psi_2) = i\hbar \partial_t (\Psi_1 + \Psi_2)$. However, as I have highlighted in my comment on Emilio's answer, in this case, a linear combination of solutions to the time-independent SE is not in general also a solution. | |
| Sep 7, 2022 at 13:03 | comment | added | basics | If you get such an eigenvalue, you should have an eigenvector associated to it. But, since any function can be written using the complete basis built with the eigenvectors associated with the eigenvalues you get from variable separation, you should now realize that there isn't such an eigenvalue | |
| Sep 7, 2022 at 12:57 | comment | added | Chris Yang | Well, my question is closer to being, given that we have obtained this set of eigenvalues $a$ and $b$ (using your example) using separation of variables, how do we be sure that there is not some eigenvalue $c$ that we cannot obtain using separation of variables? | |
| Sep 7, 2022 at 12:56 | comment | added | basics | That should be equivalent to what you wrote in your last comment to @Emilio's answer | |
| Sep 7, 2022 at 12:48 | comment | added | basics | Let's do a finite-dimensional example, trying to be more clear. Suppose we have a 3-d space with an orthogonal basis $\{\mathbf{v}^a_1$, $\mathbf{v}^a_2$, $\mathbf{v}^b_3\}$, being the eigenvectors of the an operator $H$, so that the first 2 have the same eigenvalue $H\mathbf{v}^a_1 = a \mathbf{v}^a_1$, $H\mathbf{v}^a_2 = a \mathbf{v}^a_2$, $H\mathbf{v}^b_3 = b \mathbf{v}^b_3$. What you're asking is equivalent to ask: given $\mathbf{v}^a_1$, $\mathbf{v}^a_2$ associated with the eigenvalue $a$, how can I be sure that there is no other eigenvector (having some contribute of $\mathbf{v}^b_3$)? | |
| Sep 7, 2022 at 12:46 | comment | added | basics | The results you get is a linear combination of the contributions of the Fourier series, with the multiplicative coefficients equal to the eigenvalues. | |
| Sep 7, 2022 at 12:44 | comment | added | basics | Ok, maybe I get some details I lost before. You're asking if, given a eivenvalue, the eigenfunctions you get from separations of variable are all the functions you can get. My answer is: try to do your process in the opposite direction. Given a generic function, you can write it as its Fourier series, using all the possible functions from the set of the complete basis (each of this function gives you an independent contribution, since they form an orthogonal basis). Now you can apply your operator to your function written as its Fourier series | |
| Sep 7, 2022 at 12:32 | comment | added | Chris Yang | Thanks for the reply! Please see the comment I attached to @Emilio Pisanty's answer. | |
| Sep 7, 2022 at 12:22 | history | answered | basics | CC BY-SA 4.0 |