Timeline for Form of Schrödinger equation for the probability density
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2017 at 11:59 | comment | added | Jan Bos | @J.G. I meant my reply to his comment to my question. So you can find it above. | |
| Dec 9, 2017 at 8:26 | comment | added | J.G. | @JanBox Could you link to the Zheng Liu example? | |
| Dec 9, 2017 at 2:27 | comment | added | Jan Bos | Your answer does not prove that such an equation does not exist. It just gives an example of one that does not work. Per my comment under my question in response to Zheng Liu I also am not convinced on your statement that there is more information in $\psi$ than in $\rho$. You probably could say at most that the state at point $x = x_1$ contains more information than $\rho$ at the single point $x = x_1$. | |
| Dec 7, 2017 at 20:35 | comment | added | Mauricio | Sorry for the confusion in many Europeans schools they use the probability current to calculate transmission/reflection coefficients, nevertheless these problems still need Schrodinger equation: cp3.irmp.ucl.ac.be/~maltoni/PHY1222/QM-IVa.pdf | |
| Dec 7, 2017 at 12:44 | comment | added | J.G. | @Mauricio If you come up with an example, you should probably mention it in an answer here, even if it's only "a long comment". | |
| Dec 7, 2017 at 10:06 | comment | added | Mauricio | Yeah, but if I remember correctly you can solve certain scattering introductory problems using continuity equation only and without using $\psi$. | |
| Dec 7, 2017 at 10:01 | comment | added | J.G. | Mauricio Yes, but $\mathbf{j}$ is $\psi$-dependent | |
| Dec 7, 2017 at 9:59 | comment | added | Mauricio | You can write an equation for $\rho$ and $J$ (the probability current) though. | |
| Dec 7, 2017 at 7:33 | history | answered | J.G. | CC BY-SA 3.0 |