Timeline for What is the general solution of the one-dimensional time-independent Schrodinger's equation with constant potential?
Current License: CC BY-SA 4.0
28 events
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| Feb 17, 2022 at 2:49 | answer | added | user275535 | timeline score: 1 | |
| Feb 17, 2022 at 2:31 | comment | added | user275535 | Check this out: physics.stackexchange.com/questions/513474/… | |
| Feb 17, 2022 at 1:52 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 25 characters in body; edited title
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| Feb 17, 2022 at 1:16 | answer | added | Níckolas Alves | timeline score: 1 | |
| Feb 17, 2022 at 1:15 | history | edited | Níckolas Alves | CC BY-SA 4.0 |
added 3 characters in body
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| Feb 17, 2022 at 1:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 10, 2021 at 14:52 | comment | added | Connor Behan | The TISE is a second order ODE and its solutions thus form a two-dimensional vector space. This means $P_1$ is the most general solution. $P_2$ is the same solution just with a relabelling of the parameters, $k^\prime = ik$, $C+D = A$ and $i(C-D) = B$. The only reason to use both is if you have a piecewise potential where one $U_0$ is different from the other $U_0$. | |
| Jun 10, 2021 at 11:05 | history | edited | Dipankar Mitra | CC BY-SA 4.0 |
clearified my question
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| Jun 10, 2021 at 10:51 | comment | added | Dipankar Mitra | @GiorgioP oh... Now I have understood. I missed the point that for $U_0 > E\;$ k can be imaginary. OK I am editing that. But still I have not understood which is the actual solution of TISE. | |
| Jun 10, 2021 at 8:55 | history | edited | Dipankar Mitra | CC BY-SA 4.0 |
Corrected an expression
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| Jun 10, 2021 at 8:22 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | Actually, the squared modulus could be eliminated in both cases. But, to explain why, I should write a complete answer and I am not sure I have time today. | |
| Jun 10, 2021 at 8:17 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | for most of the cases means for the values of x such that $E>U_0$. And what about the points such that $E<U_0$? So in general the solution will have a piecewise purely real or purely imaginary ${\bf k}$. The squared modulus is in order. | |
| Jun 10, 2021 at 7:40 | comment | added | Dipankar Mitra | @GiorgioP Why? As I have understood (there might be lackings in my understanding) in the first case k is used instead of k' (Actually k' = ik). And the value of k is not imaginary or complex for most of the cases. So there should no need of modulus in the first case. | |
| Jun 9, 2021 at 16:02 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | Now I understand. But this implies that the squared modulus should appear also in the first case. | |
| Jun 9, 2021 at 15:26 | vote | accept | Dipankar Mitra | ||
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| Jun 9, 2021 at 15:25 | vote | accept | Dipankar Mitra | ||
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| Jun 9, 2021 at 15:25 | vote | accept | Dipankar Mitra | ||
| Jun 9, 2021 at 15:25 | |||||
| Jun 9, 2021 at 8:50 | review | Close votes | |||
| Jun 9, 2021 at 16:44 | |||||
| Jun 9, 2021 at 7:51 | history | edited | Dipankar Mitra | CC BY-SA 4.0 |
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| Jun 9, 2021 at 7:28 | answer | added | Gandalf73 | timeline score: 0 | |
| Jun 9, 2021 at 7:24 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
deleted 2 characters in body; edited title; edited tags
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| Jun 9, 2021 at 7:18 | comment | added | Dipankar Mitra | @GiorgioP You know most of the time potential energy is less than total energy. So there is an imaginary unit hidden in k'. | |
| Jun 9, 2021 at 7:06 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | Is really the second form as you wrote? Isn't there an imaginary unit in the exponential arguments? | |
| Jun 9, 2021 at 6:56 | history | edited | Dipankar Mitra | CC BY-SA 4.0 |
fixxed grammatical mistakes
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| S Jun 9, 2021 at 6:55 | history | suggested | Cross | CC BY-SA 4.0 |
Fixed an error in the symbol used for the conjugate of ψ(x), spelling mistakes
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| Jun 9, 2021 at 6:50 | review | Suggested edits | |||
| S Jun 9, 2021 at 6:55 | |||||
| Jun 9, 2021 at 6:47 | history | edited | Vincent Thacker | CC BY-SA 4.0 |
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| Jun 9, 2021 at 6:37 | history | asked | Dipankar Mitra | CC BY-SA 4.0 |