Timeline for Deriving energy eigenfunctions of Infinite square well with nonzero bound
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2019 at 4:04 | 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 3, 2019 at 7:02 | 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. | |
| Mar 16, 2017 at 5:28 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
deleted 5 characters in body; edited tags
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| Mar 16, 2017 at 4:13 | answer | added | ZeroTheHero | timeline score: 1 | |
| Mar 15, 2017 at 14:10 | comment | added | ZeroTheHero | @garyp I know what you mean, especially if it's morning for you. | |
| Mar 15, 2017 at 14:00 | comment | added | garyp | @ZeroTheHero Yes. Typing faster than the speed of thought. | |
| Mar 15, 2017 at 13:57 | comment | added | ZeroTheHero | @garyp the higher wave vectors would come out "naturally" by enforcing the conditions at the boundaries, which could have multiple solutions. | |
| Mar 15, 2017 at 13:51 | comment | added | garyp | Your "general solution" is not. You don't allow for higher wave vectors. Also, if your solution is instead a "particular" solution for a particular eigenvalue, you should be able to find that eigenvalue from Schrodinger's eqn. Try to find the eigenvalue for your solution and see what you get. | |
| Mar 15, 2017 at 13:39 | review | First posts | |||
| Mar 15, 2017 at 13:43 | |||||
| Mar 15, 2017 at 13:39 | comment | added | ZeroTheHero | If your boundaries are at $x=\pm L/2$, why do you use a condition like $\psi(0)=0$, which is clearly NOT correct? Have you considered working backwards: starting from the solution between $x=0$ and $x=L$ and making the substitution $x\to x+L/2$ so as to shift the potential between $-L/2$ and $L/2$? | |
| Mar 15, 2017 at 13:39 | comment | added | Michael Seifert | The third unknown is usually determined by requiring $\psi(x)$ be normalized, i.e., $\int |\psi(x)|^2 dx = 1$. | |
| Mar 15, 2017 at 13:37 | history | edited | ZeroTheHero | CC BY-SA 3.0 |
fixed minor math typesetting and minor grammar.
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| Mar 15, 2017 at 13:35 | history | asked | Dogtard | CC BY-SA 3.0 |