Timeline for Calculating the probability of transmission of a wave function between two delta distributions
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 27, 2018 at 15:19 | comment | added | ndrearu | In the potential you wrote there is a minus between the deltas, so in turn the minus in front of $\delta(x-a)$ gives a $+$ in the discontinuity. | |
| Feb 27, 2018 at 15:14 | comment | added | gamma | But still I don't understand why it's $+2D\psi(a)$ instead of $-2D\psi(a)$? | |
| Feb 27, 2018 at 14:50 | history | edited | ndrearu | CC BY-SA 3.0 |
error in a formula
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| Feb 27, 2018 at 14:48 | comment | added | ndrearu | I made a typo sorry: in the last line it is a $+$ and not a $-$ sign so that it is $+2D\psi(+a)$. Probably the results you found are due to this. I will correct the answer. | |
| Feb 27, 2018 at 14:45 | comment | added | gamma | Why do you have $+2D(-a)$ when $/gamma$ is $\frac{-D\hbar^2}{m}$? When I try to solve this, I get $A = -Be^{2ika}$ can you confirm this? I get $B=D$ and $A=C$ by Equating the coefficients and I'm not quite shure if this is allowed. I'm also not shure which $\psi$ you mean after the equal sign when you say $ \frac{d\psi}{dx}_{x\rightarrow a^+}-\frac{d\psi}{dx}_{x\rightarrow a^-} = -2D\psi(-a)$ In fact I thought I can choose which Ansatz I like for that $\psi$. | |
| Feb 27, 2018 at 11:59 | history | answered | ndrearu | CC BY-SA 3.0 |