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I have 3 expressions:

$f(x)=e^{ikx}+Ae^{-ikx}; g(x)= Be^{kx}+Ce^{-kx}; h(x)=De^{ikx}$

I want to find $|A|^2$ by letting

{f'(0)\over f(0)}={g'(0)\over g(0)} and {g'(a)\over g(a)}={h'(a)\over h(a)}

I have done the calculations myself and got $|A|^2=1$. But this is not the expected answer. Could someone tell me if I have made a mistake?

Thanks.

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    I don't know quantum mechanics I am asking for any initial conditions or boundary conditions something like that.2011-10-22

2 Answers 2

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The first condition, that f'(0)/f'0)=g'(0)/g(0), is equivalent to $A=-z/w$ where $z$ and $w$ are the complex number $z=B-C-\mathrm i(B+C)$ and $w=B-C+\mathrm i(B+C)$. If $B$ and $C$ are real numbers, $w=\bar z$ hence $|A|=1$. (The only case when $w=0$ is $B=C=0$, and then $g(0)=0$ hence g'(0)/g(0) does not exist.) The second condition is irrelevant since it uses values at $a$.

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From the first condition we can conclude :

$(A=-1 \lor k=0 \lor B=C) \land (A=1 \lor k=0 \lor B=-C)$

From the second condition we have that:

$(k=0 \lor C=Be^{2ka}) \land (k=0 \lor C=-Be^{2ka})$

so...

$a)$if $k=0 \Rightarrow A$ is undetermined

$b)$if $B=\pm C=0 \Rightarrow A$ is undetermined

If we observe the second condition we can see that $C$ cannot be at same time $Be^{2ka}$ and $-Be^{2ka}$ which means that $k=0 \lor B=\pm C=0$,therefore $A$ is undetermined.