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For a finite potential barrier of magnitude $V_0$ between $x=-a$ and $x=a$ we know that the time independent schrodinger equation is $\Psi'' +\frac{2m}{\hbar}E\Psi=0$ for $x<-a$. Let $E<V_0.$ Normally we set $k_1^2=\frac{2mE}{\hbar^2}$ and get $\Psi''+k_1^2\Psi=0$ which would give $$\Psi=A_1e^{ik_1x} + B_1e^{-ik_1x}.$$ But if we set $k_2^2=\frac{-2mE}{\hbar^2}$ we get $\Psi'' - k_2^2\Psi=0$ and the solution $$\Psi=A_2e^{k_2x} + B_2e^{-k_2x}.$$ Why is the second solution incorrect, while the first one is?

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  • $\begingroup$ The first and second solution seem to be the same (v4), just written with possible imaginary $k$. $\endgroup$
    – Qmechanic
    Apr 1, 2020 at 16:32

2 Answers 2

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The difference is in the sign of $E$.

The definition $k_2^2=-2mE/\hbar^2$ with $E>0$ implies that $k$ is pure imaginary $k_2$, i.e $k_2=i k_1$ with $k_1^2=+2mE/\hbar^2>0$. Then $e^{k_2 x}= e^{ik_1 x}$.

On the other hand the definition $k_1^2=+2mE/\hbar^2$ gives $k_1$ real so again $e^{i k_1x}$.

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  • $\begingroup$ Aa now I understand. So both are correct then! $\endgroup$ Apr 1, 2020 at 16:57
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First, your Shrodinger equation seems to have a some small problems. From

$$ - \frac{\hbar^2}{2m}\Psi'' +V(x)\Psi = E\Psi$$

one get:

$$ \Psi'' + \frac{\sqrt{2m}}{\hbar} (E-V(x)) \Psi =0. $$

For your barrier, and for $0<E<V_0$, One has in the barrier ; $$ \Psi(x) = A e^{\kappa x} + B e^{-\kappa x},$$

where $\kappa=\sqrt{\frac{2m (V_0-E)}{\hbar^2}}$, and for $|x|>a$ :

$$ \Psi(x) = A e^{ikx} + B e^{-ik x},$$

with $k= \sqrt{\frac{2m E}{\hbar^2}}$,

Hence of your first solution is the only correct one, even if it is the false solution of a wrong equation.

Your problem is that if you use an assumption like $k^2=-K$ where $K$ is positive, you get an imaginary $k$ which converts real exponentials into imaginaries and reciprocally.

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