# Finite barrier. Constant including minus or not?

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 $$E Normally we set $$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?

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

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 $$e^{i k_1x}$$.

• Aa now I understand. So both are correct then! Apr 1, 2020 at 16:57

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 $$0, One has in the barrier ; $$\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.