# Qualitative Nature of the wave function under a certain potential

By putting the values of V(x) at a and point b we get, $$\frac{\partial^2\psi}{\partial x^2}=\frac{2m}{\hbar^2}(-v_1-E)\psi$$ ...(1) and $$\frac{\partial^2\psi}{\partial x^2}=\frac{2m}{\hbar^2}(-v_2-E)\psi$$...(2)

Now since E >0 we can say that the frequency of $$\psi(x)$$ is greater at point b. since $$|-v_1-E|<|-v_2-E|$$ And that is why as the particle goes from a to b its frequency goes up. But here, the correct answer is the 3rd one, where the frequency goes up but the amplitude is becoming less. I cant find any reason why that would happen and if my argument correct? • To my eyes the frequency behaves the same way in the three pictures. They're asking about the amplitude. – Javier Aug 23 '19 at 13:15

The reason for all this is because energy of the particle is proportional to the frequency of its wavefunction. The particle's kinetic energy increases when it loses potential energy by dropping into the well. This frequency-energy relationship is because $$E = hf$$ and $$\lambda =\frac{h}{p}$$ as De Broglie posited.
Amplitude has nothing to do with energy. It may affect the probability distribution of the particle since that is determined by the integral of $$\mid\Psi\mid^2$$ and more amplitude may mean more area under the curve