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?

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  • 2
    $\begingroup$ To my eyes the frequency behaves the same way in the three pictures. They're asking about the amplitude. $\endgroup$ – Javier Aug 23 '19 at 13:15

No. 2 because the frequency of the wave is greater between a and b and it increases from a to b. The wavefunction on No. 2 is also identical to the left and right of the potential well as it should be since the potential is the same and unchanging there (it's flat).

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

  • $\begingroup$ Yes, exactly. I agree to this, even this is what I thought after analyzing. But whats bugging me that they have given 3 as the correct answer! Maybe that's wrong! simply because outside the well, the wave function should be identical ! $\endgroup$ – Baibhab Bose Aug 23 '19 at 13:23
  • $\begingroup$ ah, I think it may be because the particle is moving from left to right and some of it is reflected off the potential step which is steeper on the right side. I'd have to check that though. $\endgroup$ – Andrew Aug 23 '19 at 13:53
  • $\begingroup$ Yes! This is much plausible! Please let me know if you check it! $\endgroup$ – Baibhab Bose Aug 23 '19 at 19:17

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