# quantum potential step question

amolv06
Science 20 Mar '10 22:15
1. 20 Mar '10 22:15
Sorry for asking so many questions related to class - I know most people probably don't find these interesting. That said, this is a difficult subject to understand, and I have gotten some good answers here so far.

My question is this: say when x < 0 we have a free particle. At x=0 the particle enters a constant potential field. When x<0 the wavefunction of the particle is f_1 and after it enters the field its wave function is f_2. My book asserts pictorially that f_1(0) = f_2(0), without giving any justification for this. Perhaps this should be obvious, but it is not to me. Could anyone explain this?
Baby Gauss
20 Mar '10 22:44
Originally posted by amolv06
Sorry for asking so many questions related to class - I know most people probably don't find these interesting. That said, this is a difficult subject to understand, and I have gotten some good answers here so far.

My question is this: say when x < 0 we have a free particle. At x=0 the particle enters a constant potential field. When x<0 the wavefunction ...[text shortened]... ation for this. Perhaps this should be obvious, but it is not to me. Could anyone explain this?
Think about the Physics of what's happening. If it isn't obvious again I can handwave you an explanation.
3. 23 Mar '10 06:47
It seems to be that the probability density must be continuous, otherwise things stop making sense. There shouldn't be two separate probabilities of finding a particle at x=0. But f_1(0) does not need to equal f_2(0) to represent this, only |f_1(0)|^2=|f_2(0)|^2 -- for instance, if f_1(0) = A+iB, f_2(0) = A-iB, and the probability at x=0 is still continuous. Since f_1 is generally not equal to f_2, what is the justification for the continuity of the wavefunction itself?
4. 23 Mar '10 10:36
Think about solving the Schrödinger equation. It includes a second order derivative of the wave-function. Along the lines (simplifying for convenience) of:

df/dt = -d2f/dx2 + Vf

V is discontinuous at 0, so d2f/dx2 will be as well. But if df/dx was discontinuous then d2f/dx2 would be a delta-function, and we wouldn't be able to get that to balance.

Which means, I think, that not only must the wave-function be continuous at 0, but so must its first derivative.