- #1
RedX
- 963
- 3
When solving Schrodinger's eqn. one comes across the expression:
\frac{d^2 \psi}{dx^2}=(V-E)\psi
where the mass has been chosen to make \frac{\hbar^2}{2m}=1
If V is infinity at some x, then it is said that \frac{d \psi}{dx} can have a finite jump at that x, since \frac{d^2 \psi}{dx^2}=\infty
But doesn't this assume that \psi doesn't go to zero? Then you would get: \frac{d^2 \psi}{dx^2}=(V-E)\psi=\infty*0 and it is not necessarily true that \frac{d^2 \psi}{dx^2}=\infty, hence no longer necessarily true that \frac{d \psi}{dx} can have a finite jump.
The easiest case is the infinite well. The solution at the boundary of the well does have a finite jump in its slope. However, the function \psi itself is zero at the boundaries of the well.
So in general, if you have a differential equation y'=f(x)y, where f(x) is singular at a point x_0, can you assume a finite jump in the solution y at x=x_0?
\frac{d^2 \psi}{dx^2}=(V-E)\psi
where the mass has been chosen to make \frac{\hbar^2}{2m}=1
If V is infinity at some x, then it is said that \frac{d \psi}{dx} can have a finite jump at that x, since \frac{d^2 \psi}{dx^2}=\infty
But doesn't this assume that \psi doesn't go to zero? Then you would get: \frac{d^2 \psi}{dx^2}=(V-E)\psi=\infty*0 and it is not necessarily true that \frac{d^2 \psi}{dx^2}=\infty, hence no longer necessarily true that \frac{d \psi}{dx} can have a finite jump.
The easiest case is the infinite well. The solution at the boundary of the well does have a finite jump in its slope. However, the function \psi itself is zero at the boundaries of the well.
So in general, if you have a differential equation y'=f(x)y, where f(x) is singular at a point x_0, can you assume a finite jump in the solution y at x=x_0?
