Solving Schrodinger's Equation: Finite Jumps at Singular Points?

[RedX]

When solving Schrödinger'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 $$x=x_0$$?

 

[azntoon]

No, you cannot assume a finite jump in the solution y at x=x_0. This is because the equation y'=f(x)y only holds if both y and f(x) are non-zero. If f(x) is singular at x_0, then it is not necessarily true that y will have a finite jump at x_0.