- #1
- 4,807
- 32
Is it always true that
[tex]\frac{d\psi}{dx} \rightarrow 0[/tex]
(at ±infinity)? And if so, why?
I know that for a wave function to be normalizable, we must have psi-->0 at ±infinity but as far as i can see, that does not imply that the derivative will be 0. A counter-exemple of this is a decreasing "sine-like" function with an oscillation frequency inversely proportional to its amplitude. For instance
[tex]\psi(x) = A(x)sin(x/A(x))[/tex]
for x>=0, and
[tex]\psi(x) = \psi(-x)[/tex]
for x<0., with
[tex]A(x) = e^{-x}[/tex]
This function goes to zero at ±infinity but it's derivative is wild at ±infinity.
[tex]\frac{d\psi}{dx} \rightarrow 0[/tex]
(at ±infinity)? And if so, why?
I know that for a wave function to be normalizable, we must have psi-->0 at ±infinity but as far as i can see, that does not imply that the derivative will be 0. A counter-exemple of this is a decreasing "sine-like" function with an oscillation frequency inversely proportional to its amplitude. For instance
[tex]\psi(x) = A(x)sin(x/A(x))[/tex]
for x>=0, and
[tex]\psi(x) = \psi(-x)[/tex]
for x<0., with
[tex]A(x) = e^{-x}[/tex]
This function goes to zero at ±infinity but it's derivative is wild at ±infinity.
Last edited: