QM: Does $\frac{d\psi}{dx} \rightarrow 0$ at ±infinity?

[quasar987]

Is it always true that

$$\frac{d\psi}{dx} \rightarrow 0$$

(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

$$\psi(x) = A(x)sin(x/A(x))$$

for x>=0, and

$$\psi(x) = \psi(-x)$$

for x<0., with

$$A(x) = e^{-x}$$

This function goes to zero at ±infinity but it's derivative is wild at ±infinity.

 

[inha]

Is your example even normalizable? Anyway one intuitive way of looking at this is that since infinity isn't really a well defined position i.e. inf+1 is still inf so the derivative must be zero at inf. This is a mathematical abomination though.

 

[quasar987]

Thx inha. I'd forgotten that my counter example had to be normalizable to be valid. Maybe there is a way to show that for psi normalizable, the derivative go to zero, thus getting rid of the abomination.

And another question of the same type: what is the reason that psi must go to zero at infinity faster than any power of x?

Is this also true of $d\psi/dx$? If so, I also ask why.

Thx!

 

[asrodan]

The only reason I can think of right now is that Psi has to go to zero faster than any power of x goes to infinity is so that expectation values of x. Since the wave function has no physical meaning by itself, there is the obvious requirement that it be a function such that it is possible to perform on it the operations that result in useful physical information.

The derivatives of Psi must go to zero at infinity to calculate expectation values of momentum. Although the derivatives don't need to go to zero faster than any power of x goes to infinity for <p> calculations, it may still be true.