# Question about limit of wave function as x goes to infinity

1. Sep 1, 2012

### vincent_vega

So I understand why the limit of the wave function as x goes to infinity is 0. But on pg 14 of Griffiths 2nd ed. qm for example, why does he call lim x$\rightarrow$$\infty$ ψ*$\frac{dψ}{dx}$ = 0? How can you assume that $\frac{dψ}{dx}$ doesn't blow up at x = ∞

2. Sep 2, 2012

### TSny

Hello, vincent.

You have a good question there. It is possible to construct functions ψ(x) that go to zero at infinity but such that ψ*(x)ψ'(x) is not defined at infinity. For example ψ = sin$(x^4)/\sqrt{1+x^2}$

Griffiths wants to argue that the combination ψ*(x)ψ'(x) - ψ(x)ψ*'(x) goes to zero at infinity so that the normalization of the wavefunction is time independent. For any real wavefunction, you can see that this combination is identically zero for all x. However, it is possible to construct complex valued functions ψ for which the combination does not go to zero at infinity even though ψ goes to zero at infinity. For example ψ = $e^{ix^4}/\sqrt{1+x^2}$. But this wavefunction is "pathological". For example, the expectation value of the kinetic energy operator is undefined for this function.

So, maybe the allowable wavefunctions are restricted to exclude these pathological functions. I don't know.

Last edited: Sep 2, 2012
3. Sep 2, 2012

### Oxvillian

Interesting example - I'm guessing the class of allowed wavefunctions is pretty close to the class of functions for which an inverse Fourier transform exists. Probably this one fails the bounded variation test.

4. Sep 2, 2012

### vincent_vega

thanks that makes sense.

do you have any idea why x*ψ(x)*ψ'(x) is defined to be zero at x = infinity? this also occurred in the example by griffiths. I am thrown off by the x in front of the expression

5. Sep 2, 2012

### TSny

I don't know. It seems that Griffiths is making an assumption about the behavior of the wavefunction and its derivative at infinity.