Derivatives of a normalizable wavefunction

[dyn]

Hi.
In infinite volume a normalizable wavefunction → 0 as r or x,y,z→ 0 but do all the derivatives and higher derivatives → 0 as well ?
Thanks

 

[BvU]

In infinite volume a normalizable wavefunction → 0 as r or x,y,z→ 0

What makes you think that ?

 

[stevendaryl]

Hi.
In infinite volume a normalizable wavefunction → 0 as r or x,y,z→ 0 but do all the derivatives and higher derivatives → 0 as well ?
Thanks

Do you mean as $x,y,z \rightarrow \infty$?

 

[dyn]

Sorry. Yes I meant r or x.y.z tend to infinity

 

[hilbert2]

Try to calculate the norm of $\psi (x) = \frac{\sin x^2}{x}$, and draw a graph of its derivative with Wolfram Alpha... Function is normalizable but derivative doesn't approach any limit when $x\rightarrow\infty$.

 

[dyn]

I have just found the following in Griffiths QM " the wavefunction and all its derivatives go to zero at infinity"

 

[DarMM]

I have just found the following in Griffiths QM " the wavefunction and all its derivatives go to zero at infinity"

Strictly speaking it's not true. Take the wavefunction that is a series of triangles, with the $n$-th triangle centered on the integer $n$, of height $1$ and of width $\frac{1}{n}$, its square integral converges as it equals:
$$\sum_n \frac{1}{n^2}$$

However its magnitude doesn't tend to zero as you approach infinity..

 

[hilbert2]

I have just found the following in Griffiths QM " the wavefunction and all its derivatives go to zero at infinity"

In practice you can probably assume so, but it's still possible to make artificial examples of wave functions that are not like that. I think it would be really difficult to solve the time evolution of that $\frac{\sin x^2}{x}$ function when used as the initial condition for a TDSE of a free particle, but it's likely that the time evolution would smooth it in no time to have $\psi' (x)\rightarrow 0$ when $x \rightarrow \infty$.