Proving $\Psi$ and $\Psi'$ Normalizable: A Mathematical Proof

[broegger]

I am trying to prove that

\frac{d\langle p \rangle}{dt} = \langle -\frac{\partial V}{\partial x} \rangle

I am done if I can just prove that

\left[ \Psi^*\frac{\partial^2 \Psi}{\partial x^2} \right]_{-\infty}^{\infty} = 0

\left[ \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x} \right]_{-\infty}^{\infty} = 0

My suggestion is that since \Psi is a wavefunction, it is normalizable and must approach 0 as x \rightarrow \pm\infty, and so must its derivatives. I don't know if this argument holds?

 

[vanesch]

I am trying to prove that

\langle p \rangle = \langle -\frac{\partial V}{\partial x} \rangle

I am done if I can just prove that

\left[ \Psi^*\frac{\partial^2 \Psi}{\partial x^2} \right]_{-\infty}^{\infty} = 0

\left[ \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x} \right]_{-\infty}^{\infty} = 0

My suggestion is that since \Psi is a wavefunction, it is normalizable and must approach 0 as x \rightarrow \pm\infty, and so must its derivatives. I don't know if this argument holds?

I would think that's a valid reason, no ?

cheers,
Patrick.

 

[da_willem]

I am trying to prove that

\langle p \rangle = \langle -\frac{\partial V}{\partial x} \rangle

Good luck proving something incorrect!

 

[broegger]

I see your point :) I mean the time derivative, of course.

 

[dextercioby]

That's something totally different.It is simply CORRECT...


Daniel.

 

[vanesch]

That's something totally different.It is simply CORRECT...

No, it also excaped me, but of course what is correct is:

d/dt <p> = <- dV/dx >

cheers,
Patrick.