Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Part of a homework question asks to show that for $\ell=0$ in both $\Psi_i$ and $\Psi_f$, we have $$ \int \Psi_i^\ast \vec{r} \Psi_f \; d\tau = 0 $$ for the position vector $\vec{r}$. (This is for the electron in hydrogen and the integral is over all space.) The physical interpretation of this is that since the expectation value is zero, such a transition is forbidden. I am having trouble showing the above integral is zero. Since we are asked to show this in general, and not for a special case, it seems the only thing to do is use orthogonality of the $\Psi$'s. Is this correct? Can someone nudge me in the right direction?

share|improve this question
    
Can someone fix the box after the $r$? I used \vec{} but as I usually do, but it apparently did not render here. –  unit3000-21 Apr 22 '12 at 23:10
1  
Looks fine to me... –  David Z Apr 23 '12 at 3:32
add comment

1 Answer 1

up vote 2 down vote accepted

Unless I'm missing something this is straightforward. If $\ell=0$ the wavefunction is spherically symmetric, so $\Psi_i^\ast \vec{r} \Psi_f$ is antisymmetric and automatically integrates to zero.

share|improve this answer
    
For a (electric-dipole) "forbidden" transition $i$ and $f$ have in general differnt $l$'-s, what matters is that $|l_i -l_|f$ must be exactly 1 for the integral to be non-zero. You have shown the case of $l_i=l_f=0$ only. –  Slaviks Apr 23 '12 at 8:15
    
@Slaviks: that's because the original question asked for a proof when $\ell_i = \ell_f = 0$! –  John Rennie Apr 23 '12 at 9:15
    
You're not missing anything, it really is that simple! However, I couldn't flesh out the details until late last night. I ended up converting to Cartesian coordinates, in which case $\Psi^\ast \Psi$ is even in each of $x,y$ and $z$ since $r \mapsto \sqrt{x^2+y^2+z^2}$. Thus, $\Psi^\ast x \Psi$ is odd in $x$, and similarly for $y$ and $z$. So the integral of each is zero, which means the integral of the original thing is zero. –  unit3000-21 Apr 23 '12 at 19:59
    
@JohnRennie You are totally right, I've not been careful with reading the question. –  Slaviks Apr 24 '12 at 10:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.