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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?

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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
Looks fine to me... – David Z Apr 23 '12 at 3:32
up vote 3 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.

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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

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