Using Wavefunction Parity to Simplify Integrals I have seen that $\langle \phi_{200}|r| \phi_{100} \rangle =0$ due to parity.
while on the other hand, $\langle \phi_{210}|r| \phi_{100} \rangle$ cannot be determined by the parity.
Can anyone explain please what does this mean?
Note: I know that the parity for a wavefunction is given by $P=(-1)^l$, and it implies the wavefunction is even if $P=1$, and odd if $P=-1$, and $r$ is considered as an odd function.
 A: The first thing to note is that the position vector is odd under parity, This can be expressed as
$\Pi^{\dagger}r\Pi = -r$ (1)
Then I can consider the following
$\langle\phi_{n'l'm'}|r|\phi_{nlm} \rangle$
Now let me denote $\phi_{nlm}$ as $| nlm \rangle $, Then I have
$\langle n'l'm'|r|nlm \rangle $
or equivalently
$- \langle n'l'm'|-r|nlm \rangle $
Now plugging in equation (1)
$- \langle n'l'm'|\Pi^{\dagger}r\Pi|nlm \rangle $ (2)
Now the second important point to note is that $\phi_{nlm}$ is an eigenstate of parity with eigen-value $(-1)^l$
$\Pi|nlm \rangle $ = $(-1)^l |nlm \rangle $
Plugging this in I get
$- \langle n'l'm'|(-1)^{l'}r(-1)^l|nlm \rangle $
$(-1)^{1+l'+l} \langle n'l'm'|r|nlm \rangle $
But this is equal to the original
$\langle n'l'm'|r|nlm \rangle  = (-1)^{1+l'+l} \langle n'l'm'|r|nlm \rangle$
Now since both sides must be equal either $l'+l = odd$ or $\langle n'l'm'|r|nlm \rangle = 0$
You have $\langle\phi_{200}|r|\phi_{100} \rangle$, since $0+0$ = even , hence $\langle\phi_{200}|r|\phi_{100} \rangle = 0$ by parity
whereas for $\langle\phi_{210}|r|\phi_{100} \rangle$ , $1+0$ = odd, hence it does not equal $0$ by parity and you have to actually find it.
See Griffiths chapter 6, page 314 for a more complete description of this.
