Hi I'm studying quantum mechanics with Stephen Gasiorowicz's book.
It says for the $n=2$ states of the hydrogen atom, the $l=0$ state has even parity and the $l=1$ state has odd parity.
As far as I know is, the parity transform changes the sign of the spatial component so that $r$ should be $-r$.
And if the sign of the function does not change by this transformation, it is even and if it changes, it is odd.
However, the Schrödinger equations are given by: $$\phi_{200}=(2a_{0})^{-3/2}2(1-\frac{r}{2a_{0}})e^{-r/2a_{0}}Y_{00}$$ $$\phi_{211}=(2a_{0})^{-3/2}3^{-1/2}(\frac{r}{a_0})e^{-r/2a_{0}}Y_{11}$$ $$\phi_{210}=(2a_{0})^{-3/2}3^{-1/2}(\frac{r}{a_0})e^{-r/2a_{0}}Y_{10}$$ $$\phi_{21-1}=(2a_{0})^{-3/2}3^{-1/2}(\frac{r}{a_0})e^{-r/2a_{0}}Y_{1-1}$$
For $l=0$, $\quad \phi_{200}(-r)=(2a_{0})^{-3/2}2(1+\frac{r}{2a_{0}})e^{r/2a_{0}}Y_{00} \neq \phi_{200}(r)$.
For $l=1$, $\quad \phi_{21m}(-r)=(2a_{0})^{-3/2}3^{-1/2}(\frac{-r}{a_0})e^{r/2a_{0}}Y_{1m} \neq -\phi_{21m}(r)$
But why do you say these have even parity and odd parity respectively?
Please correct me if I'm misunderstanding something.
Thank you.