# A very quick question

1. Dec 22, 2009

### andyc100

Does $$H(\underline{r})=H\psi(\underline{r})$$ ?

2. Dec 22, 2009

### andyc100

...in my text book it says that $$P\psi(\underline{r})=\psi(\underline{-r})$$ where P is the parity operator.

and that $$H(\underline{r})=H(\underline{-r})$$

Thus $$P\psi(\underline{r})=p\psi(\underline{r})$$ where p is the parity eigenvalue.

Im having difficulty getting to this myself. Could some one please show me how?

3. Dec 23, 2009

### dextercioby

This makes no sense. In the lhs you have an operator and in the rhs you have a vector. Also, in you second post, the <conclusion> is actually an assumption. The assumption is that P is an operator in Hilbert space for which the spectral equation makes sense.