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A very quick question

  1. Dec 22, 2009 #1
    Does [tex]H(\underline{r})=H\psi(\underline{r})[/tex] ?
  2. jcsd
  3. Dec 22, 2009 #2
    ...in my text book it says that [tex]P\psi(\underline{r})=\psi(\underline{-r})[/tex] where P is the parity operator.

    and that [tex]H(\underline{r})=H(\underline{-r})[/tex]

    Thus [tex]P\psi(\underline{r})=p\psi(\underline{r})[/tex] where p is the parity eigenvalue.

    Im having difficulty getting to this myself. Could some one please show me how?
  4. Dec 23, 2009 #3


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    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.
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