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This may sound like a strange question, but just to be sure: Suppose I have a general Hermitian operator in Hilbert space whose action on an eigenvector is given by $R|r\rangle = r|r\rangle$. Then, I assume that the following is true:

$\frac{1}{R}|r\rangle = \frac{1}{r}|r\rangle$ and similarly for other powers $R^{-2}|r\rangle = r^{-2}|r\rangle$

Does this follow immediately from the action of the operator, as in the case $R^2|r\rangle = RR|r\rangle = r^2|r\rangle$ or does this have to be defined?

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  • $\begingroup$ The inverse is already defined, you just have to prove it (e.g. consider the action of R times its inverse on |r>, on the one hand is |r> because the product is the identity, but it is also r R^(-1)|r>) $\endgroup$ Commented Jan 7, 2016 at 19:47
  • $\begingroup$ I don't understand your question. Both identities you ask about follow directly from $R^{-1}$ being the inverse. $\endgroup$
    – ACuriousMind
    Commented Jan 7, 2016 at 20:04
  • $\begingroup$ Observe that $R^{-1}$ might not exist... $\endgroup$
    – Phoenix87
    Commented Jan 7, 2016 at 21:30

1 Answer 1

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If you have $R|r\rangle=r|r\rangle$ then you have, where $I$ is the identity operator,

$|r\rangle=I|r\rangle=R^{-1}R|r\rangle=rR^{-1}|r\rangle$ and you immediately have

$R^{-1}|r\rangle=\frac{1}{r}|r\rangle$

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  • $\begingroup$ Note that this is no a general operator in a general Hilbert space. If $R$ is not compact and normal, it is not warranted to have a complete set of eigenvectors. Also, if there exists a vector with $0$ eigenvalue, then you wouldn't find an $P^{-1}|v\rangle$. $\endgroup$
    – mlainz
    Commented Jan 7, 2016 at 22:18
  • $\begingroup$ This is another case when it fails. Take $H$ to be the space of polynomials of real variable x. If $H p(x) = x p(x)$, there is not $H^{-1} 1$. Note that $H$ is hermitian. $\endgroup$
    – mlainz
    Commented Jan 7, 2016 at 22:38
  • $\begingroup$ Note that $H$ is hermitian, wrt., for example, $\langle p | q\rangle \int_{-1}^{1}p(x)q(x) dx$. $\endgroup$
    – mlainz
    Commented Jan 7, 2016 at 22:44
  • $\begingroup$ Good points, my approach was rather naive, although I was making the assumption that the inverse operator actually exists. At least for the finite dim case we dont have the issue of 1/0 since if 0 is an eigenvalue then the inverse doesn't exist in the first place. $\endgroup$
    – Okazaki
    Commented Jan 7, 2016 at 23:15

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