# Negative powers of operators

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?

• 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>) Commented Jan 7, 2016 at 19:47
• I don't understand your question. Both identities you ask about follow directly from $R^{-1}$ being the inverse. Commented Jan 7, 2016 at 20:04
• Observe that $R^{-1}$ might not exist... Commented Jan 7, 2016 at 21:30

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
$R^{-1}|r\rangle=\frac{1}{r}|r\rangle$
• 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 $P^{-1}|v\rangle$. Commented Jan 7, 2016 at 22:18
• 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. Commented Jan 7, 2016 at 22:38
• Note that $H$ is hermitian, wrt., for example, $\langle p | q\rangle \int_{-1}^{1}p(x)q(x) dx$. Commented Jan 7, 2016 at 22:44