Query on an operator acting on a function I have a naive question about an operator acting on a well-behaved  function. Let us say, we are talking about space translation operator acting 
on a function $\psi(x)$:
$$\hat{T(a)}\psi(x)=\psi(\hat{T}(x))=\psi(x+a)$$
For a rotation operator $\hat{R}$, the equation becomes:
$$\hat{R}\psi({\bf{r}})=\psi(\hat{R}({\bf{r}}))=\psi(\hat{R}{\bf{r}})$$
I saw these in text books. My question is that acting the operator directly on the argument ($x$ in first case and $\bf{r}$ in second case) of the function, is it a property of these linear operators? I ask this because, in my studies (Demkov 1971) on inversion in a sphere of radius $a$, I see that
$$\hat{M}\psi({\bf{r}})=\frac{a}{r}\psi(\hat{M}({\bf{r}}))=\frac{a}{r}\psi(\frac{a^2}{r^2}{\bf{r}})$$
Here, the said inversion has been denoted by $\hat{M}$. Demkov shows that $\rm{Schr\ddot{o}}dinger$ equation can be inverted w.r.t. a sphere with a transformed Hamiltonian and the inverted wave function. Here, 
$$\hat{M}\psi({\bf{r}})\neq\psi(\hat{M}({\bf{r}}))$$
Can anyone please tell me about the correct procedure? 
 A: I believe that the reason is to make the operator $\hat M$ unitary. The $\frac{a}{r}$ comes from the Jacobian of thee transformation $$\mathbf r \mapsto \frac{a^2}{r^2} \mathbf r.$$ This transformation is $$(r, \varphi, \theta)\mapsto (\frac{a^2}{r}, \varphi,\theta)$$
so clearly $r^2 \, dr = -a^2dr'$. This minus sign will be canceled by the inversion changing the orientation. Now consider $$ \langle \hat M \psi| \hat M \psi\rangle =\int r^2 dr\, d\cos\theta\, d\varphi \, (\hat M \psi) (\hat M \psi^*) = \int r^2 dr\, d\cos\theta\, d\varphi \frac{a^2}{r^2} \psi(\frac{a^2}{r^2} \mathbf r) \psi^*(\frac{a^2}{r^2}\mathbf r).$$
$$ = \int dr'\, d\cos\theta\, d\varphi\, \frac{a^4}{r^2} \psi(\mathbf r')\psi(\mathbf r').$$
Since $a^4/r^2 = r'^2$ the operator $\hat M$ is unitary.
A similar prefactor does not appear for rotations and translations since their Jacobians are unity (that is, they preserve volume).
A: I regret to write again in this thread. But I could not agree finally with the explanations that I had received. Given that the operator $\hat{M}$ transforms 
$\psi$ to $\hat{M}\psi$, $\langle\hat{M}\psi|\hat{M}\psi\rangle$ must be found 
in the transformed coordinate $\bf{r}'$. Thus, the inner product should be given by:
$$\langle\hat{M}\psi|\hat{M}\psi\rangle=\int{r'^2 dr' d(cos\theta')d\phi'(\hat{M}\psi)^*(\hat{M}\psi)}$$
Then, the above argument (that it is done to make the transformation unitary) does not seem quite correct to me. As I have told, the geometric transformation 
is not orthogonal.
To me, the factor seems more due to some general rule of transformation (with Jacobian involved, as suggested by Robin Ekman). If we agree that the wave function is like a probability amplitude, then $|\psi|^2$ is a density function. Now, for an one-to-one (bijective) transformation, there exists a specific rule: "density function in transformed coordinate = $|J|\times$ density function in previous coordinate"
(See: http://pluto.mscc.huji.ac.il/~mszucker/ASY/jac.pdf)
According to this rule and the fact that $|J|=\frac{a^2}{r^2}$ for inversion: ${\bf{r}}\rightarrow{\bf{r'}}=\frac{a^2}{r^2}{\bf{r}}$, we have
$$|\psi'({\bf{r'}})|^2=\frac{a^2}{r^2}|\psi({\bf{r'}})|^2\implies\psi'({\bf{r'}})=\frac{a}{r}\psi({\bf{r'}})$$
Please let me know if you find any inconsistency in this argument.
