Is it wrong to assume that $|{-\vec r}⟩ = - |{\vec r}⟩$? I found a problem that says the following: 

The 'even' operator is defined as: $$\Pi|{\vec r}\rangle = |{-\vec r}\rangle$$ Show that $\Pi$ is Hermitian and find $\Pi^2$ and its eigenvalues.   

All of my work is based on the assumption that I can write the following:$$\Pi|\vec{r}\rangle = -|\vec{r}\rangle$$ but I'm not sure that it's right. If it isn't, and it's incorrect to write that $|{-\vec r}\rangle=-|\vec{r}\rangle$, then I have no idea how to solve the problem, since I have no representation of the $\Pi$ operator.
 A: 
All of my work is based on the assumption that I can write the following:$$\Pi|\vec{r}\rangle = -|\vec{r}\rangle$$

This is wrong.


*

*$|{-\vec r}\rangle$ is an eigenvector of the position operator $\hat{\vec x}$ with eigenvalue $-\vec r$.

*$-|{\vec r}\rangle$ is an eigenvector of the position operator $\hat{\vec x}$ with eigenvalue $+\vec r$, since
$$
\hat{\vec x}\left(-|{\vec r}\rangle\right)
= -\hat{\vec x} |{\vec r}\rangle
= -\vec r|{\vec r}\rangle
= {\vec r}\left(-|{\vec r}\rangle\right)
.
$$
The eigenvalue is uniquely determined by the eigenvector, so the only way those two things can be true is if both eigenvalues are equal, i.e., if $-\vec r=\vec r$ (which itself implies $\vec r=\vec0$).

With this in hand:

then I have no idea how to solve the problem, since I have no representation of the $\Pi$ operator.

Yes you do $-$ you just specified
$$\Pi|{\vec r}\rangle = |{-\vec r}\rangle, \tag 1$$
which uniquely pins down the action of that operator. If you want to turn that into a specific representation in the sense of something to put on the right-hand side of an equation that starts with $\Pi= ...$, then you can put a resolution of identity on the right of $\Pi$:
\begin{align}
\Pi
& = \Pi \, \mathbb I
= \Pi \int \mathrm d\vec r \, |\vec r\rangle\langle \vec r |
=  \int \mathrm d\vec r \, \Pi|\vec r\rangle\langle \vec r |.
\end{align}
It's your job to keep working the problem from there: your specification $(1)$ gives you a specific path ahead, and then if you want to calculate $\Pi^\dagger$ you can just take the conjugate inside the integral, which you can then manipulate (using a suitable change of variables) into a form that you can match against $\Pi$ itself.

Finally, to tackle $\Pi^2$, first calculate its action on $|{\vec r}\rangle$, and then repeat the jig taken above, which should produce an extremely simple expression. You should be able to read off the eigenvalues of $\Pi^2$ directly from the result of that calculation.
A: A nice way to see what's happening is to construct an approximate "toy" state
$\newcommand{\ket}[1]{\left|#1\right>}
\ket r
$. For instance you might construct a wavefunction that's "at" a location ten units to the right of your origin by
$$
\ket{10}=\left\{
\begin{array}{cl}
1 & \text{where } 9<x<11
\\
0 & \text{elsewhere}
\end{array}
\right.
$$
You can adjust the height if you want to normalize so that $\left<r|r\right> =1$, and adjust the width if you want a more precise version of "at."  Now, how would you construct $\ket{-10}$, ten units to the left of the origin? What about $-\ket{10}$? They are different.
