Is there any operator behind probability, in quantum mechanics? In Quantum mechanics, the probability of finding a particle at position $x$ is given by $|\psi(x)|^2$, where $\psi$ is the wave function. Wonder what is the operator which gives this probability? Is probability the result of any operator acting on $\psi$?
 A: The answer is negative for two distinct reasons.
(1) In QM operator means linear operator and the map $\psi \mapsto |\psi(x)|^2$ is not linear, evidently.
(2) Wavefunctions are elements of $L^2(\mathbb R)$ and these elements are defined up to zero-measure sets. I mean that, if $\psi(x) \neq \psi'(x)$ for $x\in E$ where $E$ has zero measure, then $\psi=\psi'$ as elements of $L^2(\mathbb R)$, i.e.  (if $\int |\psi(x)|^2 dx =1$) pure quantum states. In particular every set of the form $\{x_0\}$ has always zero measure. Therefore, for any fixed $x_0$, the map $$L^2(\mathbb R) \ni \psi \mapsto |\psi(x_0)|^2$$  makes no sense. It is not defined at all, as a map associating states (if $\int |\psi(x)|^2 dx =1$) with numbers.
A: If one switches to the density operator formalism the situation changes.
Density operators $\rho $ are positive trace class operators with trace norm
$\mathrm{tr}\rho =1$. They not only describe pure states but also mixed ones
and satisfy the Liouville-von Neumann equation
\begin{equation*}
\partial _{t}\rho (t)=-i[H,\rho (t)]:=-iL\rho (t),
\end{equation*}
where $H$ is the Hamiltonian of the system and $[.,.]$ the commutator. The
right hand side defines the Liouville operator $L$ acting on elements of the
trace class. If, in your case, $\psi (x)\in \mathcal{H}$  is square
integrable with unit norm, then
\begin{equation*}
\rho =|\psi ><\psi |,
\end{equation*}
and the probability to find the particle with coordinate in $\mathcal{%
M\subset H}$ is
\begin{equation*}
E_{\mathcal{M}}=\mathrm{tr}P_{\mathcal{M}}\rho =\mathrm{tr}P_{\mathcal{M}%
}|\psi ><\psi |=\int_{\mathcal{M}}dx|\psi (x)|^{2}
\end{equation*}
Here $P_{\mathcal{M}}$ is the projector defined by the characteristic
function $\chi _{\mathcal{M}}(x)$, $\chi _{\mathcal{M}}(x)=1$ for $x\in
\mathcal{M}$ and 0 otherwise. Thus $P_{\mathcal{M}}$ is the operator
associated with the probability in coordinate space. Similarly, let $
\mathcal{N}$ be a set in momentum space and $Q_{\mathcal{N}}$ be projector
define by $\chi _{\mathcal{N}}(p)$. Then
\begin{equation*}
F_{\mathcal{N}}=\mathrm{tr}Q_{\mathcal{N}}\rho =\int_{\mathcal{N}}dp|\tilde{%
\psi}(p)|^{2},
\end{equation*}
where $\tilde{\psi}(p)$ is the Fourier transform of $\psi (x)$, is the
probability to find the particle with momentum in $\mathcal{N}$.
In more abstract approaches, such as the C$^{\ast }$-algebra
formalism, states are defined as positive linear functionals over the set of
observables. Density operators belong to this class but there are more
general ones.
A: "Probability", by itself, is a very squishy concept, and doesn't really make much sense on its own. However, if you attach a statement, as in "probability that X will happen", then you can assign an operator for it.
Let me start with a simple example, a single particle in one dimension, where you want the probability of its coordinate $x$ being between $a$ and $b$. As you know, this can be written as
$$
P(x\in[a,b])=\int_a^b|\psi(x)|^2\text dx.
$$
To bring this into an operator formalism, you express the wavefunction as a braket between the state $|\psi⟩$ and a position eigenstate $|x⟩$, to get
$$
P(x\in[a,b])=\int_a^b⟨\psi|x⟩⟨x|\psi⟩\text dx.
$$
Finally, you can 'factor out' the $\psi$'s to obtain the matrix element of an operator:
$$
P(x\in[a,b])=⟨\psi|\left(\int_a^b|x⟩⟨x|\text dx\right)|\psi⟩=⟨\psi|\hat\Pi_{[a,b]}|\psi⟩.
$$
The operator $\hat\Pi_{[a,b]}$ is called the spectral projector of the position operator $\hat x$ for the set $[a,b]\in \mathbb R$. It has the property that its expectation value in a state $|\psi⟩$ is the probability of $x$ being in $[a,b]$ in that state.
This extends to any well-behaved observable $\hat Q$ and any measurable set of real numbers $A$. In fact, the 'diagonalizability' of self-adjoint operators (including the position and momentum!) is phrased, when derived rigorously in terms of a spectral theorem, exactly in those terms. If $\hat Q$ is self-adjoint (which includes hermiticity as we know it but also some additional constraints on the domains of operators) you are not guaranteed eigenstates but rather a spectral measure. This is a function $\Pi$ which takes sets of real numbers $A$ and returns the corresponding spectral projectors $\Pi(A)$, which have the property that their expectation values are the probability that $q$ is in $A$:
$$
P(q\in A)=⟨\psi|\hat\Pi(A)|\psi⟩.
$$
If $\hat Q$ does have eigenstates, then the spectral projectors are the sum, or integral, of the individual eigenstate projectors $|q⟩⟨q|$ over the $q$'s in $A$. If $Q$ has a continuous spectrum, in fact, the individual projectors $|q⟩⟨q|$ don't make much sense on their own, and must be integrated over to give physical predictions.
This matches up with something that V. Moretti mentioned already. The probability density,
$$|\psi(x)|^2=⟨\psi|\left(|x⟩⟨x|\right)|\psi⟩,$$
is not particularly well-defined because $\psi$ can change its value at single points without affecting the state. This is OK, because probability densities are integrated over to give physical results, and these pointwise changes do not affect the total integral.
However, as it is clear from its form, the probability density is also, at least formally, the expectation value of an operator, which in this case is
$$|x⟩⟨x|.$$
Because of the above-mentioned problems, this operator is not actually that well-defined, and you need to be very careful with your states (and, in particular, restrict yourself to certain parts of a rigged Hilbert space formalism) for this to make sense. You can get a slightly sturdier definition as
$$
|x⟩⟨x|=\frac{d}{dx}\hat \Pi_{(-\infty,x]},
$$
except that now you need to worry about what it means to differentiate in operator space. The takeaway message on this is that the formal manipulations, if done correctly, do work, but if you want to make them rigorous then it gets very messy very fast. So: use your physical intuition for what quantities make sense and which ones don't, follow the  manipulation rules, and you'll be safe.
A: The theory of quantum mechanics has been developed to explain observations, i.e. measurements. Without observations it is a floating mathematical construct.
One of the postulates to connect the mathematics with reality is:
To every observable there corresponds an operator which operating on the state function will give an eigenvalue. So the question becomes : is probability an observable? and then it becomes :what is an observable.
An observable in the framework of quantum mechanics is a variable of the system under consideration which a measurement can evaluate . The energy of a single photon. The momentum of a proton. The spin of an electron. We can always measure these variables  on single particles with one observation, measurement. This is not possible with probability. It is an emergent value from a great number of measurements with the same boundary conditions: it is a normalized distribution, varying from 0 to 1, of the spread of the values found in the measurements. 
So no, there exists no quantum mechanical operator for probability,  as it is not an observable of a variable entering the quantum mechanical problem but an emergent quantity from a lot of measurements.
A: No. These cancerous probabilities come in because of the probabilistic interpretation, which also brings in all kinds of famous paradoxes that infect QM. By itself, the formalism of the theory requires just the solution of a second order differential equation to calculate $\psi(x,t)$ - a process which is completely deterministic, just like solving something coming out of Newton's second law. (I'm talking about the Schr\"odinger equation solution here, Heisenberg's matrix mechanics gives identical answers.) 
The formalism of the theory and this interpretation are absolutely independent issues. One can have another interpretation (e.g. Many-Worlds, besides others that you can find mentioned in the link) tacked on to the same formalism, which would give us a different way of making sense of these answers. But by itself, there is nothing in the formalism of QM that requires or necessitates probabilities. 
So, your answer is - no. Probability is not the result of any operator acting on $\psi$. 
