Is it possible to build eigenstates of linear combinations of $\hat{P}$ and $\hat{X}$? For the quantum harmonic oscillator, the position operator $\hat{X}$ has eigenstates saisfying $\hat{X}|x\rangle = x | x \rangle$. The momentum operator meanwhile acts like $\langle x | \hat{P} | \Psi \rangle = - i \frac{\partial}{\partial x} \langle x | \Psi \rangle$ in this eigenbasis.
My question is if we define some operator $\hat{O} = \hat{X} + \lambda \hat{P}$ (with $\lambda$ some dimensionful quantity), if it is possible to define eigenstates of $\hat{O}$? Something like $\hat{O} | o \rangle = o | o \rangle$ with $o$ built out of $x$ and $p$ eigenvalues possibly.
My guess is that you cannot do this, but I would like to understand this better.
 A: The position operator of a free particle in the Heisenberg picture is a nice example ($\hbar=1$): $$X(t)= e^{iP^2 t/2m} X e^{-i P^2 t/2m} = X+ Pt/m$$ The eigenstates of $X(t)$ are given by $|x;t \rangle := e^{iP^2 t/2m} |x\rangle$, as $$(X+Pt/m) |x;t \rangle = X(t) e^{iP^2 t/2m}| x \rangle = e^{iP^2 t/2m} X | x \rangle = x  e^{iP^2/2m} |x \rangle = x |x; t\rangle.$$ As $X$ and $X+Pt/m$ are unitarily equivalent, the spectra of $X$ a and $X+Pt/m$ are, of course, identical (namely $\mathbb{R}$).
This delivers at the same time the general solution of your question for $\lambda \in \mathbb{R}$. Replacing $t/m$ by  the parameter $\lambda$, the unitary operator $U(\lambda):= e^{-iP^2 \lambda /2}$ maps the operator $X$ into $X+\lambda P$  via $U^\dagger(\lambda) X U(\lambda) = X+ \lambda P$ with eigenstates (better: eigendistributions) $|x; \lambda \rangle = U^\dagger(\lambda) |x\rangle$, i.e. $(X+\lambda P) |x;\lambda \rangle = x |x; \lambda \rangle$.
A: 
My question is if we define some operator $\hat{O} = \hat{X} + \lambda \hat{P}$ (with $\lambda$ some dimensionful quantity), if it is possible to define eigenstates of $\hat{O}$? Something like $\hat{O} | o \rangle = o | o \rangle$ with $o$ built out of $x$ and $p$ eigenvalues possibly.

Yes, it is possible.
For example (with $\hbar=1$):
$$
\psi(x) = Ae^{\frac{i}{2\lambda}(x - x_0)^2}\;,
$$
satisfies:
$$
\left(\hat X + \lambda \hat P\right) \psi = x_0 \psi
$$

A typical example of when such a linear combination of $\hat X$ and $\hat P$ occurs is in the usual "ladder" analysis of the simple harmonic oscillator, where lowering and raising operators are typically presented as:
$$
a \propto \hat X + \frac{i}{m\omega}\hat P
$$
$$
a^\dagger \propto \hat X - \frac{i}{m\omega}\hat P\;,
$$
respectively.
Eigenstates of the lowering operator are called "coherent states."
