Can the wavefunction of a free particle always be written as some linear combination of more basic wavefunctions? In various explanations I've read, from what I've gathered all particles have a wavefunction $\Psi(\mathbf{r},t)$ where $\mathbf{r}$ is the cartesian coordinates in however many dimensions you're working in. So in regular 3d space the wavefunction is $\Psi(x,y,z,t)$.
My first question is under what conditions is the wavefunction of a particle a pure quantum state? If I take an electron and completely isolate it from the rest of the universe will it be in a pure quantum state? Is this even possible and how does it relate to the eigenfunctions that appear when you act with some operator on the wavefunction? For example when the momentum operator or the position operator act on a wavefunction we get:
$$\hat{p} = p_1_1 + p_2_2 ... + p_n_n$$
and
$$\hat{x} = x_1_1 + x_2_2 ... + x_n_n$$
Are these functions $_n$ the same for momentum and position? Does $_1$ in the momentum equation $= _1$ in the position equation? And what form do these functions have, I understand that a plane wave can be written in the form $e^{i(kx-\omega t)}$, but which of these functions are in this form? Are the eigenfunctions in this form or is the total wavefunction $\Psi$ in this form or are neither of them?
Sorry if this is a very convoluted or badly worded question, I'll happily clarify anything that doesn't make sense. I just have a very hard time connecting things in my head and unless I can make sense of this I don't feel I'll ever have a satisfactory understanding of this topic.
For reference I'm a 3rd year chemistry student so my math/physics understanding is nothing impressive but it's not totally non-existent.
Thanks for any responses in advance, I really do appreciate it.
 A: I'll try to answer your questions in the same order they were asked.

Absolute purity is impossible in real world. However, if you limit the domain of your experiment, and look for the properties of particle(s) at specific time- and distance- scale, it is possible to create a state that behaves similarly to the pure state. (In other words, if your experiment lasts very short amount of time, and happens in limited amount of space - you can minimize the effects of the external world on your particle).

Functions $\psi_i$ are different for $\hat{x}$ and $\hat{p}$ because of the uncertainty principle. These wavefunctions are eigenstates of these operators, and since they don't commute - they can't be the same.

Momentum eigenstates have a plane wave form indeed, while position eigenstates have a form of a delta-function. This is very natural when the wavefunction is viewed as a probability distribution. The eigenstate of $\hat{x}$ is a particle that sits at fixed position. Thus there is a sharp peak at that specific position in the wavefunction and zeroes everywhere else.

As for the rest, what you're looking for is called Completeness Relation. One can decompose a wavefunction into a sum of finite (or infinite) number of eigenstates of a particular operator. The operator could be $\hat{x}$, $\hat{p}$, or even the Hamiltonian $\hat{H}$ itself.
A: A basis is a basis so if you have a complete set of functions you can use it to expand anything you want, including free particle wavefunctions.  
Why you'd want to do this is another matter: this presumably would depend on the physics of your problem.  After all, choosing one basis set rather than another - say spherical rather than Cartesian - is usually done because some features are more naturally obvious in one basis set rather than in another.
A: First of all, wavefunctions don't describe particles, they describe systems. For example if you have a 2 particles system (say two electrons in Helium) the wavefunction describes the system.
Now if your system consists of a single particle and a Hamiltonian $\hat{H}$, then $\Psi _n$ is a pure state if:
$$ \hat{H} \Psi _n = E \Psi _n$$
and $\Psi$ can not be expressed as a linear combination of other pure states $\Psi _l$; in other words $\Psi$ is a state of definite energy.
For the second part of your question lets consider a general operator $\hat{A}$:
If $\hat{A}$ is the operator associated with the observable $A$, then:
$$\hat{A} \psi _n = A_n \psi_n $$
means that $A_n$ (the eigenvalues of $\hat{A}$) are the only possible results of a measurement of $A$. $\psi$ (the eigenfunctions of $\hat{A}$) are the states of definite $A$ (in your examples $A$ is position and momentum).
The set of eigenfunctions $\psi _n$ have the property that $\Psi$ can be expressed as a linear combination of the states of definite $A$.
$$\Psi = \sum _n a_n \psi_n $$ 
$$ \hat{A}\Psi = \sum _n a_n A_n\psi_n $$
If you have different operators you would have different eigenfunctions, but you still can express $\Psi$ as a linear combination of these other eigenfunctions.
