How do we translate a two particle system in bra-ket notation into a wavefunction as a function of the two particle positions? Consider the two particle system given by the following bra-ket notation
$$| \psi _1 , \psi _2 \rangle $$
where $\psi_1, \psi_2$ each describe a particle. I then want to apply the projector $\langle x \rvert$ - or some other projector, to find $\psi (x_1, x_2 )$.
Is the following true:
$$\langle x |\psi_1 , \psi _2 \rangle = \psi (x_1, x_2 ) \, ,$$
or do I need two projectors $\langle x_1 \rvert$ and $\langle x_2 \rvert$, or am I horribly off base with any of this?
 A: Given some single-particle Hilbert space $\mathcal H$ (e.g. $L^2(\mathbb R)$), the generalized position eigenvectors $|x\rangle$ form a continuous basis of the space.  Therefore, the identity operator takes the form $\mathbb I = \int dx |x\rangle\langle x|$, and any state $|\psi\rangle\in\mathcal H$ can be expanded as
$$|\psi\rangle = \mathbb I |\psi\rangle = \int dx |x\rangle\underbrace{\langle x|\psi\rangle}_{\equiv \psi(x)} = \int dx\ \psi(x) |x\rangle$$
We can construct a two-particle Hilbert space by stitching two copies of $\mathcal H$ together to form the tensor product space $\mathcal H^2 = \mathcal H \otimes \mathcal H$.  Given any choice of basis $\{\hat e_i\}$ for $\mathcal H$, the set $\{\hat e_i \otimes \hat e_j\}$ forms a basis for $\mathcal H^2$.
Therefore, the $|x\rangle$'s form a basis for $\mathcal H$, but not for $\mathcal H^2$.  If you want a basis for the latter, you need objects of the form $|x\rangle \otimes |y\rangle \equiv |x,y\rangle$.  The identity operator on $H^2$ then takes the form
$$\mathbb I = \int dx dy |x,y\rangle\langle x,y|$$
and a generic state $|\Psi\rangle \in \mathcal H^2$ can be expanded
$$|\Psi\rangle = \mathbb I |\Psi\rangle = \int dx dy |x,y\rangle\underbrace{\langle x,y|\Psi\rangle}_{\equiv \Psi(x,y)} = \int dx dy \Psi(x,y) |x,y\rangle$$
So it's not quite that you need two projectors, but rather that you need one projector which is taken from a complete basis for the space - which takes the form of a tensor product of two single-particle states.
A: You didn't say whether you had identical particles, for which you also have to take into account whether they are fermions or bosons. In this case you must act with 
$$\langle x;y| = \frac{1}{\sqrt 2}(\langle x| \langle y| \pm \langle y| \langle x|)$$ 
and your state has the form 
$$ |f;g\rangle =  \frac{1}{\sqrt 2}(|f\rangle |g \rangle \pm |g\rangle |f )\rangle$$
Then 
$$\langle x;y|f;g\rangle = \langle x|f\rangle \langle y|g\rangle \pm \langle x|g\rangle \langle y|f\rangle = f(x)g(y) \pm g(x)f(y)$$
More generally you may have an entangled state in which the wave function cannot be factorised.
A: The vector you have present is a direct product of vectors from two closed Hilbert spaces. It is of the form: $$| {\psi_1,\psi_2}\rangle= |{\psi_1}\rangle\otimes |{\psi_2}\rangle$$
Thus naturally any basis you’d want to express them in must also necessarily be a product of basis of two closed Hilbert spaces. As such:$$|{x_1,x_2}\rangle= |{x_1}\rangle\otimes |{x_2}\rangle$$
A: I hope I can clear up your conceptual confusion a little bit.
As base vectors we choose $|x_1, x_2\rangle$,
meaning that one particle is at position $x_1$,
the other at position $x_2$.
Using these base vectors we can compose
any arbitrary two-particle state $|\psi\rangle$
(I prefer to call it just $|\psi\rangle$ instead of $|\psi_1,\psi_2\rangle$):
$$|\psi\rangle=\iint\psi(x_1,x_2)|x_1,x_2\rangle dx_1\ dx_2$$
The above equation can be reversed 
to get the wave-function $\psi(x_1,x_2)$ of the state $|\psi\rangle$:
$$\psi(x_1,x_2)=\langle x_1,x_2|\psi\rangle$$
