Wavefunction of many particles Suppose that we have $N$-many identical particles, whose space-spin-coordinates are given by $x_{1}, x_{2},...x_{n}$ and whose composite system is represented by $|\Psi\rangle$. Then, according to the textbook, the wavefunction of these particles can be represented by:

$$\Psi(x_{1}, x_{2}, ... x_{n}) = \langle x_{1}, x_{2}, ... x_{2}|\Psi
 \rangle$$

What is the physical meaning of this? Whence the inner product of $|x_{1}, x_{2},... x_{n}\rangle$ and $|\Psi \rangle$? Why take the inner product to get the wavefunction?
I know it's a very elementary question, but I want to understand the physical meaning of the equation. (Also, I know this is not particularly about many-particle systems, but it's just what I happen to be looking at right now.)
 A: In very informal terms (unless you consider tensor products of rigged Hilbert spaces), the vector $|x_1,\ldots,x_n\rangle$ is the element of $\bigotimes^n H$ given by
$$|x_1\rangle\otimes\cdots\otimes|x_n\rangle$$
That is, for each particle you have a copy of $H$. Note that the statistics of the problem will reduce $\bigotimes^n H$ to a subspace. So, for bosons, you will have the subspace $S$ given by the closure of the span of all the vectors generated by symmetric tensor products of one-particle states.
Now, in order to grasp the meaning of that "inner product", suppose that your global state $|\Psi\rangle$ is the product of one-particle states, that is $|\Psi\rangle = |\psi_1\rangle\otimes\cdots\otimes|\psi_n\rangle$. The inner product gives
$$\Psi(x_1,\ldots,x_n) = \psi_1(x_1)\cdots\psi_n(x_n).$$
More generally, the expression of $\Psi$ will be an $L^2$-integrable function that can be approximated arbitrarily well with linear combinations of functions like the above.
A: You should think of $|\Psi\rangle$ as a vector and $\Psi(x)$ as the vector components. Compare this to a regular vector $\vec v$. If you have a basis $\{\vec e_i\}$ then you can decompose $\vec v$ into vector components: $\vec v=\sum_i v^i\vec e_i$. If you change the basis you get different components but the overall vector stays the same. If you call the new basis $\{\tilde e_i\}$ then $\vec v=\sum_i \tilde{ v}^i\tilde {e}_i=\sum_i v^i\vec e_i$. In the same way $|\Psi\rangle$ is a basis independent quanty while $\Psi(x)$ is in the position basis.
If $\{\vec e_i\}$ forms an orthonormal basis you can extract the components of $\vec v$
$$v_i=\vec e_i\cdot \vec v$$
In quantum mechanics this works the same, except that now the basis vectors also represent states. Also the dot product is extended to functions, so the discrete index can be replaced by a function argument: $\Psi(x)=\langle x|\Psi\rangle$. Then $|x_1,...,x_n\rangle$ just corresponds to the state where the first particle is at $x_1$, the second one at $x_2$ etc. (But you should note that because of the uncertainty principle this doesn't represent a physical state, it is just a mathematical tool)
