Wave function of many particle I read that the wave function of a system of many particles is formed from the product of the wavefunctions of the individual particles. What is the logic behind it?
 A: What you've read is inaccurate (at least in the form you're reporting it). The wavefunction of a many-particle system can be the product of single-particle wavefunctions for the individual components, but it doesn't have to be.
In general, adding particles works exactly like adding dimensions to the space of a given particle, i.e. like lifting the 1D approximation and going up to the real three-dimensional position of the particle in space, which basically looks like
$$
\text{going from }\Psi(x) \text{ to } \Psi(x,y,z).
$$
To keep things simple, if you go from one particle in 1D with position $x$ to having two particles, still in 1D, with positions $x_1$ and $x_2$, then it looks exactly the same:
$$
\text{going from }\Psi(x) \text{ to } \Psi(x_1,x_2).
$$
(Now, if the particles are bosons or fermions, then that imposes some requirements on the symmetry of $\Psi(x_1,x_2)$, namely that $\Psi(x_2,x_1) = (-1)^s \Psi(x_1,x_2)$, but that doesn't change things much.)
The problem with this, of course, is that while a wavefunction of the form $\Psi(x_1,x_2)$ might look innocent, when you actually start calculating things, it gets really ugly really fast. ODEs turn into PDEs. Integrable systems turn into non-integrable systems. Analytical solutions vanish into mist. Numerical algorithms that run in seconds now take hours. (And if you add a third particle, they take months. If you add a fourth, they take thousands of years.)
So, how can you simplify the handling of this ugly beast? Well, the same way that we simplified the handling of three-dimensional problems: by restricting our attention to wavefunctions with specific characteristics that are false in general, but which simplify the handling, and hoping that we'll be able to find enough of them to build a strong-enough framework to take on the general case. So, in the same spirit of looking for separable solutions, we look first for solutions of the form
$$
\Psi(x_1,x_2) = \psi_1(x_1)\psi_2(x_2).
$$
(And if they're bosons or fermions we symmetrize or antisymmetrize, but it's not important here.)
So, that's how you get global wavefunctions that are products of individual wavefunctions: by hoping really hard that they'll work as solutions to your problem. 


*

*Sometimes this works really well, as in Hartree-Fock methods, where you stick to single products, you try to find optimal ones, and you quite often find states that work really well. 

*Sometimes it works OK, like when you build molecular states as finite linear combinations of product states, normally involving a reasonable amount of numerical diagonalization.

*Sometimes it completely fails, and you need to reach into the bag of hard tricks that are post-Hartree-Fock methods to handle the fully entangled wavefunction $\Psi(x_1,x_2)$ in situations where molecular-orbital pictures fail or when you have highly correlated systems.


It should go without saying, but when the separable states fail to be good descriptions of your system, that's a sign to strap in because things are going to get really hard. But then again that's where a lot of interesting physics is to be found nowadays.
A: If the particles are not interacting then the Schrodinger equation for the 2-particle system will be
\begin{align}
H&=H_1+H_2\nonumber \\
H\Psi(x_1,x_2)&=
\left(-\frac{\hbar^2}{2m_1}\frac{\partial^2}{\partial x_1^2}+V_1(x_1)-
\frac{\hbar^2}{2m_2}\frac{\partial^2}{\partial x_2^2}+V_2(x_2)\right)\Psi(x_1,x_2)
=E\Psi(x_1,x_2)
\end{align}
Using the usual separation ansatz $\Psi(x_1,x_2)=\psi(x_1)\phi(x_2)$ one then obtains a pair
of independent Schrodinger equations
\begin{align}
H_1\psi(x_1)&=E_1\psi(x_1)\, ,\\
H_2\phi(x_2)&=E_2\phi(x_2)
\end{align}
with $E_1+E_2=E$.  This by assumption gives a product form.  Note that the solutions are multiplicative but the eigenvalues (i.e. the energies) are additive.
Of course if there is an interaction term $V_{12}(x_1,x_2)$ then separability is lost, although one can use the factored form as a basis set if $V_{12}$ can be treated as a perturbation.
If the particles are indistinguishable, then the solutions $\psi(x_1)\phi(x_2)$ must be additional symmetrized:
$$
\Psi_{\pm}(x_1,x_2)\sim \psi(x_1)\phi(x_2)\pm \psi(x_2)\phi(x_1)
$$
with the $+$ and $-$ signs applicable to bosons and fermions, respectively.
A: This question is not answered in standard textbooks. To me, this signals that standard quantum mechanics doesn't ask 'What is the wave function for a many-particle quantum system?' I read on the internet various work-arounds for this, but to me they seem not to really tackle the principled question 'what is the quantum mechanical general principle for constructing the wave-function for a many-particle quantum system?'
