Why the wavefunction of a set of non-interacting particles is a product of the individual wavefunctions of each particle in the set?

Question

In Quantum mechanics, before we get to Slater determinants and all that, it is said that many-particle wavefunction for the case of non-interacting particles is just a product of all the particles' individual wavefunctions. There is certainly a good reasoning to assume so: since the particles are independent, they are statistically uncorrelated.
In statistics, for several uncorrelated events, the total probability is the product of every event's probability. So, in that sense, product of wavefunctions seems like a very reasonable step. What I want to ask is: is it okay to just assume that? I think in statistics it is just one of the basic principles, and it is not to be proven. Is this the case here as well? Or is there a way to show mathematically that the many-particle wavefunction is the product?

ZeroTheHero · Answer

If the particles are not interacting then
begin{align}
H=H_1+H_2:= H_1otimes mathbb{I}+mathbb{I}otimes H_2
end{align}
and $vert psi_1ranglevertphi_2rangle$ is immediately an eigenstate with energy $E=E_1+E_2$.  Alternatively, by separation of variables, $H_1$ acts only on $x_1$ and $H_2$ on $x_2$ so that $Psi(x_1,x_2)=psi(x_1)phi(x_2)$ is also immediately a solution:
begin{align}
left(-frac{hbar^2}{2m}partial^2_{x_1}+V(x_1)right)psi(x_1)phi(x_2)
&= H_1psi(x_1)phi(x_2)=E_1psi(x_1)phi(x_2), ,
left(-frac{hbar^2}{2m}partial^2_{x_2}+V(x_2)right)psi(x_1)phi(x_2)&=
H_2psi(x_1)phi(x_2)=E_2psi(x_1)phi(x_2), ,
(H_1+H_2)psi(x_1)phi(x_2)&=(E_1+E_2)psi(x_1)phi(x_2)=Epsi(x_1)phi(x_2)
end{align}

Why the wavefunction of a set of non-interacting particles is a product of the individual wavefunctions of each particle in the set?

One Answer

Add your own answers!

Ask a Question