Does phase part get cancelled in taking probability of a 2-particle wave function?

For two classical(distinguishable) particles, the total wave function is given by the product of the individual wave functions. Does this mean that when this total wave function is multiplied with its conjugate to give the probability density, the time-dependence gets totally eliminated? Or am I missing something?

• If the two wave functions are individually time-independent, then the combined wave function will also be time independent. – cobra121 Oct 26 '18 at 15:50

The wavefunction for a system of two distinguishable particles is not necessarily given by the product of two individual wavefunctions. For example, the wavefunction could be $$\psi(x_1,x_2) = (x_1+x_2)e^{-x_1^2}e^{-x_2^2},$$ which does not have the form $$f(x_1)g(x_2)$$ for any $$f,g$$. In words, the two particles can be entangled with each other.
Even if the wavefunction is initially factorized, it will not remain factorized if the particles interact with each other. As an example, consider the usual approximate model of a (spinless) electron and a proton interacting with each other according to the Schrodinger equation $$i\frac{\partial}{\partial t}\psi(x_1,x_2,t) = \left(-\frac{\nabla_1^2}{2m_1}-\frac{\nabla_2^2}{2m_2} + V(x_1-x_2) \right)\psi(x_1,x_2,t),$$ where $$V(x_1-x_2)$$ is their mutual Coulomb attraction. Even if $$\psi(x_1,x_2,t)=f(x_1)g(x_2)$$ when $$t=0$$, this $$V$$ term in the Hamiltonian will cause $$\psi(x_1,x_2,t)$$ to be non-factorized for $$t>0$$.
For two indisinguishable fermions, the wavefunction must be antisymmetric: $$\psi(x_1,x_2)=-\psi(x_2,x_1)$$. For two indistinguishable bosons, it must be symmetric: $$\psi(x_1,x_2)=\psi(x_2,x_1)$$. For two distinguishable particles, no such symmetry is required; but it is still not factorized, in general.
• @AbhirupMukherjee If we set $V=0$ in the Schrodinger equation that I wrote above, and if the initial state is $\psi(x_1,x_2,0)=f(x_1)g(x_2)$, then the state at time $t$ will be $\psi(x_1,x_2,t)=f(x_1,t)g(x_2,t)$ with time-dependent $f,g$. The mod-squared of this, namely $|f(x_1,t)|^2|g(x_2,t)|^2$, will still be time-dependent. The exception is when $f$ and $g$ happen to be eigenstates of the Hamiltonian (energy eigenstates), in which case the mod-squared will be independent of time. – Chiral Anomaly Oct 27 '18 at 16:05