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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?

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

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  • $\begingroup$ For a pair of distinguishable non-interacting particles, the total wavefunction will be time-independent, I suppose? $\endgroup$ – Abhirup Mukherjee Oct 27 '18 at 15:44
  • $\begingroup$ @AbhirupMukherjee For a pair of disinguishable non-interacting particles, the total wavefunction will remain factorized if it is factorized initially, but it won't be time-independent. The individual factors are still time-dependent, because an individual particle can still move and undergo dispersion, etc, and these things affect the probability density as well as the wavefunction. $\endgroup$ – Chiral Anomaly Oct 27 '18 at 15:59
  • $\begingroup$ Sorry, i meant that the mod square of the wavefunction will not have any time dependence, as opposed to, say, the mod square of the superposition of the first and second eigen states of a Harmonic oscillator which is time dependent due to the cross terms.. Isn't it? $\endgroup$ – Abhirup Mukherjee Oct 27 '18 at 16:00
  • $\begingroup$ @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. $\endgroup$ – Chiral Anomaly Oct 27 '18 at 16:05

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