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In Shankar's QM book pg. 261, he said that in classical mechanics we can distinguish between identical particles by observing their histories (or previous trajectories), which are non-identical.

Why can't we do the same for identical particles in QM? We can use the Schrodinger equation to trace back the histories of each identical particle and hence distinguish them. Sure we can't pinpoint the exact trajectories of the particles, but we can distinguish them by using the state vector $|\psi(t)\rangle$. For example, particle 1 will have a history of $|\psi_1(t)\rangle$ and particle 2 will have a history of $|\psi_2(t)\rangle$.

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  • $\begingroup$ Recommended reading: pp. 1371-1377 of Cohen-Tannoudji, et. al. Quantum Mechanics Systems of Identical Particles. books.google.co.in/… $\endgroup$ Commented Jun 11, 2022 at 9:28

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When you write $| \psi_1(t) \rangle$ or $| \psi_2(t) \rangle$, in the position representation you are actually referring to one-particle wavefunctions $\psi( {\bf r}_1,t)$ and $\psi( {\bf r}_2,t)$. Actually, the exact wavefunctions for two indistinguishable particle is a symmetric or antisymmetric function of two space variables ($\psi( {\bf r}_1,{\bf r}_2,t)$). Only at a separation between the two particles larger than the typical spatial extension of the isolated particle wavefunctions can one approximate the two-particle wavefunction with a properly symmetrized or antisymmetrized product of one-particle wavefunctions. In any case (large or small distance), the probabilistic description of the particle positions hampers the possibility of assigning an individual label at all times on the base of the initial conditions.

The difference with the classical case stems from the impossibility of a symmetrized or antisymmetrized wave function to maintain a dynamic distinction between the two particles.

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