While considering a multinuclear system, I ignore the movement of nucleus and only consider the electrons to be movable. All electrons are identical. The wavefunction of the system must will depend on the spin coordinates and space coordinates of all the electrons. If the electrons are identical, why does the wavefunction becomes negative if we exchange coordinates of any 2 electrons? I thought it should be unchanged.
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2 Answers
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The wavefunction is not observable. The observable is the electron density. Thus it is not the wavefunction but the square modulus of the wavefunction that must remain unchanged on interchange of particles. As such assuming the factor introduced on particle interchange is real it must be either +1 (baryons) or -1 (fermions). This phase factor results in different statistical behaviour of the two classes of particles, The Pauli exclusion principle being at the heart of all chemistry for example, and Bose-Einstein condensates only occur for baryons. So while we can't directly observe the wavefunction, we can observe the results of these statistics. Experiment shows that both behaviours occur, and that electrons are fermions.
But for the interested who don't like this assumption there's also anyons
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It is an empirical fact. We know that the Pauli exclusion principle is true from experiments, which means that electrons or more generally wavefunctions of fermions have to change their sign under the parity transformation. A physical model of reality should respect this fact, which is why valid electronic wavefunctions are constructed to change their sign upon coordinate inversion, which can be achived by using Slater determinants for example.
