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The reason fermions and bosons behave differently is the spin-statistic theorem, a rather subtle result from relativistic quantum field theory that does not hold in all of quantum mechanics and crucially requires special relativity.

  1. The issue is actually not about fermions being in the same place (quantum objects do not have a well-defined place), but about being in the same quantum mechanical state. Any system of $N$ particles in quantum mechanics is described by a state that can be decomposed into sum of products of states of the individual particles, and in the simplest case, the non-entangled one, it just looks like $$ \lvert \psi_\text{total}\rangle = \lvert \psi_1\rangle_1 \otimes \lvert \psi_2\rangle_2\otimes\cdots\otimes \lvert \psi_N\rangle_N.$$ On these states, it makes sense to define an "exchange operator" $S_{ij}$ that switches the state of the $i$-th and $j$-th particle, e.g. $$ S_{12} (\lvert \psi_1\rangle_1\otimes \vert \psi_2\rangle_2) = \lvert \psi_2\rangle_1\otimes \lvert \psi_1\rangle_2.$$ In terms of wavefunctions, this amounts to having a wavefunction with $N$ arguments and swapping the $i$-th and $j$-th argument. If the particles are "indistinguishable", then that means that this exchanged operator should do nothing to the state, i.e. applying the exchange operator to a state of indistinguishable particles should change nothing about the physical state. Therefore, since physical states are actually rays in Hilbert space, we have that $$ S_{ij} \lvert \psi_\text{indist}\rangle = c\lvert \psi_\text{indist}\rangle$$ for some complex number $c\in\mathbb{C}$, and since intuitively exchanging particles twice should return the original state, we have that $S_{ij}^2 = 1$ and therefore $c = \pm 1$. $c=1$ means the state is symmetric under exchange and the particles are called bosons, $c=-1$ means the state is anti-symmetric under exchange and the particles are called fermions. Fermions cannot be put into the same state because $$ S_{12} (\lvert \psi\rangle_1 \otimes \lvert \psi\rangle_2) = -\lvert \psi \rangle_1 \otimes \lvert \psi\rangle_2$$ implies that $\lvert \psi\rangle_1\otimes\lvert\psi\rangle_2 = 0$, i.e. there are no non-trivial antisymmetric states with two particles in the same state.

    Interestingly, this is not the full story, since in two dimensions there can be indistinguishable particles which are neither, called anyons. For an explanation see this answer.

  2. We could simply observe that electrons etc. are fermions since they obey Pauli exclusion and leave it at that. In this ad hoc approach, the question "Why can we not put fermions into the same state?" has no deeper answer.

    However, in relativistic quantum field theory, the question does have a deeper answer (or, more concretely, we can answer the question of why certain particles are fermions and others are not). The spin-statistics theorem tells us that particles with integer spin are necessarily bosons and particles with half-integer spin are necessarily fermions, and it relies crucially on relativity - in non-relativistic quantum field theory, fermions with integer spin and bosons with half-integer spin are consistently possible. The proof of this theorem is subtle and I don't think there is consensus about which version of it is canonical, but Streater/Wightman's "PCT, Spin, Statistics and all that" is a good place to start.

The reason fermions and bosons behave differently is the spin-statistic theorem, a rather subtle result from relativistic quantum field theory that does not hold in all of quantum mechanics and crucially requires special relativity.

The reason fermions and bosons behave differently is the spin-statistic theorem, a rather subtle result from relativistic quantum field theory that does not hold in all of quantum mechanics and crucially requires special relativity.

  1. The issue is actually not about fermions being in the same place (quantum objects do not have a well-defined place), but about being in the same quantum mechanical state. Any system of $N$ particles in quantum mechanics is described by a state that can be decomposed into sum of products of states of the individual particles, and in the simplest case, the non-entangled one, it just looks like $$ \lvert \psi_\text{total}\rangle = \lvert \psi_1\rangle_1 \otimes \lvert \psi_2\rangle_2\otimes\cdots\otimes \lvert \psi_N\rangle_N.$$ On these states, it makes sense to define an "exchange operator" $S_{ij}$ that switches the state of the $i$-th and $j$-th particle, e.g. $$ S_{12} (\lvert \psi_1\rangle_1\otimes \vert \psi_2\rangle_2) = \lvert \psi_2\rangle_1\otimes \lvert \psi_1\rangle_2.$$ In terms of wavefunctions, this amounts to having a wavefunction with $N$ arguments and swapping the $i$-th and $j$-th argument. If the particles are "indistinguishable", then that means that this exchanged operator should do nothing to the state, i.e. applying the exchange operator to a state of indistinguishable particles should change nothing about the physical state. Therefore, since physical states are actually rays in Hilbert space, we have that $$ S_{ij} \lvert \psi_\text{indist}\rangle = c\lvert \psi_\text{indist}\rangle$$ for some complex number $c\in\mathbb{C}$, and since intuitively exchanging particles twice should return the original state, we have that $S_{ij}^2 = 1$ and therefore $c = \pm 1$. $c=1$ means the state is symmetric under exchange and the particles are called bosons, $c=-1$ means the state is anti-symmetric under exchange and the particles are called fermions. Fermions cannot be put into the same state because $$ S_{12} (\lvert \psi\rangle_1 \otimes \lvert \psi\rangle_2) = -\lvert \psi \rangle_1 \otimes \lvert \psi\rangle_2$$ implies that $\lvert \psi\rangle_1\otimes\lvert\psi\rangle_2 = 0$, i.e. there are no non-trivial antisymmetric states with two particles in the same state.

    Interestingly, this is not the full story, since in two dimensions there can be indistinguishable particles which are neither, called anyons. For an explanation see this answer.

  2. We could simply observe that electrons etc. are fermions since they obey Pauli exclusion and leave it at that. In this ad hoc approach, the question "Why can we not put fermions into the same state?" has no deeper answer.

    However, in relativistic quantum field theory, the question does have a deeper answer (or, more concretely, we can answer the question of why certain particles are fermions and others are not). The spin-statistics theorem tells us that particles with integer spin are necessarily bosons and particles with half-integer spin are necessarily fermions, and it relies crucially on relativity - in non-relativistic quantum field theory, fermions with integer spin and bosons with half-integer spin are consistently possible. The proof of this theorem is subtle and I don't think there is consensus about which version of it is canonical, but Streater/Wightman's "PCT, Spin, Statistics and all that" is a good place to start.

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The reason fermions and bosons behave differently is the spin-statistic theorem, a rather subtle result from relativistic quantum field theory that does not hold in all of quantum mechanics and crucially requires special relativity.