## Solution to 2008 Problem 72

 A state consisting of $n$ identical bosons labeled $1$, $2$, ..., $n$ must satisfy\begin{align*}P | \psi\left(1,2,...,n \right) \rangle = |\psi\left(1,2,...,n \right) \rangle\end{align*}where $P$ is an operator that permutes the $n$ particles. This defines a symmetric state.A state consisting of $n$ identical fermions labeled $1$, $2$, ..., $n$ must satisfy\begin{align*}P | \psi\left(1,2,...,n \right) \rangle = \left\{\begin{array}{l}|\psi\left(1,2,...,n \right) \rangle\qquad\te...where $P$ is again an operator that permutes the $n$ particles. This defines an antisymmetric state. It is only possible to antisymmetrize a collection of single-particle states when all of the single-particle states are distinct -- this is the statement of the Pauli exclusion principle. The Pauli exclusion principle applies only to multi-fermion states i.e. it is possible to symmetrize a collection of single-particle states regardless of whether there is redundancy.Collecting all of this information,\begin{itemize}\item Bosons have symmetric wavefunctions.\item Fermions have antisymmetric wavefunctions.\item Bosons do not obey the Pauli exclusion principle.\item Fermions do obey the Pauli exclusion.\end{itemize}Therefore, answer (D) is correct.